Certain Inequalities Involving the Fractional q-Integral Operators
11
Research Article Certain Inequalities Involving the Fractional -Integral Operators Dumitru Baleanu 1,2,3 and Praveen Agarwal 4 1 Department of Chemical and Materials Engineering, Faculty of Engineering, King Abdulaziz University, P.O. Box 80204, Jeddah 21589, Saudi Arabia 2 Department of Mathematics and Computer Sciences, Cankaya University, Yenimahalle, 06810 Ankara, Turkey 3 Institute of Space Sciences, P.O. Box MG-23, M˘ agurele, 76900 Bucharest, Romania 4 Department of Mathematics, Anand International College of Engineering, Jaipur 303012, India Correspondence should be addressed to Dumitru Baleanu; [email protected] Received 31 December 2013; Accepted 10 March 2014; Published 23 April 2014 Academic Editor: Soheil Salahshour Copyright © 2014 D. Baleanu and P. Agarwal. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We establish some inequalities involving Saigo fractional q-integral operator in the theory of quantum calculus by using the two parameters of deformation, 1 and 2 , whose special cases are shown to yield corresponding inequalities associated with Riemann- Liouville and Kober fractional q-integral operators, respectively. Furthermore, we also consider their relevance with other related known results. 1. Introduction and Preliminaries e fractional -calculus is the -extension of the ordinary fractional calculus (see, e.g., [1] and the references therein). e theory of -calculus operators in the recent past has been applied in the areas like ordinary fractional calculus, optimal control problems, solutions of the -difference (differential) and -integral equations, -fractional integral inequalities, - transform analysis, and many more. Fractional and -fractional integral inequalities have proved to be one of the most powerful and far-reaching tools for the development of many branches of pure and applied mathematics. ese inequalities have gained considerable popularity and importance during the past few decades due to their distinguished applications in numerical quadrature, transform theory, probability, and statistical problems, but the most useful ones are in establishing uniqueness of solu- tions in fractional boundary value problems and in fractional partial differential equations. A detailed account of such fractional integral inequalities along with their applications can be found in the research contributions by many authors (see, e.g., [2–12] and [13]; for a very recent work, see also [14]). In a recent paper, Brahim and Taf [15] investigated certain fractional integral inequalities in quantum calculus. Here we aim at establishing certain (presumable) new -integral inequalities associated with Saigo fractional -integral oper- ator, introduced by Garg and Chanchlani [16]. Results due to Brahim and Taf [15] and Sroysang [17] follow as special cases of our results, respectively. For our purpose, we also recall the following definitions (see, e.g., [12, Section 6]) and some earlier works. e -shiſted factorial (; ) is defined by (; ) := { { { { { 1, ( = 0) , −1 ∏ =0 (1 − ), ( ∈ N), (1) where , ∈ C and it is assumed that ̸ = − ( ∈ N 0 ). e -shiſted factorial for negative subscript is defined by (; ) − := 1 (1 − −1 ) (1 − −2 ) ⋅ ⋅ ⋅ (1 − − ) , ( ∈ N 0 ). (2) Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2014, Article ID 371274, 10 pages http://dx.doi.org/10.1155/2014/371274
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Transcript of Certain Inequalities Involving the Fractional q-Integral Operators
Research ArticleCertain Inequalities Involving the Fractional119902-Integral Operators
Dumitru Baleanu123 and Praveen Agarwal4
1 Department of Chemical and Materials Engineering Faculty of Engineering King Abdulaziz University PO Box 80204Jeddah 21589 Saudi Arabia
2Department of Mathematics and Computer Sciences Cankaya University Yenimahalle06810 Ankara Turkey
3 Institute of Space Sciences PO Box MG-23 Magurele 76900 Bucharest Romania4Department of Mathematics Anand International College of Engineering Jaipur 303012 India
Correspondence should be addressed to Dumitru Baleanu dumitrubaleanugmailcom
Received 31 December 2013 Accepted 10 March 2014 Published 23 April 2014
Academic Editor Soheil Salahshour
Copyright copy 2014 D Baleanu and P Agarwal This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
We establish some inequalities involving Saigo fractional q-integral operator in the theory of quantum calculus by using the twoparameters of deformation 119902
1and 119902
2 whose special cases are shown to yield corresponding inequalities associated with Riemann-
Liouville and Kober fractional q-integral operators respectively Furthermore we also consider their relevance with other relatedknown results
1 Introduction and Preliminaries
The fractional 119902-calculus is the 119902-extension of the ordinaryfractional calculus (see eg [1] and the references therein)The theory of 119902-calculus operators in the recent past has beenapplied in the areas like ordinary fractional calculus optimalcontrol problems solutions of the 119902-difference (differential)and 119902-integral equations 119902-fractional integral inequalities 119902-transform analysis and many more
Fractional and 119902-fractional integral inequalities haveproved to be one of the most powerful and far-reaching toolsfor the development of many branches of pure and appliedmathematics These inequalities have gained considerablepopularity and importance during the past few decades dueto their distinguished applications in numerical quadraturetransform theory probability and statistical problems butthe most useful ones are in establishing uniqueness of solu-tions in fractional boundary value problems and in fractionalpartial differential equations A detailed account of suchfractional integral inequalities along with their applicationscan be found in the research contributions by many authors(see eg [2ndash12] and [13] for a very recentwork see also [14])
In a recent paper Brahim and Taf [15] investigated certainfractional integral inequalities in quantum calculus Herewe aim at establishing certain (presumable) new 119902-integralinequalities associated with Saigo fractional 119902-integral oper-ator introduced by Garg and Chanchlani [16] Results due toBrahim and Taf [15] and Sroysang [17] follow as special casesof our results respectively
For our purpose we also recall the following definitions(see eg [12 Section 6]) and some earlier works
The 119902-shifted factorial (119886 119902)119899is defined by
(119886 119902)119899=
1 (119899 = 0)
119899minus1
prod
119896=0
(1 minus 119886119902119896) (119899 isin N)
(1)
where 119886 119902 isin C and it is assumed that 119886 = 119902minus119898
(119898 isin N0)
The 119902-shifted factorial for negative subscript is defined by
(119886 119902)minus119899
=1
(1 minus 119886119902minus1) (1 minus 119886119902minus2) sdot sdot sdot (1 minus 119886119902minus119899)
(119899 isin N0)
(2)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 371274 10 pageshttpdxdoiorg1011552014371274
2 Abstract and Applied Analysis
We also write
(119886 119902)infin
=
infin
prod
119896=0
(1 minus 119886119902119896) (119886 119902 isin C
10038161003816100381610038161199021003816100381610038161003816 lt 1) (3)
It follows from (1) (2) and (3) that
(119886 119902)119899=
(119886 119902)infin
(119886119902119899 119902)infin
(119899 isin Z) (4)
which can be extended to 119899 = 120572 isin C as follows
(119886 119902)120572=
(119886 119902)infin
(119886119902120572 119902)infin
(120572 isin C10038161003816100381610038161199021003816100381610038161003816 lt 1) (5)
where the principal value of 119902120572 is takenWe begin by noting that Jackson was the first to develop
119902-calculus in a systematic way The 119902-derivative of a function119891(119905) is defined by
119863119902119891 (119905) =
119889119902
119889119902119905119891 (119905) =
119891 (119902119905) minus 119891 (119905)
(119902 minus 1) 119905 (6)
It is noted that
lim119902rarr1
119863119902119891 (119905) =
119889
119889119905119891 (119905) (7)
if 119891(119905) is differentiableThe function 119865(119905) is a 119902-antiderivative of 119891(119905) if
119863119902119865(119905) = 119891(119905) It is denoted by
int119891 (119905) 119889119902119905 (8)
The Jackson integral of 119891(119905) is thus defined formally by
int119891 (119905) 119889119902119905 = (1 minus 119902) 119905
infin
sum
119895=0
119902119895119891 (119902119895119905) (9)
which can be easily generalized as follows
int119891 (119905) 119889119902119892 (119905) =
infin
sum
119895=0
119891 (119902119895119905) (119892 (119902
119895119905) minus 119892 (119902
119895+1119905)) (10)
Suppose that 0 lt 119886 lt 119887 The definite 119902-integral is definedas follows
int
119887
0
119891 (119905) 119889119902119905 = (1 minus 119902) 119887
infin
sum
119895=0
119902119895119891 (119902119895119887) (11)
int
119887
119886
119891 (119905) 119889119902119905 = int
119887
0
119891 (119905) 119889119902119905 minus int
119886
0
119891 (119905) 119889119902119905 (12)
A more general version of (11) is given by
int
119887
0
119891 (119905) 119889119902119892 (119905) =
infin
sum
119895=0
119891 (119902119895119887) (119892 (119902
119895119887) minus 119892 (119902
119895+1119887)) (13)
The classical Gamma function Γ(119911) (see eg [12 Section11]) was found by Euler while he was trying to extend thefactorial 119899 = Γ(119899 + 1) (119899 isin N
0) to real numbers The 119902-
factorial function [119899]119902 (119899 isin N
0) of 119899 defined by
[119899]119902 = 1 if 119899 = 0
[119899]119902[119899 minus 1]119902 sdot sdot sdot [2]119902[1]119902 if 119899 isin N(14)
can be rewritten as follows
(1 minus 119902)minus119899
infin
prod
119896=0
(1 minus 119902119896+1
)
(1 minus 119902119896+1+119899)
=(119902 119902)infin
(119902119899+1 119902)infin
(1 minus 119902)minus119899
= Γ119902(119899 + 1) (0 lt 119902 lt 1)
(15)
Replacing 119899 by 119886minus1 in (15) Jackson [18] defined the 119902-Gammafunction Γ
119902(119886) by
Γ119902(119886) =
(119902 119902)infin
(119902119886 119902)infin
(1 minus 119902)1minus119886
(0 lt 119902 lt 1) (16)
The 119902-analogue of (119905 minus 119886)119899 is defined by the polynomial
(119905 minus 119886)119899
119902=
1 (119899 = 0)
(119905 minus 119886) (119905 minus 119902119886) sdot sdot sdot (119905 minus 119902119899minus1
119886) (119899 isin N0)
= 119905119899(119886
119905 119902)119899
(119899 isin N)
(17)
Definition 1 Let R(120572) gt 0 120573 and 120578 be real or complexnumbers Then a 119902-analogue of Saigorsquos fractional integral119868120572120573120578
119902is given for |120591119905| lt 1 by (see [16 page 172 equation
(21)])
119868120572120573120578
119902119891 (119905) =
119905minus120573minus1
Γ119902(120572)
times int
119905
0
(119902120591
119905 119902)120572minus1
infin
sum
119898=0
(119902120572+120573
119902)119898(119902minus120578 119902)119898
(119902120572 119902)119898(119902 119902)119898
sdot 119902(120578minus120573)119898
(minus1)119898119902minus(119898
2 )
times (120591
119905minus 1)
119898
119902
119891 (120591) 119889119902120591
(18)
The integral operator 119868120572120573120578119902
includes both the 119902-analoguesof the Riemann-Liouville and Erdelyi-Kober fractional inte-gral operators given by the following definitions
Abstract and Applied Analysis 3
Definition 2 A 119902-analogue of Riemann-Liouville fractionalintegral operator of a function 119891(119905) of an order 120572 is given by(see [1])
119868120572
119902119891 (119905) (= 119868
120572minus120572120578
119902119891 (119905))
=119905120572minus1
Γ119902(120572)
int
119905
0
(119902120591
119905 119902)120572minus1
119891 (120591) 119889119902120591
(120572 gt 0 0 lt 119902 lt 1)
(19)
where (119886 119902)120572is given by (5)
Definition 3 A 119902-analogue of the Erdelyi-Kober fractionalintegral operator for 120572 gt 0 120578 isin R and 0 lt 119902 lt 1 is givenby (see [1])
119868120578120572
119902119891 (119905) (= 119868
1205720120578
119902119891 (119905))
=119905minus120578minus1
Γ119902(120572)
int
119905
0
(119902120591
119905 119902)120572minus1
120591120578119891 (120591) 119889
119902120591
(120572 gt 0 0 lt 119902 lt 1)
(20)
2 Saigo Fractional 119902-Integral Inequalities
In this section we establish certain fractional 119902-integralinequalities some of which are (presumably) new ones Forour purpose we begin with providing comparison propertiesfor the fractional 119902-integral operators asserted by the follow-ing lemma
Lemma 4 (Choi and Agarwal [13]) Let 0 lt 119902 lt 1 and 119891
[0infin) rarr R be a continuous function with 119891(119905) ge 0 for all119905 isin [0infin) Then one has the following inequalities
(i) the Saigo fractional 119902-integral operator of the function119891(119905) in (18)
119868120572120573120578
119902119891 (119905) ge 0 (21)
for all 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 gt 0 and 120578 lt 0
(ii) the 119902-analogue of Riemann-Liouville fractional integraloperator of the function 119891(119905) of an order 120572 in (19)
119868120572
119902119891 (119905) ge 0 (22)
for all 120572 gt 0
(iii) the 119902-analogue of Erdelyi-Kober fractional integraloperator of the function 119891(119905) in (20)
119868120578120572
119902119891 (119905) ge 0 (23)
for all 120572 gt 0 and 120578 isin R
Proof Applying (11) to the 119902-integral in (18) we have
119868120572120573120578
119902119891 (119905) =
119905minus120573minus1
Γ119902(120572)
times
infin
sum
119898=0
(119902120572+120573
119902)119898(119902minus120578 119902)119898
(119902120572 119902)119898(119902 119902)119898
times 119902(120578minus120573)119898
(minus1)119898119902minus(119898
2 ) sdot (1 minus 119902) 119905
times
infin
sum
119895=0
119902119895(119902119895+1
119902)120572minus1
(119902119895minus 1)119898
119902119891 (119902119895119905)
(24)
It is easy to see that
Γ119902(120572) gt 0 (119902
120572 119902)119898gt 0
(119902119895+1
119902)120572minus1
=
(119902119895+1
119902)infin
(119902120572+119895 119902)infin
gt 0
(25)
for all 120572 gt 0 and 119895119898 isin N0 Next for simplicity
ℎ (119895119898 119902) = (minus1)119898(119902119895minus 1)119898
119902(119895 119898 isin N
0) (26)
Then we claim that ℎ(119895119898 119902) ge 0 for all 119895 119898 isin N0 We find
from (17) that
ℎ (119895119898 119902) = (minus1)119898119902119895119898(1
119902119895 119902)
119898
= 119902119895119898(minus1)119898
119898minus1
prod
119896=0
(1 minus 119902119896minus119895
)
(27)
It is easy to see that if119898 gt 119895 then ℎ(119895 119898 119902) = 0 On the otherhand if119898 le 119895 then we have
ℎ (119895119898 119902) = 119902119895119898
119898minus1
prod
119896=0
(119902119896minus119895
minus 1) gt 0 (28)
since 119896 minus 119895 lt 0 for all 119896 with 0 le 119896 le 119898 minus 1 lt 119898 le 119895Finally we find that under the given conditions each term inthe double series of (24) is nonnegative This completes theproof of (21) The other two inequalities in (22) and (23) maybe easily proved
For convenience and simplicity we define the followingfunctionH by
F (120591 119905 119906 (120591) 120572 120573 120578 119902) =119905minus120573minus1
Γ119902(120572)
(119902120591
119905 119902)120572minus1
sdot
infin
sum
119898=0
(119902120572+120573
119902)119898(119902minus120578 119902)119898
(119902120572 119902)119898(119902 119902)119898
times119902(120578minus120573)119898
(minus1)119898119902minus(119898
2 )
times (120591
119905minus 1)
119898
119902
119906 (120591)
(29)
4 Abstract and Applied Analysis
where 119905 gt 0 0 le 120591 le 119905 120572 gt 0 120573 120578 isin R with 120572 + 120573 gt 0
and 120578 lt 0 0 lt 119902 lt 1 119906 [0infin) rarr [0infin) is a continuousfunction As in the process of Lemma 4 it is seen that
F (120591 119905 119906 (120591) 120572 120573 120578 119902) ge 0 (30)
under the conditions given in (29)Here we present six 119902-integral inequalities involving the
Saigo fractional 119902-integral operator (18) stated inTheorems 5to 12 below
Theorem 5 Let 1199021 1199022isin (0 119905) 119891 119892 and ℎ be three continuous
and synchronous functions on [0infin) and 119906 [0infin) rarr
[0infin) be a continuous function Then the following inequalityholds true For 119905 isin (0infin)
119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
119906 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
ge 119868120574120575120577
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905)
(31)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Let 119891 119892 and ℎ be three continuous and synchronousfunctions on [0infin) Then for all 120591 120588 ge 0 we have
((119891 (120591) minus 119891 (120588)) (119892 (120591) minus 119892 (120588)) (ℎ (120591) + ℎ (120588))) ge 0 (32)
which implies that
119891 (120591) 119892 (120591) ℎ (120591) + 119891 (120588) 119892 (120588) ℎ (120588)
+ 119891 (120591) 119892 (120591) ℎ (120588) + 119891 (120588) 119892 (120588) ℎ (120591)
ge 119891 (120591) 119892 (120588) ℎ (120591) + 119891 (120591) 119892 (120588) ℎ (120588)
+ 119891 (120588) 119892 (120591) ℎ (120591) + 119891 (120588) 119892 (120591) ℎ (120588)
(33)
Multiplying both sides of (33) by F(120591 119905 119906(120591) 120572 120573 120578 1199021)
in (29) together with (30) and taking 119902-integration of
the resulting inequality with respect to 120591 from 0 to 119905 with theaid of Definition 1 we get
119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
+ ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119891 (120588) 119892 (120588) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119891 (120588) 119892 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905)
ge 119892 (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905)
+ 119892 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905)
+ 119891 (120588) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119891 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
(34)
Next multiplying both sides of (34) by F(120588 119905 119906(120588)
120574 120575 120577 1199022) in (29) together with (30) taking 119902-integration of
the resulting inequality with respect to 120588 from 0 to 119905 andusing Definition 1 we are led to the desired result (31) Thiscompletes the proof of Theorem 5
Theorem 6 Let 1199021 1199022isin (0 119905)119891 119892 and ℎ be three continuous
and synchronous functions on [0infin) and 119906 V [0infin) rarr
[0infin) be a continuous function Then the following inequalityholds true For 119905 isin (0infin)
119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905)
+ 119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
V (119905) ℎ (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
V (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
ge 119868120574120575120577
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905)
+ 119868120574120575120577
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
V (119905) 119891 (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
V (119905) 119892 (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) ℎ (119905)
(35)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof To prove the above result multiplying both sides of(34) by F(120588 119905 V(120588) 120574 120575 120577 119902
2) in (29) together with (30)
and taking the 119902-integration of the resulting inequality withrespect to 120588 from 0 to 119905with the aid of Definition 1 we get thedesired result (35)
Remark 7 It may be noted that the inequalities in (31) and(35) are reversed if functions 119891 119892 and ℎ are asynchronousIt is also easily seen that the special case 119906 = V of (35) inTheorem 6 reduces to that in Theorem 5
Abstract and Applied Analysis 5
Theorem 8 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be a
continuous function and 119891 119892 and ℎ are be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(36)
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905)
+ 119868120572120573120578
1199021
119906 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119892 (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119868120574120575120577
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119892 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574120575120577
1199022
119906 (119905) ℎ (119905)
minus119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
119906 (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(37)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Let 119891 119892 and ℎ be three continuous and synchronousfunctions on [0infin) Then for all 120591 120588 ge 0 we have from (36)
1003816100381610038161003816119891 (120591) minus 119891 (120588)1003816100381610038161003816 le (Ψ minus 120595)
1003816100381610038161003816119892 (120591) minus 119892 (120588)1003816100381610038161003816 le (Φ minus 120601)
1003816100381610038161003816ℎ (120591) minus ℎ (120588)1003816100381610038161003816 le (Ω minus 120596)
(38)
which implies that
1003816100381610038161003816(119891 (120591) minus 119891 (120588)) (119892 (120591) minus 119892 (120588)) (ℎ (120591) minus ℎ (120588))1003816100381610038161003816
le (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(39)
Let us define a function
A (120591 120588) = 119891 (120591) 119892 (120591) ℎ (120591) + 119891 (120588) 119892 (120588) ℎ (120591)
+ 119891 (120591) 119892 (120588) ℎ (120588) + 119891 (120588) 119892 (120591) ℎ (120588)
minus 119891 (120591) 119892 (120588) ℎ (120591) minus 119891 (120588) 119892 (120588) ℎ (120588)
minus 119891 (120591) 119892 (120591) ℎ (120588) minus 119891 (120588) 119892 (120591) ℎ (120591)
(40)
Multiplying both sides of (40) by F(120591 119905 119906(120591) 120572 120573 120578 1199021) in
(29) together with (30) taking 119902-integration of the resultinginequality with respect to 120591 from 0 to 119905 and using (18) we get
int
119905
0
F (120591 119905 119906 (120591) 120572 120573 120578 1199021)A (120591 120588) 119889
1199021
120591
= 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
+ 119891 (120588) 119892 (120588) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119892 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905)
+ 119891 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
minus ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905)
minus 119892 (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905)
minus 119891 (120588) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus 119891 (120588) 119892 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905)
(41)
Next multiplying both sides of (41) by F(120588 119905 119906(120588)
120574 120575 120577 1199022) in (29) together with (30) taking 119902-integration of
the resulting inequality with respect to 120588 from 0 to 119905 andusing (18) we get
int
119905
0
int
119905
0
F (120591 119905 119906 (120591) 120572 120573 120578 1199021)F (120588 119905 119906 (120588) 120574 120575 120577 119902
2)
timesA (120591 120588) 1198891199021
1205911198891199022
120588
= 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905)
+ 119868120572120573120578
1199021
119906 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119892 (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119868120574120575120577
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119892 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574120575120577
1199022
119906 (119905) ℎ (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
(42)
Finally by using (39) on to (42) we arrive at the desired result(37) involved inTheorem 8 after a little simplification
6 Abstract and Applied Analysis
Theorem 9 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(43)
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
V (119905)
+ 119868120572120573120578
1199021
119906 (119905) ℎ (119905) 119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119892 (119905) 119868120574120575120577
1199022
V (119905) 119891 (119905) ℎ (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119868120574120575120577
1199022
119892 (119905) ℎ (119905) V (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
V (119905) 119891 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574120575120577
1199022
V (119905) 119892 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574120575120577
1199022
V (119905) ℎ (119905)
minus119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
V (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(44)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Multiplying both sides of (41) by F(120588 119905 V(120588)120574 120575 120577 119902
2) in (29) together with (30) taking the 119902-integration
of the resulting inequality with respect to 120588 from 0 to 119905
with the aid of Definition 1 and then applying (39) on theresulting inequality we get the desired result (44)
Remark 10 It is easily seen that the special case 119906 = V of (44)inTheorem 9 reduces to that in Theorem 8
Theorem 11 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be a
continuous function and f g and h be three Lipschitzian func-tions on [0infin) with the constants 119871
1 1198712 and 119871
3 respectively
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
119906 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574120575120577
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905)
minus 119868120574120575120577
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
minus 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574120575120577
1199022
119906 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572120573120578
1199021
1205913119906 (119905) 119868
120574120575120577
1199022
119906 (119905)
+ 3119868120572120573120578
1199021
120591119906 (119905) 119868120574120575120577
1199022
1205882119906 (119905)
minus 3119868120572120573120578
1199021
1205912119906 (119905) 119868
120574120575120577
1199022
120588119906 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
1205883119906 (119905)]
(45)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Let us define the following relations for all 120591 120588 isin
[0infin)
1003816100381610038161003816119891 (120591) minus 119891 (120588)1003816100381610038161003816 le 1198711(120591 minus 120588)
1003816100381610038161003816119892 (120591) minus 119892 (120588)1003816100381610038161003816 le 1198712(120591 minus 120588)
1003816100381610038161003816ℎ (120591) minus ℎ (120588)1003816100381610038161003816 le 1198713(120591 minus 120588)
(46)
which implies that
1003816100381610038161003816A (120591 120588)1003816100381610038161003816 le 119871111987121198713(120591 minus 120588)
3
(47)
whereA(120591 120588) is given by (40) Then by taking
B (120591 120588) = 119871111987121198713(120591 minus 120588)
3
(48)
First multiplying both sides of (48) by F(120591 119905 119906(120591)
120572 120573 120578 1199021) and F(120588 119905 119906(120588) 120574 120575 120577 119902
2) respectively in (29)
together with (30) taking the 119902-integration of the resultinginequality with respect to 120591 and 120588 from 0 to 119905 with the aid ofDefinition 1 and then applying (39) and (47) on the resultinginequality we get the desired result (45) This completes theproof of Theorem 11
Theorem 12 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3
Abstract and Applied Analysis 7
respectively Then the following inequality holds true For 119905 isin
(0infin)
10038161003816100381610038161003816119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905)
+ 119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
V (119905) ℎ (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
V (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574120575120577
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905)
minus 119868120574120575120577
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
minus 119868120574120575120577
1199022
V (119905) 119891 (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574120575120577
1199022
V (119905) 119892 (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572120573120578
1199021
1205913119906 (119905) 119868
120574120575120577
1199022
V (119905)
+ 3119868120572120573120578
1199021
120591119906 (119905) 119868120574120575120577
1199022
1205882V (119905)
minus 3119868120572120573120578
1199021
1205912119906 (119905) 119868
120574120575120577
1199022
120588V (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
1205883V (119905)]
(49)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Multiplying both sides of (48) by F(120591 119905 119906(120591)
120572 120573 120578 1199021) and F(120588 119905 V(120588) 120574 120575 120577 119902
2) respectively in (29)
together with (30) taking the 119902-integration of the resultinginequality with respect to 120591 and 120588 from 0 to 119905 with the aid ofDefinition 1 and then applying (39) and (47) on the resultinginequality we get the desired result (49) This completes theproof of Theorem 12
Remark 13 It is easily seen that the special case 119906 = V of (49)inTheorem 12 reduces to that in Theorem 11
3 Special Cases and Concluding Remarks
In this section we consider some consequences of the mainresults derived in the preceding sections Following Garg andChanchlani [16] the operator (18) would reduce immediatelyto the extensively investigated 119902-analogue of Erdelyi-Koberand Riemann-Liouville type fractional integral operators in(19) and (20) respectively (see also [1])
For example if we consider120573 = minus120572 and 120575 = minus120574 (and V = 119906
additionally forTheorem 6) andmake use of the relation (20)Theorems 5 and 6 provide respectively the known fractionalintegral inequalities due to Sroysang [17]
Here we derive certain (presumably) new integralinequalities by setting 120573 = minus120572 and 120575 = minus120574 in Theorems 8to 12 respectively and applying the integral operator (19) tothe resulting inequalities we obtain four integral inequalities
involving 119902-Riemann-Liouville fractional integral operatorsstated in Corollaries 14 to 17 respectively below
Corollary 14 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(50)
Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120574
1199022
119906 (119905)
+ 119868120572
1199021
119906 (119905) ℎ (119905) 119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120572
1199021
119906 (119905) 119892 (119905) 119868120574
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120572
1199021
119906 (119905) 119891 (119905) 119868120574
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574
1199022
119906 (119905) 119891 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574
1199022
119906 (119905) 119892 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574
1199022
119906 (119905) ℎ (119905)
minus119868120572
1199021
119906 (119905) 119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572
1199021
119906 (119905) 119868120574
1199022
119906 (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(51)
for all 120572 gt 0 and 120574 gt 0
Corollary 15 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(52)
Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120572
1199021
119906 (119905) 119891 (119905) (119905) ℎ (119905) 119868120574
1199022
V (119905)
+ 119868120572
1199021
119906 (119905) ℎ (119905) 119868120574
1199022
V (119905) 119891 (119905) 119892 (119905)
+ 119868120572
1199021
119906 (119905) 119892 (119905) 119868120574
1199022
V (119905) 119891 (119905) ℎ (119905)
+ 119868120572
1199021
119906 (119905) 119891 (119905) 119868120574
1199022
119892 (119905) ℎ (119905) V (119905)
minus 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574
1199022
V (119905) 119891 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574
1199022
V (119905) 119892 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574
1199022
V (119905) ℎ (119905)
minus119868120572
1199021
119906 (119905) 119868120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572
1199021
119906 (119905) 119868120574
1199022
V (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(53)
for all 120572 gt 0 and 120574 gt 0
8 Abstract and Applied Analysis
Corollary 16 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572
1199021
119906 (119905)
+ 119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120572
1199021
119906 (119905) ℎ (119905)
+ 119868120574
1199022
119906 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574
1199022
119906 (119905) 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119891 (119905)
minus 119868120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119892 (119905)
minus 119868120574
1199022
119906 (119905) 119891 (119905) 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574
1199022
119906 (119905) 119892 (119905) 119868120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572
1199021
1205913119906 (119905) 119868
120574
1199022
119906 (119905)
+ 3119868120572
1199021
120591119906 (119905) 119868120574
1199022
1205882119906 (119905)
minus 3119868120572
1199021
1205912119906 (119905) 119868
120574
1199022
120588119906 (119905)
minus 119868120572
1199021
119906 (119905) 119868120574
1199022
1205883119906 (119905) ]
(54)
for all 120572 gt 0 and 120574 gt 0
Corollary 17 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572
1199021
119906 (119905)
+ 119868120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120572
1199021
119906 (119905) ℎ (119905)
+ 119868120574
1199022
V (119905) ℎ (119905) 1198681205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574
1199022
V (119905) 1198681205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205721199021
119906 (119905) 119891 (119905)
minus 119868120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119892 (119905)
minus 119868120574
1199022
V (119905) 119891 (119905) 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574
1199022
V (119905) 119892 (119905) 1198681205721199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572
1199021
1205913119906 (119905) 119868
120574
1199022
V (119905)
+ 3119868120572
1199021
120591119906 (119905) 119868120574
1199022
1205882V (119905)
minus 3119868120572
1199021
1205912119906 (119905) 119868
120574
1199022
120588V (119905)
minus119868120572
1199021
119906 (119905) 119868120574
1199022
1205883V (119905)]
(55)
for all 120572 gt 0 and 120574 gt 0
Furthermore if one sets 120573 = 0 = 120575 in Theorems 5 to12 and applying the integral operator (20) to the resulting
inequalities one obtains the following (presumably) newinequalities involving 119902-analogue of the Erdelyi-Kober typefractional integral operators asserted by Corollaries 18 to 23below
Corollary 18 Let 1199021 1199022
isin (0 119905) 119891 119892 and ℎ be threecontinuous and synchronous functions on [0infin) and 119906
[0infin) rarr [0infin) be a continuous functionThen the followinginequality holds true For 119905 isin (0infin)
119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
ge 119868120577120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)
(56)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 19 Let 1199021 1199022
isin (0 119905) 119891 119892 and ℎ be threecontinuous and synchronous functions on [0infin) and 119906 V
[0infin) rarr [0infin) be a continuous functionThen the followinginequality holds true For 119905 isin (0infin)
119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
ge 119868120577120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) 119892 (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) ℎ (119905)
(57)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 20 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(58)
Abstract and Applied Analysis 9
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905)
+ 119868120578120572
1199021
119906 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
1199021
119906 (119905) 119892 (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120578120572
1199021
119906 (119905) 119891 (119905) 119868120577120574
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119892 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120577120574
1199022
119906 (119905) ℎ (119905)
minus119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
119906 (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(59)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 21 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(60)
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
V (119905)
+ 119868120578120572
1199021
119906 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
1199021
119906 (119905) 119892 (119905) 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905)
+ 119868120578120572
1199021
119906 (119905) 119891 (119905) 119868120577120574
1199022
119892 (119905) ℎ (119905) V (119905)
minus 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119891 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119892 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120577120574
1199022
V (119905) ℎ (119905)
minus119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
V (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(61)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 22 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3
respectively Then the following inequality holds true For 119905 isin
(0infin)
10038161003816100381610038161003816119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120577120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905)
minus 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
minus 119868120577120574
1199022
119906 (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120577120574
1199022
119906 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120578120572
1199021
1205913119906 (119905) 119868
120574120575120577
1199022
119906 (119905)
+ 3119868120578120572
1199021
120591119906 (119905) 119868120577120574
1199022
1205882119906 (119905)
minus 3119868120578120572
1199021
1205912119906 (119905) 119868
120577120574
1199022
120588119906 (119905)
minus 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
1205883119906 (119905)]
(62)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 23 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120577120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905)
minus 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
minus 119868120577120574
1199022
V (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120577120574
1199022
V (119905) 119892 (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
10 Abstract and Applied Analysis
le 119871111987121198713[119868120578120572
1199021
1205913119906 (119905) 119868
120577120574
1199022
V (119905)
+ 3119868120578120572
1199021
120591119906 (119905) 119868120577120574
1199022
1205882V (119905)
minus 3119868120578120572
1199021
1205912119906 (119905) 119868
120577120574
1199022
120588V (119905)
minus 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
1205883V (119905)]
(63)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
It is also noted that if we consider 119892(119905) as a unity and120573 = minus120572 and 120575 = minus120574 in Theorems 5 and 6 respectively andmake use of the relation (19) thenwe get the known fractionalintegral inequalities due to Brahim and Taf [15]
We conclude our present investigation by remarkingfurther that the results obtained here are general in characterand useful in deriving various inequalities involving frac-tional and 119902-fractional integral operators Moreover they areexpected to find some applications for establishing unique-ness of solutions in fractional boundary value problems andin fractional partial differential equations in the theory ofquantum calculus
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors should express their deep thanks for the editorialand reviewer boardmembers for accepting this paper in theirreputable journal
References
[1] R P Agrawal ldquoCertain fractional q-integrals and q-derivativesrdquoMathematical Proceedings of the Cambridge Philosophical Soci-ety vol 66 no 2 pp 365ndash370 1969
[2] R P Agarwal Difference Equations and Inequalities MarcelDekker 1992
[3] G A AnastassiouAdvances on Fractional Inequalities SpringerBriefs in Mathematics Springer New York NY USA 2011
[4] G E Andrews q-Series Their Development and Applications inAnalysis NumberTheory Combinatorics Physics and ComputerAlgebra American Mathematical Society 1986
[5] S Belarbi and Z Dahmani ldquoOn some new fractional integralinequalitiesrdquo Journal of Inequalities in Pure and Applied Mathe-matics vol 10 no 3 article 86 pp 1ndash5 2009
[6] Z Dahmani ldquoNew inequalities in fractional integralsrdquo Interna-tional Journal of Nonlinear Science vol 9 no 4 pp 493ndash4972010
[7] Z Dahmani O Mechouar and S Brahami ldquoCertain inequal-ities related to the Chebyshevrsquos functional involving a typeRiemann-Liouville operatorrdquo Bulletin of Mathematical Analysisand Applications vol 3 no 4 pp 38ndash44 2011
[8] S S Dragomir ldquoSome integral inequalities of Gruss typerdquoIndian Journal of Pure and Applied Mathematics vol 31 no 4pp 397ndash415 2000
[9] G Gasper and M Rahman Basic Hypergeometric Series Cam-bridge University Press Cambridge UK 2004
[10] S L Kalla and A Rao ldquoOn Gruss type inequality for hyperge-ometric fractional integralsrdquo Le Matematiche vol 66 no 1 pp57ndash64 2011
[11] H Ogunmez and U M Ozkan ldquoFractional quantum integralinequalitiesrdquo Journal of Inequalities and Applications vol 2011Article ID 787939 7 pages 2011
[12] H M Srivastava and J Choi Zeta and q-Zeta Functions andAssociated Series and Integrals Elsevier Science New York NYUSA 2012
[13] J Choi and P Agarwal ldquoCertain inequalities associated withSaigo fractional integrals and their q-analoguesrdquo Abstract andApplied Analysis vol 2014 Article ID 5792609 11 pages 2014
[14] D Baleanu S D Purohit and P Agarwal ldquoOn fractional inte-gral inequalities involving hypergeometric operatorsrdquo ChineseJournal of Mathematics vol 2014 Article ID 609476 5 pages2014
[15] K Brahim and S Taf ldquoSome fractional integral inequalities inquantum calculusrdquo Journal of Fractional Calculus and Applica-tions vol 4 no 2 pp 1ndash8 2013
[16] M Garg and L Chanchlani ldquoq-analogues of Saigorsquos fractionalcalculus operatorsrdquoBulletin ofMathematical Analysis andAppli-cations vol 3 no 4 pp 169ndash179 2011
[17] B Sroysang ldquoA study on a new fractional integral inequalities inquantum calculusrdquo Advanced Studies inTheoretical Physics vol7 no 14 pp 689ndash692 2013
[18] F H Jackson ldquoOn q-definite integralsrdquoTheQuarterly Journal ofPure and Applied Mathematics vol 41 pp 193ndash203 1910
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2 Abstract and Applied Analysis
We also write
(119886 119902)infin
=
infin
prod
119896=0
(1 minus 119886119902119896) (119886 119902 isin C
10038161003816100381610038161199021003816100381610038161003816 lt 1) (3)
It follows from (1) (2) and (3) that
(119886 119902)119899=
(119886 119902)infin
(119886119902119899 119902)infin
(119899 isin Z) (4)
which can be extended to 119899 = 120572 isin C as follows
(119886 119902)120572=
(119886 119902)infin
(119886119902120572 119902)infin
(120572 isin C10038161003816100381610038161199021003816100381610038161003816 lt 1) (5)
where the principal value of 119902120572 is takenWe begin by noting that Jackson was the first to develop
119902-calculus in a systematic way The 119902-derivative of a function119891(119905) is defined by
119863119902119891 (119905) =
119889119902
119889119902119905119891 (119905) =
119891 (119902119905) minus 119891 (119905)
(119902 minus 1) 119905 (6)
It is noted that
lim119902rarr1
119863119902119891 (119905) =
119889
119889119905119891 (119905) (7)
if 119891(119905) is differentiableThe function 119865(119905) is a 119902-antiderivative of 119891(119905) if
119863119902119865(119905) = 119891(119905) It is denoted by
int119891 (119905) 119889119902119905 (8)
The Jackson integral of 119891(119905) is thus defined formally by
int119891 (119905) 119889119902119905 = (1 minus 119902) 119905
infin
sum
119895=0
119902119895119891 (119902119895119905) (9)
which can be easily generalized as follows
int119891 (119905) 119889119902119892 (119905) =
infin
sum
119895=0
119891 (119902119895119905) (119892 (119902
119895119905) minus 119892 (119902
119895+1119905)) (10)
Suppose that 0 lt 119886 lt 119887 The definite 119902-integral is definedas follows
int
119887
0
119891 (119905) 119889119902119905 = (1 minus 119902) 119887
infin
sum
119895=0
119902119895119891 (119902119895119887) (11)
int
119887
119886
119891 (119905) 119889119902119905 = int
119887
0
119891 (119905) 119889119902119905 minus int
119886
0
119891 (119905) 119889119902119905 (12)
A more general version of (11) is given by
int
119887
0
119891 (119905) 119889119902119892 (119905) =
infin
sum
119895=0
119891 (119902119895119887) (119892 (119902
119895119887) minus 119892 (119902
119895+1119887)) (13)
The classical Gamma function Γ(119911) (see eg [12 Section11]) was found by Euler while he was trying to extend thefactorial 119899 = Γ(119899 + 1) (119899 isin N
0) to real numbers The 119902-
factorial function [119899]119902 (119899 isin N
0) of 119899 defined by
[119899]119902 = 1 if 119899 = 0
[119899]119902[119899 minus 1]119902 sdot sdot sdot [2]119902[1]119902 if 119899 isin N(14)
can be rewritten as follows
(1 minus 119902)minus119899
infin
prod
119896=0
(1 minus 119902119896+1
)
(1 minus 119902119896+1+119899)
=(119902 119902)infin
(119902119899+1 119902)infin
(1 minus 119902)minus119899
= Γ119902(119899 + 1) (0 lt 119902 lt 1)
(15)
Replacing 119899 by 119886minus1 in (15) Jackson [18] defined the 119902-Gammafunction Γ
119902(119886) by
Γ119902(119886) =
(119902 119902)infin
(119902119886 119902)infin
(1 minus 119902)1minus119886
(0 lt 119902 lt 1) (16)
The 119902-analogue of (119905 minus 119886)119899 is defined by the polynomial
(119905 minus 119886)119899
119902=
1 (119899 = 0)
(119905 minus 119886) (119905 minus 119902119886) sdot sdot sdot (119905 minus 119902119899minus1
119886) (119899 isin N0)
= 119905119899(119886
119905 119902)119899
(119899 isin N)
(17)
Definition 1 Let R(120572) gt 0 120573 and 120578 be real or complexnumbers Then a 119902-analogue of Saigorsquos fractional integral119868120572120573120578
119902is given for |120591119905| lt 1 by (see [16 page 172 equation
(21)])
119868120572120573120578
119902119891 (119905) =
119905minus120573minus1
Γ119902(120572)
times int
119905
0
(119902120591
119905 119902)120572minus1
infin
sum
119898=0
(119902120572+120573
119902)119898(119902minus120578 119902)119898
(119902120572 119902)119898(119902 119902)119898
sdot 119902(120578minus120573)119898
(minus1)119898119902minus(119898
2 )
times (120591
119905minus 1)
119898
119902
119891 (120591) 119889119902120591
(18)
The integral operator 119868120572120573120578119902
includes both the 119902-analoguesof the Riemann-Liouville and Erdelyi-Kober fractional inte-gral operators given by the following definitions
Abstract and Applied Analysis 3
Definition 2 A 119902-analogue of Riemann-Liouville fractionalintegral operator of a function 119891(119905) of an order 120572 is given by(see [1])
119868120572
119902119891 (119905) (= 119868
120572minus120572120578
119902119891 (119905))
=119905120572minus1
Γ119902(120572)
int
119905
0
(119902120591
119905 119902)120572minus1
119891 (120591) 119889119902120591
(120572 gt 0 0 lt 119902 lt 1)
(19)
where (119886 119902)120572is given by (5)
Definition 3 A 119902-analogue of the Erdelyi-Kober fractionalintegral operator for 120572 gt 0 120578 isin R and 0 lt 119902 lt 1 is givenby (see [1])
119868120578120572
119902119891 (119905) (= 119868
1205720120578
119902119891 (119905))
=119905minus120578minus1
Γ119902(120572)
int
119905
0
(119902120591
119905 119902)120572minus1
120591120578119891 (120591) 119889
119902120591
(120572 gt 0 0 lt 119902 lt 1)
(20)
2 Saigo Fractional 119902-Integral Inequalities
In this section we establish certain fractional 119902-integralinequalities some of which are (presumably) new ones Forour purpose we begin with providing comparison propertiesfor the fractional 119902-integral operators asserted by the follow-ing lemma
Lemma 4 (Choi and Agarwal [13]) Let 0 lt 119902 lt 1 and 119891
[0infin) rarr R be a continuous function with 119891(119905) ge 0 for all119905 isin [0infin) Then one has the following inequalities
(i) the Saigo fractional 119902-integral operator of the function119891(119905) in (18)
119868120572120573120578
119902119891 (119905) ge 0 (21)
for all 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 gt 0 and 120578 lt 0
(ii) the 119902-analogue of Riemann-Liouville fractional integraloperator of the function 119891(119905) of an order 120572 in (19)
119868120572
119902119891 (119905) ge 0 (22)
for all 120572 gt 0
(iii) the 119902-analogue of Erdelyi-Kober fractional integraloperator of the function 119891(119905) in (20)
119868120578120572
119902119891 (119905) ge 0 (23)
for all 120572 gt 0 and 120578 isin R
Proof Applying (11) to the 119902-integral in (18) we have
119868120572120573120578
119902119891 (119905) =
119905minus120573minus1
Γ119902(120572)
times
infin
sum
119898=0
(119902120572+120573
119902)119898(119902minus120578 119902)119898
(119902120572 119902)119898(119902 119902)119898
times 119902(120578minus120573)119898
(minus1)119898119902minus(119898
2 ) sdot (1 minus 119902) 119905
times
infin
sum
119895=0
119902119895(119902119895+1
119902)120572minus1
(119902119895minus 1)119898
119902119891 (119902119895119905)
(24)
It is easy to see that
Γ119902(120572) gt 0 (119902
120572 119902)119898gt 0
(119902119895+1
119902)120572minus1
=
(119902119895+1
119902)infin
(119902120572+119895 119902)infin
gt 0
(25)
for all 120572 gt 0 and 119895119898 isin N0 Next for simplicity
ℎ (119895119898 119902) = (minus1)119898(119902119895minus 1)119898
119902(119895 119898 isin N
0) (26)
Then we claim that ℎ(119895119898 119902) ge 0 for all 119895 119898 isin N0 We find
from (17) that
ℎ (119895119898 119902) = (minus1)119898119902119895119898(1
119902119895 119902)
119898
= 119902119895119898(minus1)119898
119898minus1
prod
119896=0
(1 minus 119902119896minus119895
)
(27)
It is easy to see that if119898 gt 119895 then ℎ(119895 119898 119902) = 0 On the otherhand if119898 le 119895 then we have
ℎ (119895119898 119902) = 119902119895119898
119898minus1
prod
119896=0
(119902119896minus119895
minus 1) gt 0 (28)
since 119896 minus 119895 lt 0 for all 119896 with 0 le 119896 le 119898 minus 1 lt 119898 le 119895Finally we find that under the given conditions each term inthe double series of (24) is nonnegative This completes theproof of (21) The other two inequalities in (22) and (23) maybe easily proved
For convenience and simplicity we define the followingfunctionH by
F (120591 119905 119906 (120591) 120572 120573 120578 119902) =119905minus120573minus1
Γ119902(120572)
(119902120591
119905 119902)120572minus1
sdot
infin
sum
119898=0
(119902120572+120573
119902)119898(119902minus120578 119902)119898
(119902120572 119902)119898(119902 119902)119898
times119902(120578minus120573)119898
(minus1)119898119902minus(119898
2 )
times (120591
119905minus 1)
119898
119902
119906 (120591)
(29)
4 Abstract and Applied Analysis
where 119905 gt 0 0 le 120591 le 119905 120572 gt 0 120573 120578 isin R with 120572 + 120573 gt 0
and 120578 lt 0 0 lt 119902 lt 1 119906 [0infin) rarr [0infin) is a continuousfunction As in the process of Lemma 4 it is seen that
F (120591 119905 119906 (120591) 120572 120573 120578 119902) ge 0 (30)
under the conditions given in (29)Here we present six 119902-integral inequalities involving the
Saigo fractional 119902-integral operator (18) stated inTheorems 5to 12 below
Theorem 5 Let 1199021 1199022isin (0 119905) 119891 119892 and ℎ be three continuous
and synchronous functions on [0infin) and 119906 [0infin) rarr
[0infin) be a continuous function Then the following inequalityholds true For 119905 isin (0infin)
119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
119906 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
ge 119868120574120575120577
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905)
(31)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Let 119891 119892 and ℎ be three continuous and synchronousfunctions on [0infin) Then for all 120591 120588 ge 0 we have
((119891 (120591) minus 119891 (120588)) (119892 (120591) minus 119892 (120588)) (ℎ (120591) + ℎ (120588))) ge 0 (32)
which implies that
119891 (120591) 119892 (120591) ℎ (120591) + 119891 (120588) 119892 (120588) ℎ (120588)
+ 119891 (120591) 119892 (120591) ℎ (120588) + 119891 (120588) 119892 (120588) ℎ (120591)
ge 119891 (120591) 119892 (120588) ℎ (120591) + 119891 (120591) 119892 (120588) ℎ (120588)
+ 119891 (120588) 119892 (120591) ℎ (120591) + 119891 (120588) 119892 (120591) ℎ (120588)
(33)
Multiplying both sides of (33) by F(120591 119905 119906(120591) 120572 120573 120578 1199021)
in (29) together with (30) and taking 119902-integration of
the resulting inequality with respect to 120591 from 0 to 119905 with theaid of Definition 1 we get
119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
+ ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119891 (120588) 119892 (120588) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119891 (120588) 119892 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905)
ge 119892 (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905)
+ 119892 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905)
+ 119891 (120588) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119891 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
(34)
Next multiplying both sides of (34) by F(120588 119905 119906(120588)
120574 120575 120577 1199022) in (29) together with (30) taking 119902-integration of
the resulting inequality with respect to 120588 from 0 to 119905 andusing Definition 1 we are led to the desired result (31) Thiscompletes the proof of Theorem 5
Theorem 6 Let 1199021 1199022isin (0 119905)119891 119892 and ℎ be three continuous
and synchronous functions on [0infin) and 119906 V [0infin) rarr
[0infin) be a continuous function Then the following inequalityholds true For 119905 isin (0infin)
119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905)
+ 119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
V (119905) ℎ (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
V (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
ge 119868120574120575120577
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905)
+ 119868120574120575120577
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
V (119905) 119891 (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
V (119905) 119892 (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) ℎ (119905)
(35)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof To prove the above result multiplying both sides of(34) by F(120588 119905 V(120588) 120574 120575 120577 119902
2) in (29) together with (30)
and taking the 119902-integration of the resulting inequality withrespect to 120588 from 0 to 119905with the aid of Definition 1 we get thedesired result (35)
Remark 7 It may be noted that the inequalities in (31) and(35) are reversed if functions 119891 119892 and ℎ are asynchronousIt is also easily seen that the special case 119906 = V of (35) inTheorem 6 reduces to that in Theorem 5
Abstract and Applied Analysis 5
Theorem 8 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be a
continuous function and 119891 119892 and ℎ are be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(36)
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905)
+ 119868120572120573120578
1199021
119906 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119892 (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119868120574120575120577
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119892 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574120575120577
1199022
119906 (119905) ℎ (119905)
minus119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
119906 (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(37)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Let 119891 119892 and ℎ be three continuous and synchronousfunctions on [0infin) Then for all 120591 120588 ge 0 we have from (36)
1003816100381610038161003816119891 (120591) minus 119891 (120588)1003816100381610038161003816 le (Ψ minus 120595)
1003816100381610038161003816119892 (120591) minus 119892 (120588)1003816100381610038161003816 le (Φ minus 120601)
1003816100381610038161003816ℎ (120591) minus ℎ (120588)1003816100381610038161003816 le (Ω minus 120596)
(38)
which implies that
1003816100381610038161003816(119891 (120591) minus 119891 (120588)) (119892 (120591) minus 119892 (120588)) (ℎ (120591) minus ℎ (120588))1003816100381610038161003816
le (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(39)
Let us define a function
A (120591 120588) = 119891 (120591) 119892 (120591) ℎ (120591) + 119891 (120588) 119892 (120588) ℎ (120591)
+ 119891 (120591) 119892 (120588) ℎ (120588) + 119891 (120588) 119892 (120591) ℎ (120588)
minus 119891 (120591) 119892 (120588) ℎ (120591) minus 119891 (120588) 119892 (120588) ℎ (120588)
minus 119891 (120591) 119892 (120591) ℎ (120588) minus 119891 (120588) 119892 (120591) ℎ (120591)
(40)
Multiplying both sides of (40) by F(120591 119905 119906(120591) 120572 120573 120578 1199021) in
(29) together with (30) taking 119902-integration of the resultinginequality with respect to 120591 from 0 to 119905 and using (18) we get
int
119905
0
F (120591 119905 119906 (120591) 120572 120573 120578 1199021)A (120591 120588) 119889
1199021
120591
= 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
+ 119891 (120588) 119892 (120588) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119892 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905)
+ 119891 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
minus ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905)
minus 119892 (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905)
minus 119891 (120588) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus 119891 (120588) 119892 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905)
(41)
Next multiplying both sides of (41) by F(120588 119905 119906(120588)
120574 120575 120577 1199022) in (29) together with (30) taking 119902-integration of
the resulting inequality with respect to 120588 from 0 to 119905 andusing (18) we get
int
119905
0
int
119905
0
F (120591 119905 119906 (120591) 120572 120573 120578 1199021)F (120588 119905 119906 (120588) 120574 120575 120577 119902
2)
timesA (120591 120588) 1198891199021
1205911198891199022
120588
= 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905)
+ 119868120572120573120578
1199021
119906 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119892 (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119868120574120575120577
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119892 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574120575120577
1199022
119906 (119905) ℎ (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
(42)
Finally by using (39) on to (42) we arrive at the desired result(37) involved inTheorem 8 after a little simplification
6 Abstract and Applied Analysis
Theorem 9 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(43)
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
V (119905)
+ 119868120572120573120578
1199021
119906 (119905) ℎ (119905) 119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119892 (119905) 119868120574120575120577
1199022
V (119905) 119891 (119905) ℎ (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119868120574120575120577
1199022
119892 (119905) ℎ (119905) V (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
V (119905) 119891 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574120575120577
1199022
V (119905) 119892 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574120575120577
1199022
V (119905) ℎ (119905)
minus119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
V (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(44)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Multiplying both sides of (41) by F(120588 119905 V(120588)120574 120575 120577 119902
2) in (29) together with (30) taking the 119902-integration
of the resulting inequality with respect to 120588 from 0 to 119905
with the aid of Definition 1 and then applying (39) on theresulting inequality we get the desired result (44)
Remark 10 It is easily seen that the special case 119906 = V of (44)inTheorem 9 reduces to that in Theorem 8
Theorem 11 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be a
continuous function and f g and h be three Lipschitzian func-tions on [0infin) with the constants 119871
1 1198712 and 119871
3 respectively
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
119906 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574120575120577
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905)
minus 119868120574120575120577
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
minus 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574120575120577
1199022
119906 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572120573120578
1199021
1205913119906 (119905) 119868
120574120575120577
1199022
119906 (119905)
+ 3119868120572120573120578
1199021
120591119906 (119905) 119868120574120575120577
1199022
1205882119906 (119905)
minus 3119868120572120573120578
1199021
1205912119906 (119905) 119868
120574120575120577
1199022
120588119906 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
1205883119906 (119905)]
(45)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Let us define the following relations for all 120591 120588 isin
[0infin)
1003816100381610038161003816119891 (120591) minus 119891 (120588)1003816100381610038161003816 le 1198711(120591 minus 120588)
1003816100381610038161003816119892 (120591) minus 119892 (120588)1003816100381610038161003816 le 1198712(120591 minus 120588)
1003816100381610038161003816ℎ (120591) minus ℎ (120588)1003816100381610038161003816 le 1198713(120591 minus 120588)
(46)
which implies that
1003816100381610038161003816A (120591 120588)1003816100381610038161003816 le 119871111987121198713(120591 minus 120588)
3
(47)
whereA(120591 120588) is given by (40) Then by taking
B (120591 120588) = 119871111987121198713(120591 minus 120588)
3
(48)
First multiplying both sides of (48) by F(120591 119905 119906(120591)
120572 120573 120578 1199021) and F(120588 119905 119906(120588) 120574 120575 120577 119902
2) respectively in (29)
together with (30) taking the 119902-integration of the resultinginequality with respect to 120591 and 120588 from 0 to 119905 with the aid ofDefinition 1 and then applying (39) and (47) on the resultinginequality we get the desired result (45) This completes theproof of Theorem 11
Theorem 12 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3
Abstract and Applied Analysis 7
respectively Then the following inequality holds true For 119905 isin
(0infin)
10038161003816100381610038161003816119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905)
+ 119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
V (119905) ℎ (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
V (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574120575120577
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905)
minus 119868120574120575120577
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
minus 119868120574120575120577
1199022
V (119905) 119891 (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574120575120577
1199022
V (119905) 119892 (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572120573120578
1199021
1205913119906 (119905) 119868
120574120575120577
1199022
V (119905)
+ 3119868120572120573120578
1199021
120591119906 (119905) 119868120574120575120577
1199022
1205882V (119905)
minus 3119868120572120573120578
1199021
1205912119906 (119905) 119868
120574120575120577
1199022
120588V (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
1205883V (119905)]
(49)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Multiplying both sides of (48) by F(120591 119905 119906(120591)
120572 120573 120578 1199021) and F(120588 119905 V(120588) 120574 120575 120577 119902
2) respectively in (29)
together with (30) taking the 119902-integration of the resultinginequality with respect to 120591 and 120588 from 0 to 119905 with the aid ofDefinition 1 and then applying (39) and (47) on the resultinginequality we get the desired result (49) This completes theproof of Theorem 12
Remark 13 It is easily seen that the special case 119906 = V of (49)inTheorem 12 reduces to that in Theorem 11
3 Special Cases and Concluding Remarks
In this section we consider some consequences of the mainresults derived in the preceding sections Following Garg andChanchlani [16] the operator (18) would reduce immediatelyto the extensively investigated 119902-analogue of Erdelyi-Koberand Riemann-Liouville type fractional integral operators in(19) and (20) respectively (see also [1])
For example if we consider120573 = minus120572 and 120575 = minus120574 (and V = 119906
additionally forTheorem 6) andmake use of the relation (20)Theorems 5 and 6 provide respectively the known fractionalintegral inequalities due to Sroysang [17]
Here we derive certain (presumably) new integralinequalities by setting 120573 = minus120572 and 120575 = minus120574 in Theorems 8to 12 respectively and applying the integral operator (19) tothe resulting inequalities we obtain four integral inequalities
involving 119902-Riemann-Liouville fractional integral operatorsstated in Corollaries 14 to 17 respectively below
Corollary 14 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(50)
Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120574
1199022
119906 (119905)
+ 119868120572
1199021
119906 (119905) ℎ (119905) 119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120572
1199021
119906 (119905) 119892 (119905) 119868120574
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120572
1199021
119906 (119905) 119891 (119905) 119868120574
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574
1199022
119906 (119905) 119891 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574
1199022
119906 (119905) 119892 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574
1199022
119906 (119905) ℎ (119905)
minus119868120572
1199021
119906 (119905) 119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572
1199021
119906 (119905) 119868120574
1199022
119906 (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(51)
for all 120572 gt 0 and 120574 gt 0
Corollary 15 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(52)
Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120572
1199021
119906 (119905) 119891 (119905) (119905) ℎ (119905) 119868120574
1199022
V (119905)
+ 119868120572
1199021
119906 (119905) ℎ (119905) 119868120574
1199022
V (119905) 119891 (119905) 119892 (119905)
+ 119868120572
1199021
119906 (119905) 119892 (119905) 119868120574
1199022
V (119905) 119891 (119905) ℎ (119905)
+ 119868120572
1199021
119906 (119905) 119891 (119905) 119868120574
1199022
119892 (119905) ℎ (119905) V (119905)
minus 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574
1199022
V (119905) 119891 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574
1199022
V (119905) 119892 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574
1199022
V (119905) ℎ (119905)
minus119868120572
1199021
119906 (119905) 119868120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572
1199021
119906 (119905) 119868120574
1199022
V (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(53)
for all 120572 gt 0 and 120574 gt 0
8 Abstract and Applied Analysis
Corollary 16 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572
1199021
119906 (119905)
+ 119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120572
1199021
119906 (119905) ℎ (119905)
+ 119868120574
1199022
119906 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574
1199022
119906 (119905) 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119891 (119905)
minus 119868120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119892 (119905)
minus 119868120574
1199022
119906 (119905) 119891 (119905) 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574
1199022
119906 (119905) 119892 (119905) 119868120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572
1199021
1205913119906 (119905) 119868
120574
1199022
119906 (119905)
+ 3119868120572
1199021
120591119906 (119905) 119868120574
1199022
1205882119906 (119905)
minus 3119868120572
1199021
1205912119906 (119905) 119868
120574
1199022
120588119906 (119905)
minus 119868120572
1199021
119906 (119905) 119868120574
1199022
1205883119906 (119905) ]
(54)
for all 120572 gt 0 and 120574 gt 0
Corollary 17 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572
1199021
119906 (119905)
+ 119868120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120572
1199021
119906 (119905) ℎ (119905)
+ 119868120574
1199022
V (119905) ℎ (119905) 1198681205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574
1199022
V (119905) 1198681205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205721199021
119906 (119905) 119891 (119905)
minus 119868120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119892 (119905)
minus 119868120574
1199022
V (119905) 119891 (119905) 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574
1199022
V (119905) 119892 (119905) 1198681205721199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572
1199021
1205913119906 (119905) 119868
120574
1199022
V (119905)
+ 3119868120572
1199021
120591119906 (119905) 119868120574
1199022
1205882V (119905)
minus 3119868120572
1199021
1205912119906 (119905) 119868
120574
1199022
120588V (119905)
minus119868120572
1199021
119906 (119905) 119868120574
1199022
1205883V (119905)]
(55)
for all 120572 gt 0 and 120574 gt 0
Furthermore if one sets 120573 = 0 = 120575 in Theorems 5 to12 and applying the integral operator (20) to the resulting
inequalities one obtains the following (presumably) newinequalities involving 119902-analogue of the Erdelyi-Kober typefractional integral operators asserted by Corollaries 18 to 23below
Corollary 18 Let 1199021 1199022
isin (0 119905) 119891 119892 and ℎ be threecontinuous and synchronous functions on [0infin) and 119906
[0infin) rarr [0infin) be a continuous functionThen the followinginequality holds true For 119905 isin (0infin)
119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
ge 119868120577120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)
(56)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 19 Let 1199021 1199022
isin (0 119905) 119891 119892 and ℎ be threecontinuous and synchronous functions on [0infin) and 119906 V
[0infin) rarr [0infin) be a continuous functionThen the followinginequality holds true For 119905 isin (0infin)
119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
ge 119868120577120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) 119892 (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) ℎ (119905)
(57)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 20 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(58)
Abstract and Applied Analysis 9
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905)
+ 119868120578120572
1199021
119906 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
1199021
119906 (119905) 119892 (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120578120572
1199021
119906 (119905) 119891 (119905) 119868120577120574
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119892 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120577120574
1199022
119906 (119905) ℎ (119905)
minus119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
119906 (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(59)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 21 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(60)
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
V (119905)
+ 119868120578120572
1199021
119906 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
1199021
119906 (119905) 119892 (119905) 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905)
+ 119868120578120572
1199021
119906 (119905) 119891 (119905) 119868120577120574
1199022
119892 (119905) ℎ (119905) V (119905)
minus 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119891 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119892 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120577120574
1199022
V (119905) ℎ (119905)
minus119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
V (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(61)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 22 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3
respectively Then the following inequality holds true For 119905 isin
(0infin)
10038161003816100381610038161003816119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120577120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905)
minus 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
minus 119868120577120574
1199022
119906 (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120577120574
1199022
119906 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120578120572
1199021
1205913119906 (119905) 119868
120574120575120577
1199022
119906 (119905)
+ 3119868120578120572
1199021
120591119906 (119905) 119868120577120574
1199022
1205882119906 (119905)
minus 3119868120578120572
1199021
1205912119906 (119905) 119868
120577120574
1199022
120588119906 (119905)
minus 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
1205883119906 (119905)]
(62)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 23 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120577120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905)
minus 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
minus 119868120577120574
1199022
V (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120577120574
1199022
V (119905) 119892 (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
10 Abstract and Applied Analysis
le 119871111987121198713[119868120578120572
1199021
1205913119906 (119905) 119868
120577120574
1199022
V (119905)
+ 3119868120578120572
1199021
120591119906 (119905) 119868120577120574
1199022
1205882V (119905)
minus 3119868120578120572
1199021
1205912119906 (119905) 119868
120577120574
1199022
120588V (119905)
minus 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
1205883V (119905)]
(63)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
It is also noted that if we consider 119892(119905) as a unity and120573 = minus120572 and 120575 = minus120574 in Theorems 5 and 6 respectively andmake use of the relation (19) thenwe get the known fractionalintegral inequalities due to Brahim and Taf [15]
We conclude our present investigation by remarkingfurther that the results obtained here are general in characterand useful in deriving various inequalities involving frac-tional and 119902-fractional integral operators Moreover they areexpected to find some applications for establishing unique-ness of solutions in fractional boundary value problems andin fractional partial differential equations in the theory ofquantum calculus
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors should express their deep thanks for the editorialand reviewer boardmembers for accepting this paper in theirreputable journal
References
[1] R P Agrawal ldquoCertain fractional q-integrals and q-derivativesrdquoMathematical Proceedings of the Cambridge Philosophical Soci-ety vol 66 no 2 pp 365ndash370 1969
[2] R P Agarwal Difference Equations and Inequalities MarcelDekker 1992
[3] G A AnastassiouAdvances on Fractional Inequalities SpringerBriefs in Mathematics Springer New York NY USA 2011
[4] G E Andrews q-Series Their Development and Applications inAnalysis NumberTheory Combinatorics Physics and ComputerAlgebra American Mathematical Society 1986
[5] S Belarbi and Z Dahmani ldquoOn some new fractional integralinequalitiesrdquo Journal of Inequalities in Pure and Applied Mathe-matics vol 10 no 3 article 86 pp 1ndash5 2009
[6] Z Dahmani ldquoNew inequalities in fractional integralsrdquo Interna-tional Journal of Nonlinear Science vol 9 no 4 pp 493ndash4972010
[7] Z Dahmani O Mechouar and S Brahami ldquoCertain inequal-ities related to the Chebyshevrsquos functional involving a typeRiemann-Liouville operatorrdquo Bulletin of Mathematical Analysisand Applications vol 3 no 4 pp 38ndash44 2011
[8] S S Dragomir ldquoSome integral inequalities of Gruss typerdquoIndian Journal of Pure and Applied Mathematics vol 31 no 4pp 397ndash415 2000
[9] G Gasper and M Rahman Basic Hypergeometric Series Cam-bridge University Press Cambridge UK 2004
[10] S L Kalla and A Rao ldquoOn Gruss type inequality for hyperge-ometric fractional integralsrdquo Le Matematiche vol 66 no 1 pp57ndash64 2011
[11] H Ogunmez and U M Ozkan ldquoFractional quantum integralinequalitiesrdquo Journal of Inequalities and Applications vol 2011Article ID 787939 7 pages 2011
[12] H M Srivastava and J Choi Zeta and q-Zeta Functions andAssociated Series and Integrals Elsevier Science New York NYUSA 2012
[13] J Choi and P Agarwal ldquoCertain inequalities associated withSaigo fractional integrals and their q-analoguesrdquo Abstract andApplied Analysis vol 2014 Article ID 5792609 11 pages 2014
[14] D Baleanu S D Purohit and P Agarwal ldquoOn fractional inte-gral inequalities involving hypergeometric operatorsrdquo ChineseJournal of Mathematics vol 2014 Article ID 609476 5 pages2014
[15] K Brahim and S Taf ldquoSome fractional integral inequalities inquantum calculusrdquo Journal of Fractional Calculus and Applica-tions vol 4 no 2 pp 1ndash8 2013
[16] M Garg and L Chanchlani ldquoq-analogues of Saigorsquos fractionalcalculus operatorsrdquoBulletin ofMathematical Analysis andAppli-cations vol 3 no 4 pp 169ndash179 2011
[17] B Sroysang ldquoA study on a new fractional integral inequalities inquantum calculusrdquo Advanced Studies inTheoretical Physics vol7 no 14 pp 689ndash692 2013
[18] F H Jackson ldquoOn q-definite integralsrdquoTheQuarterly Journal ofPure and Applied Mathematics vol 41 pp 193ndash203 1910
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Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 3
Definition 2 A 119902-analogue of Riemann-Liouville fractionalintegral operator of a function 119891(119905) of an order 120572 is given by(see [1])
119868120572
119902119891 (119905) (= 119868
120572minus120572120578
119902119891 (119905))
=119905120572minus1
Γ119902(120572)
int
119905
0
(119902120591
119905 119902)120572minus1
119891 (120591) 119889119902120591
(120572 gt 0 0 lt 119902 lt 1)
(19)
where (119886 119902)120572is given by (5)
Definition 3 A 119902-analogue of the Erdelyi-Kober fractionalintegral operator for 120572 gt 0 120578 isin R and 0 lt 119902 lt 1 is givenby (see [1])
119868120578120572
119902119891 (119905) (= 119868
1205720120578
119902119891 (119905))
=119905minus120578minus1
Γ119902(120572)
int
119905
0
(119902120591
119905 119902)120572minus1
120591120578119891 (120591) 119889
119902120591
(120572 gt 0 0 lt 119902 lt 1)
(20)
2 Saigo Fractional 119902-Integral Inequalities
In this section we establish certain fractional 119902-integralinequalities some of which are (presumably) new ones Forour purpose we begin with providing comparison propertiesfor the fractional 119902-integral operators asserted by the follow-ing lemma
Lemma 4 (Choi and Agarwal [13]) Let 0 lt 119902 lt 1 and 119891
[0infin) rarr R be a continuous function with 119891(119905) ge 0 for all119905 isin [0infin) Then one has the following inequalities
(i) the Saigo fractional 119902-integral operator of the function119891(119905) in (18)
119868120572120573120578
119902119891 (119905) ge 0 (21)
for all 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 gt 0 and 120578 lt 0
(ii) the 119902-analogue of Riemann-Liouville fractional integraloperator of the function 119891(119905) of an order 120572 in (19)
119868120572
119902119891 (119905) ge 0 (22)
for all 120572 gt 0
(iii) the 119902-analogue of Erdelyi-Kober fractional integraloperator of the function 119891(119905) in (20)
119868120578120572
119902119891 (119905) ge 0 (23)
for all 120572 gt 0 and 120578 isin R
Proof Applying (11) to the 119902-integral in (18) we have
119868120572120573120578
119902119891 (119905) =
119905minus120573minus1
Γ119902(120572)
times
infin
sum
119898=0
(119902120572+120573
119902)119898(119902minus120578 119902)119898
(119902120572 119902)119898(119902 119902)119898
times 119902(120578minus120573)119898
(minus1)119898119902minus(119898
2 ) sdot (1 minus 119902) 119905
times
infin
sum
119895=0
119902119895(119902119895+1
119902)120572minus1
(119902119895minus 1)119898
119902119891 (119902119895119905)
(24)
It is easy to see that
Γ119902(120572) gt 0 (119902
120572 119902)119898gt 0
(119902119895+1
119902)120572minus1
=
(119902119895+1
119902)infin
(119902120572+119895 119902)infin
gt 0
(25)
for all 120572 gt 0 and 119895119898 isin N0 Next for simplicity
ℎ (119895119898 119902) = (minus1)119898(119902119895minus 1)119898
119902(119895 119898 isin N
0) (26)
Then we claim that ℎ(119895119898 119902) ge 0 for all 119895 119898 isin N0 We find
from (17) that
ℎ (119895119898 119902) = (minus1)119898119902119895119898(1
119902119895 119902)
119898
= 119902119895119898(minus1)119898
119898minus1
prod
119896=0
(1 minus 119902119896minus119895
)
(27)
It is easy to see that if119898 gt 119895 then ℎ(119895 119898 119902) = 0 On the otherhand if119898 le 119895 then we have
ℎ (119895119898 119902) = 119902119895119898
119898minus1
prod
119896=0
(119902119896minus119895
minus 1) gt 0 (28)
since 119896 minus 119895 lt 0 for all 119896 with 0 le 119896 le 119898 minus 1 lt 119898 le 119895Finally we find that under the given conditions each term inthe double series of (24) is nonnegative This completes theproof of (21) The other two inequalities in (22) and (23) maybe easily proved
For convenience and simplicity we define the followingfunctionH by
F (120591 119905 119906 (120591) 120572 120573 120578 119902) =119905minus120573minus1
Γ119902(120572)
(119902120591
119905 119902)120572minus1
sdot
infin
sum
119898=0
(119902120572+120573
119902)119898(119902minus120578 119902)119898
(119902120572 119902)119898(119902 119902)119898
times119902(120578minus120573)119898
(minus1)119898119902minus(119898
2 )
times (120591
119905minus 1)
119898
119902
119906 (120591)
(29)
4 Abstract and Applied Analysis
where 119905 gt 0 0 le 120591 le 119905 120572 gt 0 120573 120578 isin R with 120572 + 120573 gt 0
and 120578 lt 0 0 lt 119902 lt 1 119906 [0infin) rarr [0infin) is a continuousfunction As in the process of Lemma 4 it is seen that
F (120591 119905 119906 (120591) 120572 120573 120578 119902) ge 0 (30)
under the conditions given in (29)Here we present six 119902-integral inequalities involving the
Saigo fractional 119902-integral operator (18) stated inTheorems 5to 12 below
Theorem 5 Let 1199021 1199022isin (0 119905) 119891 119892 and ℎ be three continuous
and synchronous functions on [0infin) and 119906 [0infin) rarr
[0infin) be a continuous function Then the following inequalityholds true For 119905 isin (0infin)
119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
119906 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
ge 119868120574120575120577
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905)
(31)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Let 119891 119892 and ℎ be three continuous and synchronousfunctions on [0infin) Then for all 120591 120588 ge 0 we have
((119891 (120591) minus 119891 (120588)) (119892 (120591) minus 119892 (120588)) (ℎ (120591) + ℎ (120588))) ge 0 (32)
which implies that
119891 (120591) 119892 (120591) ℎ (120591) + 119891 (120588) 119892 (120588) ℎ (120588)
+ 119891 (120591) 119892 (120591) ℎ (120588) + 119891 (120588) 119892 (120588) ℎ (120591)
ge 119891 (120591) 119892 (120588) ℎ (120591) + 119891 (120591) 119892 (120588) ℎ (120588)
+ 119891 (120588) 119892 (120591) ℎ (120591) + 119891 (120588) 119892 (120591) ℎ (120588)
(33)
Multiplying both sides of (33) by F(120591 119905 119906(120591) 120572 120573 120578 1199021)
in (29) together with (30) and taking 119902-integration of
the resulting inequality with respect to 120591 from 0 to 119905 with theaid of Definition 1 we get
119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
+ ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119891 (120588) 119892 (120588) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119891 (120588) 119892 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905)
ge 119892 (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905)
+ 119892 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905)
+ 119891 (120588) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119891 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
(34)
Next multiplying both sides of (34) by F(120588 119905 119906(120588)
120574 120575 120577 1199022) in (29) together with (30) taking 119902-integration of
the resulting inequality with respect to 120588 from 0 to 119905 andusing Definition 1 we are led to the desired result (31) Thiscompletes the proof of Theorem 5
Theorem 6 Let 1199021 1199022isin (0 119905)119891 119892 and ℎ be three continuous
and synchronous functions on [0infin) and 119906 V [0infin) rarr
[0infin) be a continuous function Then the following inequalityholds true For 119905 isin (0infin)
119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905)
+ 119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
V (119905) ℎ (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
V (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
ge 119868120574120575120577
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905)
+ 119868120574120575120577
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
V (119905) 119891 (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
V (119905) 119892 (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) ℎ (119905)
(35)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof To prove the above result multiplying both sides of(34) by F(120588 119905 V(120588) 120574 120575 120577 119902
2) in (29) together with (30)
and taking the 119902-integration of the resulting inequality withrespect to 120588 from 0 to 119905with the aid of Definition 1 we get thedesired result (35)
Remark 7 It may be noted that the inequalities in (31) and(35) are reversed if functions 119891 119892 and ℎ are asynchronousIt is also easily seen that the special case 119906 = V of (35) inTheorem 6 reduces to that in Theorem 5
Abstract and Applied Analysis 5
Theorem 8 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be a
continuous function and 119891 119892 and ℎ are be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(36)
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905)
+ 119868120572120573120578
1199021
119906 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119892 (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119868120574120575120577
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119892 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574120575120577
1199022
119906 (119905) ℎ (119905)
minus119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
119906 (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(37)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Let 119891 119892 and ℎ be three continuous and synchronousfunctions on [0infin) Then for all 120591 120588 ge 0 we have from (36)
1003816100381610038161003816119891 (120591) minus 119891 (120588)1003816100381610038161003816 le (Ψ minus 120595)
1003816100381610038161003816119892 (120591) minus 119892 (120588)1003816100381610038161003816 le (Φ minus 120601)
1003816100381610038161003816ℎ (120591) minus ℎ (120588)1003816100381610038161003816 le (Ω minus 120596)
(38)
which implies that
1003816100381610038161003816(119891 (120591) minus 119891 (120588)) (119892 (120591) minus 119892 (120588)) (ℎ (120591) minus ℎ (120588))1003816100381610038161003816
le (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(39)
Let us define a function
A (120591 120588) = 119891 (120591) 119892 (120591) ℎ (120591) + 119891 (120588) 119892 (120588) ℎ (120591)
+ 119891 (120591) 119892 (120588) ℎ (120588) + 119891 (120588) 119892 (120591) ℎ (120588)
minus 119891 (120591) 119892 (120588) ℎ (120591) minus 119891 (120588) 119892 (120588) ℎ (120588)
minus 119891 (120591) 119892 (120591) ℎ (120588) minus 119891 (120588) 119892 (120591) ℎ (120591)
(40)
Multiplying both sides of (40) by F(120591 119905 119906(120591) 120572 120573 120578 1199021) in
(29) together with (30) taking 119902-integration of the resultinginequality with respect to 120591 from 0 to 119905 and using (18) we get
int
119905
0
F (120591 119905 119906 (120591) 120572 120573 120578 1199021)A (120591 120588) 119889
1199021
120591
= 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
+ 119891 (120588) 119892 (120588) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119892 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905)
+ 119891 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
minus ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905)
minus 119892 (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905)
minus 119891 (120588) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus 119891 (120588) 119892 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905)
(41)
Next multiplying both sides of (41) by F(120588 119905 119906(120588)
120574 120575 120577 1199022) in (29) together with (30) taking 119902-integration of
the resulting inequality with respect to 120588 from 0 to 119905 andusing (18) we get
int
119905
0
int
119905
0
F (120591 119905 119906 (120591) 120572 120573 120578 1199021)F (120588 119905 119906 (120588) 120574 120575 120577 119902
2)
timesA (120591 120588) 1198891199021
1205911198891199022
120588
= 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905)
+ 119868120572120573120578
1199021
119906 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119892 (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119868120574120575120577
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119892 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574120575120577
1199022
119906 (119905) ℎ (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
(42)
Finally by using (39) on to (42) we arrive at the desired result(37) involved inTheorem 8 after a little simplification
6 Abstract and Applied Analysis
Theorem 9 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(43)
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
V (119905)
+ 119868120572120573120578
1199021
119906 (119905) ℎ (119905) 119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119892 (119905) 119868120574120575120577
1199022
V (119905) 119891 (119905) ℎ (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119868120574120575120577
1199022
119892 (119905) ℎ (119905) V (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
V (119905) 119891 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574120575120577
1199022
V (119905) 119892 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574120575120577
1199022
V (119905) ℎ (119905)
minus119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
V (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(44)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Multiplying both sides of (41) by F(120588 119905 V(120588)120574 120575 120577 119902
2) in (29) together with (30) taking the 119902-integration
of the resulting inequality with respect to 120588 from 0 to 119905
with the aid of Definition 1 and then applying (39) on theresulting inequality we get the desired result (44)
Remark 10 It is easily seen that the special case 119906 = V of (44)inTheorem 9 reduces to that in Theorem 8
Theorem 11 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be a
continuous function and f g and h be three Lipschitzian func-tions on [0infin) with the constants 119871
1 1198712 and 119871
3 respectively
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
119906 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574120575120577
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905)
minus 119868120574120575120577
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
minus 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574120575120577
1199022
119906 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572120573120578
1199021
1205913119906 (119905) 119868
120574120575120577
1199022
119906 (119905)
+ 3119868120572120573120578
1199021
120591119906 (119905) 119868120574120575120577
1199022
1205882119906 (119905)
minus 3119868120572120573120578
1199021
1205912119906 (119905) 119868
120574120575120577
1199022
120588119906 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
1205883119906 (119905)]
(45)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Let us define the following relations for all 120591 120588 isin
[0infin)
1003816100381610038161003816119891 (120591) minus 119891 (120588)1003816100381610038161003816 le 1198711(120591 minus 120588)
1003816100381610038161003816119892 (120591) minus 119892 (120588)1003816100381610038161003816 le 1198712(120591 minus 120588)
1003816100381610038161003816ℎ (120591) minus ℎ (120588)1003816100381610038161003816 le 1198713(120591 minus 120588)
(46)
which implies that
1003816100381610038161003816A (120591 120588)1003816100381610038161003816 le 119871111987121198713(120591 minus 120588)
3
(47)
whereA(120591 120588) is given by (40) Then by taking
B (120591 120588) = 119871111987121198713(120591 minus 120588)
3
(48)
First multiplying both sides of (48) by F(120591 119905 119906(120591)
120572 120573 120578 1199021) and F(120588 119905 119906(120588) 120574 120575 120577 119902
2) respectively in (29)
together with (30) taking the 119902-integration of the resultinginequality with respect to 120591 and 120588 from 0 to 119905 with the aid ofDefinition 1 and then applying (39) and (47) on the resultinginequality we get the desired result (45) This completes theproof of Theorem 11
Theorem 12 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3
Abstract and Applied Analysis 7
respectively Then the following inequality holds true For 119905 isin
(0infin)
10038161003816100381610038161003816119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905)
+ 119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
V (119905) ℎ (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
V (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574120575120577
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905)
minus 119868120574120575120577
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
minus 119868120574120575120577
1199022
V (119905) 119891 (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574120575120577
1199022
V (119905) 119892 (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572120573120578
1199021
1205913119906 (119905) 119868
120574120575120577
1199022
V (119905)
+ 3119868120572120573120578
1199021
120591119906 (119905) 119868120574120575120577
1199022
1205882V (119905)
minus 3119868120572120573120578
1199021
1205912119906 (119905) 119868
120574120575120577
1199022
120588V (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
1205883V (119905)]
(49)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Multiplying both sides of (48) by F(120591 119905 119906(120591)
120572 120573 120578 1199021) and F(120588 119905 V(120588) 120574 120575 120577 119902
2) respectively in (29)
together with (30) taking the 119902-integration of the resultinginequality with respect to 120591 and 120588 from 0 to 119905 with the aid ofDefinition 1 and then applying (39) and (47) on the resultinginequality we get the desired result (49) This completes theproof of Theorem 12
Remark 13 It is easily seen that the special case 119906 = V of (49)inTheorem 12 reduces to that in Theorem 11
3 Special Cases and Concluding Remarks
In this section we consider some consequences of the mainresults derived in the preceding sections Following Garg andChanchlani [16] the operator (18) would reduce immediatelyto the extensively investigated 119902-analogue of Erdelyi-Koberand Riemann-Liouville type fractional integral operators in(19) and (20) respectively (see also [1])
For example if we consider120573 = minus120572 and 120575 = minus120574 (and V = 119906
additionally forTheorem 6) andmake use of the relation (20)Theorems 5 and 6 provide respectively the known fractionalintegral inequalities due to Sroysang [17]
Here we derive certain (presumably) new integralinequalities by setting 120573 = minus120572 and 120575 = minus120574 in Theorems 8to 12 respectively and applying the integral operator (19) tothe resulting inequalities we obtain four integral inequalities
involving 119902-Riemann-Liouville fractional integral operatorsstated in Corollaries 14 to 17 respectively below
Corollary 14 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(50)
Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120574
1199022
119906 (119905)
+ 119868120572
1199021
119906 (119905) ℎ (119905) 119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120572
1199021
119906 (119905) 119892 (119905) 119868120574
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120572
1199021
119906 (119905) 119891 (119905) 119868120574
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574
1199022
119906 (119905) 119891 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574
1199022
119906 (119905) 119892 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574
1199022
119906 (119905) ℎ (119905)
minus119868120572
1199021
119906 (119905) 119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572
1199021
119906 (119905) 119868120574
1199022
119906 (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(51)
for all 120572 gt 0 and 120574 gt 0
Corollary 15 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(52)
Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120572
1199021
119906 (119905) 119891 (119905) (119905) ℎ (119905) 119868120574
1199022
V (119905)
+ 119868120572
1199021
119906 (119905) ℎ (119905) 119868120574
1199022
V (119905) 119891 (119905) 119892 (119905)
+ 119868120572
1199021
119906 (119905) 119892 (119905) 119868120574
1199022
V (119905) 119891 (119905) ℎ (119905)
+ 119868120572
1199021
119906 (119905) 119891 (119905) 119868120574
1199022
119892 (119905) ℎ (119905) V (119905)
minus 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574
1199022
V (119905) 119891 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574
1199022
V (119905) 119892 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574
1199022
V (119905) ℎ (119905)
minus119868120572
1199021
119906 (119905) 119868120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572
1199021
119906 (119905) 119868120574
1199022
V (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(53)
for all 120572 gt 0 and 120574 gt 0
8 Abstract and Applied Analysis
Corollary 16 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572
1199021
119906 (119905)
+ 119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120572
1199021
119906 (119905) ℎ (119905)
+ 119868120574
1199022
119906 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574
1199022
119906 (119905) 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119891 (119905)
minus 119868120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119892 (119905)
minus 119868120574
1199022
119906 (119905) 119891 (119905) 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574
1199022
119906 (119905) 119892 (119905) 119868120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572
1199021
1205913119906 (119905) 119868
120574
1199022
119906 (119905)
+ 3119868120572
1199021
120591119906 (119905) 119868120574
1199022
1205882119906 (119905)
minus 3119868120572
1199021
1205912119906 (119905) 119868
120574
1199022
120588119906 (119905)
minus 119868120572
1199021
119906 (119905) 119868120574
1199022
1205883119906 (119905) ]
(54)
for all 120572 gt 0 and 120574 gt 0
Corollary 17 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572
1199021
119906 (119905)
+ 119868120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120572
1199021
119906 (119905) ℎ (119905)
+ 119868120574
1199022
V (119905) ℎ (119905) 1198681205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574
1199022
V (119905) 1198681205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205721199021
119906 (119905) 119891 (119905)
minus 119868120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119892 (119905)
minus 119868120574
1199022
V (119905) 119891 (119905) 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574
1199022
V (119905) 119892 (119905) 1198681205721199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572
1199021
1205913119906 (119905) 119868
120574
1199022
V (119905)
+ 3119868120572
1199021
120591119906 (119905) 119868120574
1199022
1205882V (119905)
minus 3119868120572
1199021
1205912119906 (119905) 119868
120574
1199022
120588V (119905)
minus119868120572
1199021
119906 (119905) 119868120574
1199022
1205883V (119905)]
(55)
for all 120572 gt 0 and 120574 gt 0
Furthermore if one sets 120573 = 0 = 120575 in Theorems 5 to12 and applying the integral operator (20) to the resulting
inequalities one obtains the following (presumably) newinequalities involving 119902-analogue of the Erdelyi-Kober typefractional integral operators asserted by Corollaries 18 to 23below
Corollary 18 Let 1199021 1199022
isin (0 119905) 119891 119892 and ℎ be threecontinuous and synchronous functions on [0infin) and 119906
[0infin) rarr [0infin) be a continuous functionThen the followinginequality holds true For 119905 isin (0infin)
119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
ge 119868120577120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)
(56)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 19 Let 1199021 1199022
isin (0 119905) 119891 119892 and ℎ be threecontinuous and synchronous functions on [0infin) and 119906 V
[0infin) rarr [0infin) be a continuous functionThen the followinginequality holds true For 119905 isin (0infin)
119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
ge 119868120577120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) 119892 (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) ℎ (119905)
(57)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 20 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(58)
Abstract and Applied Analysis 9
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905)
+ 119868120578120572
1199021
119906 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
1199021
119906 (119905) 119892 (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120578120572
1199021
119906 (119905) 119891 (119905) 119868120577120574
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119892 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120577120574
1199022
119906 (119905) ℎ (119905)
minus119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
119906 (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(59)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 21 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(60)
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
V (119905)
+ 119868120578120572
1199021
119906 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
1199021
119906 (119905) 119892 (119905) 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905)
+ 119868120578120572
1199021
119906 (119905) 119891 (119905) 119868120577120574
1199022
119892 (119905) ℎ (119905) V (119905)
minus 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119891 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119892 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120577120574
1199022
V (119905) ℎ (119905)
minus119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
V (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(61)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 22 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3
respectively Then the following inequality holds true For 119905 isin
(0infin)
10038161003816100381610038161003816119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120577120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905)
minus 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
minus 119868120577120574
1199022
119906 (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120577120574
1199022
119906 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120578120572
1199021
1205913119906 (119905) 119868
120574120575120577
1199022
119906 (119905)
+ 3119868120578120572
1199021
120591119906 (119905) 119868120577120574
1199022
1205882119906 (119905)
minus 3119868120578120572
1199021
1205912119906 (119905) 119868
120577120574
1199022
120588119906 (119905)
minus 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
1205883119906 (119905)]
(62)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 23 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120577120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905)
minus 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
minus 119868120577120574
1199022
V (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120577120574
1199022
V (119905) 119892 (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
10 Abstract and Applied Analysis
le 119871111987121198713[119868120578120572
1199021
1205913119906 (119905) 119868
120577120574
1199022
V (119905)
+ 3119868120578120572
1199021
120591119906 (119905) 119868120577120574
1199022
1205882V (119905)
minus 3119868120578120572
1199021
1205912119906 (119905) 119868
120577120574
1199022
120588V (119905)
minus 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
1205883V (119905)]
(63)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
It is also noted that if we consider 119892(119905) as a unity and120573 = minus120572 and 120575 = minus120574 in Theorems 5 and 6 respectively andmake use of the relation (19) thenwe get the known fractionalintegral inequalities due to Brahim and Taf [15]
We conclude our present investigation by remarkingfurther that the results obtained here are general in characterand useful in deriving various inequalities involving frac-tional and 119902-fractional integral operators Moreover they areexpected to find some applications for establishing unique-ness of solutions in fractional boundary value problems andin fractional partial differential equations in the theory ofquantum calculus
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors should express their deep thanks for the editorialand reviewer boardmembers for accepting this paper in theirreputable journal
References
[1] R P Agrawal ldquoCertain fractional q-integrals and q-derivativesrdquoMathematical Proceedings of the Cambridge Philosophical Soci-ety vol 66 no 2 pp 365ndash370 1969
[2] R P Agarwal Difference Equations and Inequalities MarcelDekker 1992
[3] G A AnastassiouAdvances on Fractional Inequalities SpringerBriefs in Mathematics Springer New York NY USA 2011
[4] G E Andrews q-Series Their Development and Applications inAnalysis NumberTheory Combinatorics Physics and ComputerAlgebra American Mathematical Society 1986
[5] S Belarbi and Z Dahmani ldquoOn some new fractional integralinequalitiesrdquo Journal of Inequalities in Pure and Applied Mathe-matics vol 10 no 3 article 86 pp 1ndash5 2009
[6] Z Dahmani ldquoNew inequalities in fractional integralsrdquo Interna-tional Journal of Nonlinear Science vol 9 no 4 pp 493ndash4972010
[7] Z Dahmani O Mechouar and S Brahami ldquoCertain inequal-ities related to the Chebyshevrsquos functional involving a typeRiemann-Liouville operatorrdquo Bulletin of Mathematical Analysisand Applications vol 3 no 4 pp 38ndash44 2011
[8] S S Dragomir ldquoSome integral inequalities of Gruss typerdquoIndian Journal of Pure and Applied Mathematics vol 31 no 4pp 397ndash415 2000
[9] G Gasper and M Rahman Basic Hypergeometric Series Cam-bridge University Press Cambridge UK 2004
[10] S L Kalla and A Rao ldquoOn Gruss type inequality for hyperge-ometric fractional integralsrdquo Le Matematiche vol 66 no 1 pp57ndash64 2011
[11] H Ogunmez and U M Ozkan ldquoFractional quantum integralinequalitiesrdquo Journal of Inequalities and Applications vol 2011Article ID 787939 7 pages 2011
[12] H M Srivastava and J Choi Zeta and q-Zeta Functions andAssociated Series and Integrals Elsevier Science New York NYUSA 2012
[13] J Choi and P Agarwal ldquoCertain inequalities associated withSaigo fractional integrals and their q-analoguesrdquo Abstract andApplied Analysis vol 2014 Article ID 5792609 11 pages 2014
[14] D Baleanu S D Purohit and P Agarwal ldquoOn fractional inte-gral inequalities involving hypergeometric operatorsrdquo ChineseJournal of Mathematics vol 2014 Article ID 609476 5 pages2014
[15] K Brahim and S Taf ldquoSome fractional integral inequalities inquantum calculusrdquo Journal of Fractional Calculus and Applica-tions vol 4 no 2 pp 1ndash8 2013
[16] M Garg and L Chanchlani ldquoq-analogues of Saigorsquos fractionalcalculus operatorsrdquoBulletin ofMathematical Analysis andAppli-cations vol 3 no 4 pp 169ndash179 2011
[17] B Sroysang ldquoA study on a new fractional integral inequalities inquantum calculusrdquo Advanced Studies inTheoretical Physics vol7 no 14 pp 689ndash692 2013
[18] F H Jackson ldquoOn q-definite integralsrdquoTheQuarterly Journal ofPure and Applied Mathematics vol 41 pp 193ndash203 1910
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Stochastic AnalysisInternational Journal of
4 Abstract and Applied Analysis
where 119905 gt 0 0 le 120591 le 119905 120572 gt 0 120573 120578 isin R with 120572 + 120573 gt 0
and 120578 lt 0 0 lt 119902 lt 1 119906 [0infin) rarr [0infin) is a continuousfunction As in the process of Lemma 4 it is seen that
F (120591 119905 119906 (120591) 120572 120573 120578 119902) ge 0 (30)
under the conditions given in (29)Here we present six 119902-integral inequalities involving the
Saigo fractional 119902-integral operator (18) stated inTheorems 5to 12 below
Theorem 5 Let 1199021 1199022isin (0 119905) 119891 119892 and ℎ be three continuous
and synchronous functions on [0infin) and 119906 [0infin) rarr
[0infin) be a continuous function Then the following inequalityholds true For 119905 isin (0infin)
119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
119906 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
ge 119868120574120575120577
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905)
(31)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Let 119891 119892 and ℎ be three continuous and synchronousfunctions on [0infin) Then for all 120591 120588 ge 0 we have
((119891 (120591) minus 119891 (120588)) (119892 (120591) minus 119892 (120588)) (ℎ (120591) + ℎ (120588))) ge 0 (32)
which implies that
119891 (120591) 119892 (120591) ℎ (120591) + 119891 (120588) 119892 (120588) ℎ (120588)
+ 119891 (120591) 119892 (120591) ℎ (120588) + 119891 (120588) 119892 (120588) ℎ (120591)
ge 119891 (120591) 119892 (120588) ℎ (120591) + 119891 (120591) 119892 (120588) ℎ (120588)
+ 119891 (120588) 119892 (120591) ℎ (120591) + 119891 (120588) 119892 (120591) ℎ (120588)
(33)
Multiplying both sides of (33) by F(120591 119905 119906(120591) 120572 120573 120578 1199021)
in (29) together with (30) and taking 119902-integration of
the resulting inequality with respect to 120591 from 0 to 119905 with theaid of Definition 1 we get
119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
+ ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119891 (120588) 119892 (120588) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119891 (120588) 119892 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905)
ge 119892 (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905)
+ 119892 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905)
+ 119891 (120588) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119891 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
(34)
Next multiplying both sides of (34) by F(120588 119905 119906(120588)
120574 120575 120577 1199022) in (29) together with (30) taking 119902-integration of
the resulting inequality with respect to 120588 from 0 to 119905 andusing Definition 1 we are led to the desired result (31) Thiscompletes the proof of Theorem 5
Theorem 6 Let 1199021 1199022isin (0 119905)119891 119892 and ℎ be three continuous
and synchronous functions on [0infin) and 119906 V [0infin) rarr
[0infin) be a continuous function Then the following inequalityholds true For 119905 isin (0infin)
119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905)
+ 119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
V (119905) ℎ (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
V (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
ge 119868120574120575120577
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905)
+ 119868120574120575120577
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
V (119905) 119891 (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
V (119905) 119892 (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) ℎ (119905)
(35)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof To prove the above result multiplying both sides of(34) by F(120588 119905 V(120588) 120574 120575 120577 119902
2) in (29) together with (30)
and taking the 119902-integration of the resulting inequality withrespect to 120588 from 0 to 119905with the aid of Definition 1 we get thedesired result (35)
Remark 7 It may be noted that the inequalities in (31) and(35) are reversed if functions 119891 119892 and ℎ are asynchronousIt is also easily seen that the special case 119906 = V of (35) inTheorem 6 reduces to that in Theorem 5
Abstract and Applied Analysis 5
Theorem 8 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be a
continuous function and 119891 119892 and ℎ are be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(36)
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905)
+ 119868120572120573120578
1199021
119906 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119892 (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119868120574120575120577
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119892 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574120575120577
1199022
119906 (119905) ℎ (119905)
minus119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
119906 (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(37)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Let 119891 119892 and ℎ be three continuous and synchronousfunctions on [0infin) Then for all 120591 120588 ge 0 we have from (36)
1003816100381610038161003816119891 (120591) minus 119891 (120588)1003816100381610038161003816 le (Ψ minus 120595)
1003816100381610038161003816119892 (120591) minus 119892 (120588)1003816100381610038161003816 le (Φ minus 120601)
1003816100381610038161003816ℎ (120591) minus ℎ (120588)1003816100381610038161003816 le (Ω minus 120596)
(38)
which implies that
1003816100381610038161003816(119891 (120591) minus 119891 (120588)) (119892 (120591) minus 119892 (120588)) (ℎ (120591) minus ℎ (120588))1003816100381610038161003816
le (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(39)
Let us define a function
A (120591 120588) = 119891 (120591) 119892 (120591) ℎ (120591) + 119891 (120588) 119892 (120588) ℎ (120591)
+ 119891 (120591) 119892 (120588) ℎ (120588) + 119891 (120588) 119892 (120591) ℎ (120588)
minus 119891 (120591) 119892 (120588) ℎ (120591) minus 119891 (120588) 119892 (120588) ℎ (120588)
minus 119891 (120591) 119892 (120591) ℎ (120588) minus 119891 (120588) 119892 (120591) ℎ (120591)
(40)
Multiplying both sides of (40) by F(120591 119905 119906(120591) 120572 120573 120578 1199021) in
(29) together with (30) taking 119902-integration of the resultinginequality with respect to 120591 from 0 to 119905 and using (18) we get
int
119905
0
F (120591 119905 119906 (120591) 120572 120573 120578 1199021)A (120591 120588) 119889
1199021
120591
= 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
+ 119891 (120588) 119892 (120588) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119892 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905)
+ 119891 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
minus ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905)
minus 119892 (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905)
minus 119891 (120588) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus 119891 (120588) 119892 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905)
(41)
Next multiplying both sides of (41) by F(120588 119905 119906(120588)
120574 120575 120577 1199022) in (29) together with (30) taking 119902-integration of
the resulting inequality with respect to 120588 from 0 to 119905 andusing (18) we get
int
119905
0
int
119905
0
F (120591 119905 119906 (120591) 120572 120573 120578 1199021)F (120588 119905 119906 (120588) 120574 120575 120577 119902
2)
timesA (120591 120588) 1198891199021
1205911198891199022
120588
= 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905)
+ 119868120572120573120578
1199021
119906 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119892 (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119868120574120575120577
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119892 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574120575120577
1199022
119906 (119905) ℎ (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
(42)
Finally by using (39) on to (42) we arrive at the desired result(37) involved inTheorem 8 after a little simplification
6 Abstract and Applied Analysis
Theorem 9 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(43)
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
V (119905)
+ 119868120572120573120578
1199021
119906 (119905) ℎ (119905) 119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119892 (119905) 119868120574120575120577
1199022
V (119905) 119891 (119905) ℎ (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119868120574120575120577
1199022
119892 (119905) ℎ (119905) V (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
V (119905) 119891 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574120575120577
1199022
V (119905) 119892 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574120575120577
1199022
V (119905) ℎ (119905)
minus119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
V (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(44)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Multiplying both sides of (41) by F(120588 119905 V(120588)120574 120575 120577 119902
2) in (29) together with (30) taking the 119902-integration
of the resulting inequality with respect to 120588 from 0 to 119905
with the aid of Definition 1 and then applying (39) on theresulting inequality we get the desired result (44)
Remark 10 It is easily seen that the special case 119906 = V of (44)inTheorem 9 reduces to that in Theorem 8
Theorem 11 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be a
continuous function and f g and h be three Lipschitzian func-tions on [0infin) with the constants 119871
1 1198712 and 119871
3 respectively
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
119906 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574120575120577
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905)
minus 119868120574120575120577
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
minus 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574120575120577
1199022
119906 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572120573120578
1199021
1205913119906 (119905) 119868
120574120575120577
1199022
119906 (119905)
+ 3119868120572120573120578
1199021
120591119906 (119905) 119868120574120575120577
1199022
1205882119906 (119905)
minus 3119868120572120573120578
1199021
1205912119906 (119905) 119868
120574120575120577
1199022
120588119906 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
1205883119906 (119905)]
(45)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Let us define the following relations for all 120591 120588 isin
[0infin)
1003816100381610038161003816119891 (120591) minus 119891 (120588)1003816100381610038161003816 le 1198711(120591 minus 120588)
1003816100381610038161003816119892 (120591) minus 119892 (120588)1003816100381610038161003816 le 1198712(120591 minus 120588)
1003816100381610038161003816ℎ (120591) minus ℎ (120588)1003816100381610038161003816 le 1198713(120591 minus 120588)
(46)
which implies that
1003816100381610038161003816A (120591 120588)1003816100381610038161003816 le 119871111987121198713(120591 minus 120588)
3
(47)
whereA(120591 120588) is given by (40) Then by taking
B (120591 120588) = 119871111987121198713(120591 minus 120588)
3
(48)
First multiplying both sides of (48) by F(120591 119905 119906(120591)
120572 120573 120578 1199021) and F(120588 119905 119906(120588) 120574 120575 120577 119902
2) respectively in (29)
together with (30) taking the 119902-integration of the resultinginequality with respect to 120591 and 120588 from 0 to 119905 with the aid ofDefinition 1 and then applying (39) and (47) on the resultinginequality we get the desired result (45) This completes theproof of Theorem 11
Theorem 12 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3
Abstract and Applied Analysis 7
respectively Then the following inequality holds true For 119905 isin
(0infin)
10038161003816100381610038161003816119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905)
+ 119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
V (119905) ℎ (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
V (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574120575120577
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905)
minus 119868120574120575120577
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
minus 119868120574120575120577
1199022
V (119905) 119891 (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574120575120577
1199022
V (119905) 119892 (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572120573120578
1199021
1205913119906 (119905) 119868
120574120575120577
1199022
V (119905)
+ 3119868120572120573120578
1199021
120591119906 (119905) 119868120574120575120577
1199022
1205882V (119905)
minus 3119868120572120573120578
1199021
1205912119906 (119905) 119868
120574120575120577
1199022
120588V (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
1205883V (119905)]
(49)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Multiplying both sides of (48) by F(120591 119905 119906(120591)
120572 120573 120578 1199021) and F(120588 119905 V(120588) 120574 120575 120577 119902
2) respectively in (29)
together with (30) taking the 119902-integration of the resultinginequality with respect to 120591 and 120588 from 0 to 119905 with the aid ofDefinition 1 and then applying (39) and (47) on the resultinginequality we get the desired result (49) This completes theproof of Theorem 12
Remark 13 It is easily seen that the special case 119906 = V of (49)inTheorem 12 reduces to that in Theorem 11
3 Special Cases and Concluding Remarks
In this section we consider some consequences of the mainresults derived in the preceding sections Following Garg andChanchlani [16] the operator (18) would reduce immediatelyto the extensively investigated 119902-analogue of Erdelyi-Koberand Riemann-Liouville type fractional integral operators in(19) and (20) respectively (see also [1])
For example if we consider120573 = minus120572 and 120575 = minus120574 (and V = 119906
additionally forTheorem 6) andmake use of the relation (20)Theorems 5 and 6 provide respectively the known fractionalintegral inequalities due to Sroysang [17]
Here we derive certain (presumably) new integralinequalities by setting 120573 = minus120572 and 120575 = minus120574 in Theorems 8to 12 respectively and applying the integral operator (19) tothe resulting inequalities we obtain four integral inequalities
involving 119902-Riemann-Liouville fractional integral operatorsstated in Corollaries 14 to 17 respectively below
Corollary 14 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(50)
Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120574
1199022
119906 (119905)
+ 119868120572
1199021
119906 (119905) ℎ (119905) 119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120572
1199021
119906 (119905) 119892 (119905) 119868120574
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120572
1199021
119906 (119905) 119891 (119905) 119868120574
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574
1199022
119906 (119905) 119891 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574
1199022
119906 (119905) 119892 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574
1199022
119906 (119905) ℎ (119905)
minus119868120572
1199021
119906 (119905) 119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572
1199021
119906 (119905) 119868120574
1199022
119906 (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(51)
for all 120572 gt 0 and 120574 gt 0
Corollary 15 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(52)
Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120572
1199021
119906 (119905) 119891 (119905) (119905) ℎ (119905) 119868120574
1199022
V (119905)
+ 119868120572
1199021
119906 (119905) ℎ (119905) 119868120574
1199022
V (119905) 119891 (119905) 119892 (119905)
+ 119868120572
1199021
119906 (119905) 119892 (119905) 119868120574
1199022
V (119905) 119891 (119905) ℎ (119905)
+ 119868120572
1199021
119906 (119905) 119891 (119905) 119868120574
1199022
119892 (119905) ℎ (119905) V (119905)
minus 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574
1199022
V (119905) 119891 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574
1199022
V (119905) 119892 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574
1199022
V (119905) ℎ (119905)
minus119868120572
1199021
119906 (119905) 119868120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572
1199021
119906 (119905) 119868120574
1199022
V (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(53)
for all 120572 gt 0 and 120574 gt 0
8 Abstract and Applied Analysis
Corollary 16 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572
1199021
119906 (119905)
+ 119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120572
1199021
119906 (119905) ℎ (119905)
+ 119868120574
1199022
119906 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574
1199022
119906 (119905) 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119891 (119905)
minus 119868120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119892 (119905)
minus 119868120574
1199022
119906 (119905) 119891 (119905) 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574
1199022
119906 (119905) 119892 (119905) 119868120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572
1199021
1205913119906 (119905) 119868
120574
1199022
119906 (119905)
+ 3119868120572
1199021
120591119906 (119905) 119868120574
1199022
1205882119906 (119905)
minus 3119868120572
1199021
1205912119906 (119905) 119868
120574
1199022
120588119906 (119905)
minus 119868120572
1199021
119906 (119905) 119868120574
1199022
1205883119906 (119905) ]
(54)
for all 120572 gt 0 and 120574 gt 0
Corollary 17 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572
1199021
119906 (119905)
+ 119868120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120572
1199021
119906 (119905) ℎ (119905)
+ 119868120574
1199022
V (119905) ℎ (119905) 1198681205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574
1199022
V (119905) 1198681205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205721199021
119906 (119905) 119891 (119905)
minus 119868120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119892 (119905)
minus 119868120574
1199022
V (119905) 119891 (119905) 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574
1199022
V (119905) 119892 (119905) 1198681205721199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572
1199021
1205913119906 (119905) 119868
120574
1199022
V (119905)
+ 3119868120572
1199021
120591119906 (119905) 119868120574
1199022
1205882V (119905)
minus 3119868120572
1199021
1205912119906 (119905) 119868
120574
1199022
120588V (119905)
minus119868120572
1199021
119906 (119905) 119868120574
1199022
1205883V (119905)]
(55)
for all 120572 gt 0 and 120574 gt 0
Furthermore if one sets 120573 = 0 = 120575 in Theorems 5 to12 and applying the integral operator (20) to the resulting
inequalities one obtains the following (presumably) newinequalities involving 119902-analogue of the Erdelyi-Kober typefractional integral operators asserted by Corollaries 18 to 23below
Corollary 18 Let 1199021 1199022
isin (0 119905) 119891 119892 and ℎ be threecontinuous and synchronous functions on [0infin) and 119906
[0infin) rarr [0infin) be a continuous functionThen the followinginequality holds true For 119905 isin (0infin)
119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
ge 119868120577120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)
(56)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 19 Let 1199021 1199022
isin (0 119905) 119891 119892 and ℎ be threecontinuous and synchronous functions on [0infin) and 119906 V
[0infin) rarr [0infin) be a continuous functionThen the followinginequality holds true For 119905 isin (0infin)
119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
ge 119868120577120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) 119892 (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) ℎ (119905)
(57)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 20 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(58)
Abstract and Applied Analysis 9
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905)
+ 119868120578120572
1199021
119906 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
1199021
119906 (119905) 119892 (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120578120572
1199021
119906 (119905) 119891 (119905) 119868120577120574
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119892 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120577120574
1199022
119906 (119905) ℎ (119905)
minus119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
119906 (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(59)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 21 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(60)
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
V (119905)
+ 119868120578120572
1199021
119906 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
1199021
119906 (119905) 119892 (119905) 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905)
+ 119868120578120572
1199021
119906 (119905) 119891 (119905) 119868120577120574
1199022
119892 (119905) ℎ (119905) V (119905)
minus 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119891 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119892 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120577120574
1199022
V (119905) ℎ (119905)
minus119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
V (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(61)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 22 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3
respectively Then the following inequality holds true For 119905 isin
(0infin)
10038161003816100381610038161003816119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120577120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905)
minus 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
minus 119868120577120574
1199022
119906 (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120577120574
1199022
119906 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120578120572
1199021
1205913119906 (119905) 119868
120574120575120577
1199022
119906 (119905)
+ 3119868120578120572
1199021
120591119906 (119905) 119868120577120574
1199022
1205882119906 (119905)
minus 3119868120578120572
1199021
1205912119906 (119905) 119868
120577120574
1199022
120588119906 (119905)
minus 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
1205883119906 (119905)]
(62)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 23 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120577120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905)
minus 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
minus 119868120577120574
1199022
V (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120577120574
1199022
V (119905) 119892 (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
10 Abstract and Applied Analysis
le 119871111987121198713[119868120578120572
1199021
1205913119906 (119905) 119868
120577120574
1199022
V (119905)
+ 3119868120578120572
1199021
120591119906 (119905) 119868120577120574
1199022
1205882V (119905)
minus 3119868120578120572
1199021
1205912119906 (119905) 119868
120577120574
1199022
120588V (119905)
minus 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
1205883V (119905)]
(63)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
It is also noted that if we consider 119892(119905) as a unity and120573 = minus120572 and 120575 = minus120574 in Theorems 5 and 6 respectively andmake use of the relation (19) thenwe get the known fractionalintegral inequalities due to Brahim and Taf [15]
We conclude our present investigation by remarkingfurther that the results obtained here are general in characterand useful in deriving various inequalities involving frac-tional and 119902-fractional integral operators Moreover they areexpected to find some applications for establishing unique-ness of solutions in fractional boundary value problems andin fractional partial differential equations in the theory ofquantum calculus
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors should express their deep thanks for the editorialand reviewer boardmembers for accepting this paper in theirreputable journal
References
[1] R P Agrawal ldquoCertain fractional q-integrals and q-derivativesrdquoMathematical Proceedings of the Cambridge Philosophical Soci-ety vol 66 no 2 pp 365ndash370 1969
[2] R P Agarwal Difference Equations and Inequalities MarcelDekker 1992
[3] G A AnastassiouAdvances on Fractional Inequalities SpringerBriefs in Mathematics Springer New York NY USA 2011
[4] G E Andrews q-Series Their Development and Applications inAnalysis NumberTheory Combinatorics Physics and ComputerAlgebra American Mathematical Society 1986
[5] S Belarbi and Z Dahmani ldquoOn some new fractional integralinequalitiesrdquo Journal of Inequalities in Pure and Applied Mathe-matics vol 10 no 3 article 86 pp 1ndash5 2009
[6] Z Dahmani ldquoNew inequalities in fractional integralsrdquo Interna-tional Journal of Nonlinear Science vol 9 no 4 pp 493ndash4972010
[7] Z Dahmani O Mechouar and S Brahami ldquoCertain inequal-ities related to the Chebyshevrsquos functional involving a typeRiemann-Liouville operatorrdquo Bulletin of Mathematical Analysisand Applications vol 3 no 4 pp 38ndash44 2011
[8] S S Dragomir ldquoSome integral inequalities of Gruss typerdquoIndian Journal of Pure and Applied Mathematics vol 31 no 4pp 397ndash415 2000
[9] G Gasper and M Rahman Basic Hypergeometric Series Cam-bridge University Press Cambridge UK 2004
[10] S L Kalla and A Rao ldquoOn Gruss type inequality for hyperge-ometric fractional integralsrdquo Le Matematiche vol 66 no 1 pp57ndash64 2011
[11] H Ogunmez and U M Ozkan ldquoFractional quantum integralinequalitiesrdquo Journal of Inequalities and Applications vol 2011Article ID 787939 7 pages 2011
[12] H M Srivastava and J Choi Zeta and q-Zeta Functions andAssociated Series and Integrals Elsevier Science New York NYUSA 2012
[13] J Choi and P Agarwal ldquoCertain inequalities associated withSaigo fractional integrals and their q-analoguesrdquo Abstract andApplied Analysis vol 2014 Article ID 5792609 11 pages 2014
[14] D Baleanu S D Purohit and P Agarwal ldquoOn fractional inte-gral inequalities involving hypergeometric operatorsrdquo ChineseJournal of Mathematics vol 2014 Article ID 609476 5 pages2014
[15] K Brahim and S Taf ldquoSome fractional integral inequalities inquantum calculusrdquo Journal of Fractional Calculus and Applica-tions vol 4 no 2 pp 1ndash8 2013
[16] M Garg and L Chanchlani ldquoq-analogues of Saigorsquos fractionalcalculus operatorsrdquoBulletin ofMathematical Analysis andAppli-cations vol 3 no 4 pp 169ndash179 2011
[17] B Sroysang ldquoA study on a new fractional integral inequalities inquantum calculusrdquo Advanced Studies inTheoretical Physics vol7 no 14 pp 689ndash692 2013
[18] F H Jackson ldquoOn q-definite integralsrdquoTheQuarterly Journal ofPure and Applied Mathematics vol 41 pp 193ndash203 1910
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Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 5
Theorem 8 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be a
continuous function and 119891 119892 and ℎ are be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(36)
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905)
+ 119868120572120573120578
1199021
119906 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119892 (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119868120574120575120577
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119892 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574120575120577
1199022
119906 (119905) ℎ (119905)
minus119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
119906 (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(37)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Let 119891 119892 and ℎ be three continuous and synchronousfunctions on [0infin) Then for all 120591 120588 ge 0 we have from (36)
1003816100381610038161003816119891 (120591) minus 119891 (120588)1003816100381610038161003816 le (Ψ minus 120595)
1003816100381610038161003816119892 (120591) minus 119892 (120588)1003816100381610038161003816 le (Φ minus 120601)
1003816100381610038161003816ℎ (120591) minus ℎ (120588)1003816100381610038161003816 le (Ω minus 120596)
(38)
which implies that
1003816100381610038161003816(119891 (120591) minus 119891 (120588)) (119892 (120591) minus 119892 (120588)) (ℎ (120591) minus ℎ (120588))1003816100381610038161003816
le (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(39)
Let us define a function
A (120591 120588) = 119891 (120591) 119892 (120591) ℎ (120591) + 119891 (120588) 119892 (120588) ℎ (120591)
+ 119891 (120591) 119892 (120588) ℎ (120588) + 119891 (120588) 119892 (120591) ℎ (120588)
minus 119891 (120591) 119892 (120588) ℎ (120591) minus 119891 (120588) 119892 (120588) ℎ (120588)
minus 119891 (120591) 119892 (120591) ℎ (120588) minus 119891 (120588) 119892 (120591) ℎ (120591)
(40)
Multiplying both sides of (40) by F(120591 119905 119906(120591) 120572 120573 120578 1199021) in
(29) together with (30) taking 119902-integration of the resultinginequality with respect to 120591 from 0 to 119905 and using (18) we get
int
119905
0
F (120591 119905 119906 (120591) 120572 120573 120578 1199021)A (120591 120588) 119889
1199021
120591
= 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
+ 119891 (120588) 119892 (120588) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119892 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905)
+ 119891 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
minus ℎ (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905)
minus 119892 (120588) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905)
minus 119891 (120588) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus 119891 (120588) 119892 (120588) ℎ (120588) 119868120572120573120578
1199021
119906 (119905)
(41)
Next multiplying both sides of (41) by F(120588 119905 119906(120588)
120574 120575 120577 1199022) in (29) together with (30) taking 119902-integration of
the resulting inequality with respect to 120588 from 0 to 119905 andusing (18) we get
int
119905
0
int
119905
0
F (120591 119905 119906 (120591) 120572 120573 120578 1199021)F (120588 119905 119906 (120588) 120574 120575 120577 119902
2)
timesA (120591 120588) 1198891199021
1205911198891199022
120588
= 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905)
+ 119868120572120573120578
1199021
119906 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119892 (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119868120574120575120577
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574120575120577
1199022
119906 (119905) 119892 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574120575120577
1199022
119906 (119905) ℎ (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
(42)
Finally by using (39) on to (42) we arrive at the desired result(37) involved inTheorem 8 after a little simplification
6 Abstract and Applied Analysis
Theorem 9 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(43)
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
V (119905)
+ 119868120572120573120578
1199021
119906 (119905) ℎ (119905) 119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119892 (119905) 119868120574120575120577
1199022
V (119905) 119891 (119905) ℎ (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119868120574120575120577
1199022
119892 (119905) ℎ (119905) V (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
V (119905) 119891 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574120575120577
1199022
V (119905) 119892 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574120575120577
1199022
V (119905) ℎ (119905)
minus119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
V (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(44)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Multiplying both sides of (41) by F(120588 119905 V(120588)120574 120575 120577 119902
2) in (29) together with (30) taking the 119902-integration
of the resulting inequality with respect to 120588 from 0 to 119905
with the aid of Definition 1 and then applying (39) on theresulting inequality we get the desired result (44)
Remark 10 It is easily seen that the special case 119906 = V of (44)inTheorem 9 reduces to that in Theorem 8
Theorem 11 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be a
continuous function and f g and h be three Lipschitzian func-tions on [0infin) with the constants 119871
1 1198712 and 119871
3 respectively
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
119906 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574120575120577
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905)
minus 119868120574120575120577
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
minus 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574120575120577
1199022
119906 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572120573120578
1199021
1205913119906 (119905) 119868
120574120575120577
1199022
119906 (119905)
+ 3119868120572120573120578
1199021
120591119906 (119905) 119868120574120575120577
1199022
1205882119906 (119905)
minus 3119868120572120573120578
1199021
1205912119906 (119905) 119868
120574120575120577
1199022
120588119906 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
1205883119906 (119905)]
(45)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Let us define the following relations for all 120591 120588 isin
[0infin)
1003816100381610038161003816119891 (120591) minus 119891 (120588)1003816100381610038161003816 le 1198711(120591 minus 120588)
1003816100381610038161003816119892 (120591) minus 119892 (120588)1003816100381610038161003816 le 1198712(120591 minus 120588)
1003816100381610038161003816ℎ (120591) minus ℎ (120588)1003816100381610038161003816 le 1198713(120591 minus 120588)
(46)
which implies that
1003816100381610038161003816A (120591 120588)1003816100381610038161003816 le 119871111987121198713(120591 minus 120588)
3
(47)
whereA(120591 120588) is given by (40) Then by taking
B (120591 120588) = 119871111987121198713(120591 minus 120588)
3
(48)
First multiplying both sides of (48) by F(120591 119905 119906(120591)
120572 120573 120578 1199021) and F(120588 119905 119906(120588) 120574 120575 120577 119902
2) respectively in (29)
together with (30) taking the 119902-integration of the resultinginequality with respect to 120591 and 120588 from 0 to 119905 with the aid ofDefinition 1 and then applying (39) and (47) on the resultinginequality we get the desired result (45) This completes theproof of Theorem 11
Theorem 12 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3
Abstract and Applied Analysis 7
respectively Then the following inequality holds true For 119905 isin
(0infin)
10038161003816100381610038161003816119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905)
+ 119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
V (119905) ℎ (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
V (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574120575120577
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905)
minus 119868120574120575120577
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
minus 119868120574120575120577
1199022
V (119905) 119891 (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574120575120577
1199022
V (119905) 119892 (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572120573120578
1199021
1205913119906 (119905) 119868
120574120575120577
1199022
V (119905)
+ 3119868120572120573120578
1199021
120591119906 (119905) 119868120574120575120577
1199022
1205882V (119905)
minus 3119868120572120573120578
1199021
1205912119906 (119905) 119868
120574120575120577
1199022
120588V (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
1205883V (119905)]
(49)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Multiplying both sides of (48) by F(120591 119905 119906(120591)
120572 120573 120578 1199021) and F(120588 119905 V(120588) 120574 120575 120577 119902
2) respectively in (29)
together with (30) taking the 119902-integration of the resultinginequality with respect to 120591 and 120588 from 0 to 119905 with the aid ofDefinition 1 and then applying (39) and (47) on the resultinginequality we get the desired result (49) This completes theproof of Theorem 12
Remark 13 It is easily seen that the special case 119906 = V of (49)inTheorem 12 reduces to that in Theorem 11
3 Special Cases and Concluding Remarks
In this section we consider some consequences of the mainresults derived in the preceding sections Following Garg andChanchlani [16] the operator (18) would reduce immediatelyto the extensively investigated 119902-analogue of Erdelyi-Koberand Riemann-Liouville type fractional integral operators in(19) and (20) respectively (see also [1])
For example if we consider120573 = minus120572 and 120575 = minus120574 (and V = 119906
additionally forTheorem 6) andmake use of the relation (20)Theorems 5 and 6 provide respectively the known fractionalintegral inequalities due to Sroysang [17]
Here we derive certain (presumably) new integralinequalities by setting 120573 = minus120572 and 120575 = minus120574 in Theorems 8to 12 respectively and applying the integral operator (19) tothe resulting inequalities we obtain four integral inequalities
involving 119902-Riemann-Liouville fractional integral operatorsstated in Corollaries 14 to 17 respectively below
Corollary 14 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(50)
Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120574
1199022
119906 (119905)
+ 119868120572
1199021
119906 (119905) ℎ (119905) 119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120572
1199021
119906 (119905) 119892 (119905) 119868120574
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120572
1199021
119906 (119905) 119891 (119905) 119868120574
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574
1199022
119906 (119905) 119891 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574
1199022
119906 (119905) 119892 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574
1199022
119906 (119905) ℎ (119905)
minus119868120572
1199021
119906 (119905) 119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572
1199021
119906 (119905) 119868120574
1199022
119906 (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(51)
for all 120572 gt 0 and 120574 gt 0
Corollary 15 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(52)
Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120572
1199021
119906 (119905) 119891 (119905) (119905) ℎ (119905) 119868120574
1199022
V (119905)
+ 119868120572
1199021
119906 (119905) ℎ (119905) 119868120574
1199022
V (119905) 119891 (119905) 119892 (119905)
+ 119868120572
1199021
119906 (119905) 119892 (119905) 119868120574
1199022
V (119905) 119891 (119905) ℎ (119905)
+ 119868120572
1199021
119906 (119905) 119891 (119905) 119868120574
1199022
119892 (119905) ℎ (119905) V (119905)
minus 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574
1199022
V (119905) 119891 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574
1199022
V (119905) 119892 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574
1199022
V (119905) ℎ (119905)
minus119868120572
1199021
119906 (119905) 119868120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572
1199021
119906 (119905) 119868120574
1199022
V (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(53)
for all 120572 gt 0 and 120574 gt 0
8 Abstract and Applied Analysis
Corollary 16 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572
1199021
119906 (119905)
+ 119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120572
1199021
119906 (119905) ℎ (119905)
+ 119868120574
1199022
119906 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574
1199022
119906 (119905) 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119891 (119905)
minus 119868120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119892 (119905)
minus 119868120574
1199022
119906 (119905) 119891 (119905) 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574
1199022
119906 (119905) 119892 (119905) 119868120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572
1199021
1205913119906 (119905) 119868
120574
1199022
119906 (119905)
+ 3119868120572
1199021
120591119906 (119905) 119868120574
1199022
1205882119906 (119905)
minus 3119868120572
1199021
1205912119906 (119905) 119868
120574
1199022
120588119906 (119905)
minus 119868120572
1199021
119906 (119905) 119868120574
1199022
1205883119906 (119905) ]
(54)
for all 120572 gt 0 and 120574 gt 0
Corollary 17 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572
1199021
119906 (119905)
+ 119868120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120572
1199021
119906 (119905) ℎ (119905)
+ 119868120574
1199022
V (119905) ℎ (119905) 1198681205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574
1199022
V (119905) 1198681205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205721199021
119906 (119905) 119891 (119905)
minus 119868120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119892 (119905)
minus 119868120574
1199022
V (119905) 119891 (119905) 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574
1199022
V (119905) 119892 (119905) 1198681205721199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572
1199021
1205913119906 (119905) 119868
120574
1199022
V (119905)
+ 3119868120572
1199021
120591119906 (119905) 119868120574
1199022
1205882V (119905)
minus 3119868120572
1199021
1205912119906 (119905) 119868
120574
1199022
120588V (119905)
minus119868120572
1199021
119906 (119905) 119868120574
1199022
1205883V (119905)]
(55)
for all 120572 gt 0 and 120574 gt 0
Furthermore if one sets 120573 = 0 = 120575 in Theorems 5 to12 and applying the integral operator (20) to the resulting
inequalities one obtains the following (presumably) newinequalities involving 119902-analogue of the Erdelyi-Kober typefractional integral operators asserted by Corollaries 18 to 23below
Corollary 18 Let 1199021 1199022
isin (0 119905) 119891 119892 and ℎ be threecontinuous and synchronous functions on [0infin) and 119906
[0infin) rarr [0infin) be a continuous functionThen the followinginequality holds true For 119905 isin (0infin)
119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
ge 119868120577120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)
(56)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 19 Let 1199021 1199022
isin (0 119905) 119891 119892 and ℎ be threecontinuous and synchronous functions on [0infin) and 119906 V
[0infin) rarr [0infin) be a continuous functionThen the followinginequality holds true For 119905 isin (0infin)
119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
ge 119868120577120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) 119892 (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) ℎ (119905)
(57)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 20 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(58)
Abstract and Applied Analysis 9
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905)
+ 119868120578120572
1199021
119906 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
1199021
119906 (119905) 119892 (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120578120572
1199021
119906 (119905) 119891 (119905) 119868120577120574
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119892 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120577120574
1199022
119906 (119905) ℎ (119905)
minus119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
119906 (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(59)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 21 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(60)
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
V (119905)
+ 119868120578120572
1199021
119906 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
1199021
119906 (119905) 119892 (119905) 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905)
+ 119868120578120572
1199021
119906 (119905) 119891 (119905) 119868120577120574
1199022
119892 (119905) ℎ (119905) V (119905)
minus 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119891 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119892 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120577120574
1199022
V (119905) ℎ (119905)
minus119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
V (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(61)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 22 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3
respectively Then the following inequality holds true For 119905 isin
(0infin)
10038161003816100381610038161003816119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120577120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905)
minus 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
minus 119868120577120574
1199022
119906 (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120577120574
1199022
119906 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120578120572
1199021
1205913119906 (119905) 119868
120574120575120577
1199022
119906 (119905)
+ 3119868120578120572
1199021
120591119906 (119905) 119868120577120574
1199022
1205882119906 (119905)
minus 3119868120578120572
1199021
1205912119906 (119905) 119868
120577120574
1199022
120588119906 (119905)
minus 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
1205883119906 (119905)]
(62)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 23 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120577120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905)
minus 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
minus 119868120577120574
1199022
V (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120577120574
1199022
V (119905) 119892 (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
10 Abstract and Applied Analysis
le 119871111987121198713[119868120578120572
1199021
1205913119906 (119905) 119868
120577120574
1199022
V (119905)
+ 3119868120578120572
1199021
120591119906 (119905) 119868120577120574
1199022
1205882V (119905)
minus 3119868120578120572
1199021
1205912119906 (119905) 119868
120577120574
1199022
120588V (119905)
minus 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
1205883V (119905)]
(63)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
It is also noted that if we consider 119892(119905) as a unity and120573 = minus120572 and 120575 = minus120574 in Theorems 5 and 6 respectively andmake use of the relation (19) thenwe get the known fractionalintegral inequalities due to Brahim and Taf [15]
We conclude our present investigation by remarkingfurther that the results obtained here are general in characterand useful in deriving various inequalities involving frac-tional and 119902-fractional integral operators Moreover they areexpected to find some applications for establishing unique-ness of solutions in fractional boundary value problems andin fractional partial differential equations in the theory ofquantum calculus
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors should express their deep thanks for the editorialand reviewer boardmembers for accepting this paper in theirreputable journal
References
[1] R P Agrawal ldquoCertain fractional q-integrals and q-derivativesrdquoMathematical Proceedings of the Cambridge Philosophical Soci-ety vol 66 no 2 pp 365ndash370 1969
[2] R P Agarwal Difference Equations and Inequalities MarcelDekker 1992
[3] G A AnastassiouAdvances on Fractional Inequalities SpringerBriefs in Mathematics Springer New York NY USA 2011
[4] G E Andrews q-Series Their Development and Applications inAnalysis NumberTheory Combinatorics Physics and ComputerAlgebra American Mathematical Society 1986
[5] S Belarbi and Z Dahmani ldquoOn some new fractional integralinequalitiesrdquo Journal of Inequalities in Pure and Applied Mathe-matics vol 10 no 3 article 86 pp 1ndash5 2009
[6] Z Dahmani ldquoNew inequalities in fractional integralsrdquo Interna-tional Journal of Nonlinear Science vol 9 no 4 pp 493ndash4972010
[7] Z Dahmani O Mechouar and S Brahami ldquoCertain inequal-ities related to the Chebyshevrsquos functional involving a typeRiemann-Liouville operatorrdquo Bulletin of Mathematical Analysisand Applications vol 3 no 4 pp 38ndash44 2011
[8] S S Dragomir ldquoSome integral inequalities of Gruss typerdquoIndian Journal of Pure and Applied Mathematics vol 31 no 4pp 397ndash415 2000
[9] G Gasper and M Rahman Basic Hypergeometric Series Cam-bridge University Press Cambridge UK 2004
[10] S L Kalla and A Rao ldquoOn Gruss type inequality for hyperge-ometric fractional integralsrdquo Le Matematiche vol 66 no 1 pp57ndash64 2011
[11] H Ogunmez and U M Ozkan ldquoFractional quantum integralinequalitiesrdquo Journal of Inequalities and Applications vol 2011Article ID 787939 7 pages 2011
[12] H M Srivastava and J Choi Zeta and q-Zeta Functions andAssociated Series and Integrals Elsevier Science New York NYUSA 2012
[13] J Choi and P Agarwal ldquoCertain inequalities associated withSaigo fractional integrals and their q-analoguesrdquo Abstract andApplied Analysis vol 2014 Article ID 5792609 11 pages 2014
[14] D Baleanu S D Purohit and P Agarwal ldquoOn fractional inte-gral inequalities involving hypergeometric operatorsrdquo ChineseJournal of Mathematics vol 2014 Article ID 609476 5 pages2014
[15] K Brahim and S Taf ldquoSome fractional integral inequalities inquantum calculusrdquo Journal of Fractional Calculus and Applica-tions vol 4 no 2 pp 1ndash8 2013
[16] M Garg and L Chanchlani ldquoq-analogues of Saigorsquos fractionalcalculus operatorsrdquoBulletin ofMathematical Analysis andAppli-cations vol 3 no 4 pp 169ndash179 2011
[17] B Sroysang ldquoA study on a new fractional integral inequalities inquantum calculusrdquo Advanced Studies inTheoretical Physics vol7 no 14 pp 689ndash692 2013
[18] F H Jackson ldquoOn q-definite integralsrdquoTheQuarterly Journal ofPure and Applied Mathematics vol 41 pp 193ndash203 1910
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Stochastic AnalysisInternational Journal of
6 Abstract and Applied Analysis
Theorem 9 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(43)
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
V (119905)
+ 119868120572120573120578
1199021
119906 (119905) ℎ (119905) 119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119892 (119905) 119868120574120575120577
1199022
V (119905) 119891 (119905) ℎ (119905)
+ 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119868120574120575120577
1199022
119892 (119905) ℎ (119905) V (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574120575120577
1199022
V (119905) 119891 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574120575120577
1199022
V (119905) 119892 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574120575120577
1199022
V (119905) ℎ (119905)
minus119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
V (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(44)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Multiplying both sides of (41) by F(120588 119905 V(120588)120574 120575 120577 119902
2) in (29) together with (30) taking the 119902-integration
of the resulting inequality with respect to 120588 from 0 to 119905
with the aid of Definition 1 and then applying (39) on theresulting inequality we get the desired result (44)
Remark 10 It is easily seen that the special case 119906 = V of (44)inTheorem 9 reduces to that in Theorem 8
Theorem 11 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be a
continuous function and f g and h be three Lipschitzian func-tions on [0infin) with the constants 119871
1 1198712 and 119871
3 respectively
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
119906 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
119906 (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574120575120577
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905)
minus 119868120574120575120577
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
minus 119868120574120575120577
1199022
119906 (119905) 119891 (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574120575120577
1199022
119906 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572120573120578
1199021
1205913119906 (119905) 119868
120574120575120577
1199022
119906 (119905)
+ 3119868120572120573120578
1199021
120591119906 (119905) 119868120574120575120577
1199022
1205882119906 (119905)
minus 3119868120572120573120578
1199021
1205912119906 (119905) 119868
120574120575120577
1199022
120588119906 (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
1205883119906 (119905)]
(45)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Let us define the following relations for all 120591 120588 isin
[0infin)
1003816100381610038161003816119891 (120591) minus 119891 (120588)1003816100381610038161003816 le 1198711(120591 minus 120588)
1003816100381610038161003816119892 (120591) minus 119892 (120588)1003816100381610038161003816 le 1198712(120591 minus 120588)
1003816100381610038161003816ℎ (120591) minus ℎ (120588)1003816100381610038161003816 le 1198713(120591 minus 120588)
(46)
which implies that
1003816100381610038161003816A (120591 120588)1003816100381610038161003816 le 119871111987121198713(120591 minus 120588)
3
(47)
whereA(120591 120588) is given by (40) Then by taking
B (120591 120588) = 119871111987121198713(120591 minus 120588)
3
(48)
First multiplying both sides of (48) by F(120591 119905 119906(120591)
120572 120573 120578 1199021) and F(120588 119905 119906(120588) 120574 120575 120577 119902
2) respectively in (29)
together with (30) taking the 119902-integration of the resultinginequality with respect to 120591 and 120588 from 0 to 119905 with the aid ofDefinition 1 and then applying (39) and (47) on the resultinginequality we get the desired result (45) This completes theproof of Theorem 11
Theorem 12 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3
Abstract and Applied Analysis 7
respectively Then the following inequality holds true For 119905 isin
(0infin)
10038161003816100381610038161003816119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905)
+ 119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
V (119905) ℎ (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
V (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574120575120577
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905)
minus 119868120574120575120577
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
minus 119868120574120575120577
1199022
V (119905) 119891 (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574120575120577
1199022
V (119905) 119892 (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572120573120578
1199021
1205913119906 (119905) 119868
120574120575120577
1199022
V (119905)
+ 3119868120572120573120578
1199021
120591119906 (119905) 119868120574120575120577
1199022
1205882V (119905)
minus 3119868120572120573120578
1199021
1205912119906 (119905) 119868
120574120575120577
1199022
120588V (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
1205883V (119905)]
(49)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Multiplying both sides of (48) by F(120591 119905 119906(120591)
120572 120573 120578 1199021) and F(120588 119905 V(120588) 120574 120575 120577 119902
2) respectively in (29)
together with (30) taking the 119902-integration of the resultinginequality with respect to 120591 and 120588 from 0 to 119905 with the aid ofDefinition 1 and then applying (39) and (47) on the resultinginequality we get the desired result (49) This completes theproof of Theorem 12
Remark 13 It is easily seen that the special case 119906 = V of (49)inTheorem 12 reduces to that in Theorem 11
3 Special Cases and Concluding Remarks
In this section we consider some consequences of the mainresults derived in the preceding sections Following Garg andChanchlani [16] the operator (18) would reduce immediatelyto the extensively investigated 119902-analogue of Erdelyi-Koberand Riemann-Liouville type fractional integral operators in(19) and (20) respectively (see also [1])
For example if we consider120573 = minus120572 and 120575 = minus120574 (and V = 119906
additionally forTheorem 6) andmake use of the relation (20)Theorems 5 and 6 provide respectively the known fractionalintegral inequalities due to Sroysang [17]
Here we derive certain (presumably) new integralinequalities by setting 120573 = minus120572 and 120575 = minus120574 in Theorems 8to 12 respectively and applying the integral operator (19) tothe resulting inequalities we obtain four integral inequalities
involving 119902-Riemann-Liouville fractional integral operatorsstated in Corollaries 14 to 17 respectively below
Corollary 14 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(50)
Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120574
1199022
119906 (119905)
+ 119868120572
1199021
119906 (119905) ℎ (119905) 119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120572
1199021
119906 (119905) 119892 (119905) 119868120574
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120572
1199021
119906 (119905) 119891 (119905) 119868120574
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574
1199022
119906 (119905) 119891 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574
1199022
119906 (119905) 119892 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574
1199022
119906 (119905) ℎ (119905)
minus119868120572
1199021
119906 (119905) 119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572
1199021
119906 (119905) 119868120574
1199022
119906 (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(51)
for all 120572 gt 0 and 120574 gt 0
Corollary 15 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(52)
Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120572
1199021
119906 (119905) 119891 (119905) (119905) ℎ (119905) 119868120574
1199022
V (119905)
+ 119868120572
1199021
119906 (119905) ℎ (119905) 119868120574
1199022
V (119905) 119891 (119905) 119892 (119905)
+ 119868120572
1199021
119906 (119905) 119892 (119905) 119868120574
1199022
V (119905) 119891 (119905) ℎ (119905)
+ 119868120572
1199021
119906 (119905) 119891 (119905) 119868120574
1199022
119892 (119905) ℎ (119905) V (119905)
minus 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574
1199022
V (119905) 119891 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574
1199022
V (119905) 119892 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574
1199022
V (119905) ℎ (119905)
minus119868120572
1199021
119906 (119905) 119868120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572
1199021
119906 (119905) 119868120574
1199022
V (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(53)
for all 120572 gt 0 and 120574 gt 0
8 Abstract and Applied Analysis
Corollary 16 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572
1199021
119906 (119905)
+ 119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120572
1199021
119906 (119905) ℎ (119905)
+ 119868120574
1199022
119906 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574
1199022
119906 (119905) 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119891 (119905)
minus 119868120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119892 (119905)
minus 119868120574
1199022
119906 (119905) 119891 (119905) 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574
1199022
119906 (119905) 119892 (119905) 119868120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572
1199021
1205913119906 (119905) 119868
120574
1199022
119906 (119905)
+ 3119868120572
1199021
120591119906 (119905) 119868120574
1199022
1205882119906 (119905)
minus 3119868120572
1199021
1205912119906 (119905) 119868
120574
1199022
120588119906 (119905)
minus 119868120572
1199021
119906 (119905) 119868120574
1199022
1205883119906 (119905) ]
(54)
for all 120572 gt 0 and 120574 gt 0
Corollary 17 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572
1199021
119906 (119905)
+ 119868120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120572
1199021
119906 (119905) ℎ (119905)
+ 119868120574
1199022
V (119905) ℎ (119905) 1198681205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574
1199022
V (119905) 1198681205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205721199021
119906 (119905) 119891 (119905)
minus 119868120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119892 (119905)
minus 119868120574
1199022
V (119905) 119891 (119905) 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574
1199022
V (119905) 119892 (119905) 1198681205721199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572
1199021
1205913119906 (119905) 119868
120574
1199022
V (119905)
+ 3119868120572
1199021
120591119906 (119905) 119868120574
1199022
1205882V (119905)
minus 3119868120572
1199021
1205912119906 (119905) 119868
120574
1199022
120588V (119905)
minus119868120572
1199021
119906 (119905) 119868120574
1199022
1205883V (119905)]
(55)
for all 120572 gt 0 and 120574 gt 0
Furthermore if one sets 120573 = 0 = 120575 in Theorems 5 to12 and applying the integral operator (20) to the resulting
inequalities one obtains the following (presumably) newinequalities involving 119902-analogue of the Erdelyi-Kober typefractional integral operators asserted by Corollaries 18 to 23below
Corollary 18 Let 1199021 1199022
isin (0 119905) 119891 119892 and ℎ be threecontinuous and synchronous functions on [0infin) and 119906
[0infin) rarr [0infin) be a continuous functionThen the followinginequality holds true For 119905 isin (0infin)
119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
ge 119868120577120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)
(56)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 19 Let 1199021 1199022
isin (0 119905) 119891 119892 and ℎ be threecontinuous and synchronous functions on [0infin) and 119906 V
[0infin) rarr [0infin) be a continuous functionThen the followinginequality holds true For 119905 isin (0infin)
119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
ge 119868120577120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) 119892 (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) ℎ (119905)
(57)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 20 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(58)
Abstract and Applied Analysis 9
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905)
+ 119868120578120572
1199021
119906 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
1199021
119906 (119905) 119892 (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120578120572
1199021
119906 (119905) 119891 (119905) 119868120577120574
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119892 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120577120574
1199022
119906 (119905) ℎ (119905)
minus119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
119906 (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(59)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 21 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(60)
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
V (119905)
+ 119868120578120572
1199021
119906 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
1199021
119906 (119905) 119892 (119905) 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905)
+ 119868120578120572
1199021
119906 (119905) 119891 (119905) 119868120577120574
1199022
119892 (119905) ℎ (119905) V (119905)
minus 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119891 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119892 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120577120574
1199022
V (119905) ℎ (119905)
minus119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
V (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(61)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 22 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3
respectively Then the following inequality holds true For 119905 isin
(0infin)
10038161003816100381610038161003816119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120577120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905)
minus 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
minus 119868120577120574
1199022
119906 (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120577120574
1199022
119906 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120578120572
1199021
1205913119906 (119905) 119868
120574120575120577
1199022
119906 (119905)
+ 3119868120578120572
1199021
120591119906 (119905) 119868120577120574
1199022
1205882119906 (119905)
minus 3119868120578120572
1199021
1205912119906 (119905) 119868
120577120574
1199022
120588119906 (119905)
minus 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
1205883119906 (119905)]
(62)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 23 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120577120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905)
minus 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
minus 119868120577120574
1199022
V (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120577120574
1199022
V (119905) 119892 (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
10 Abstract and Applied Analysis
le 119871111987121198713[119868120578120572
1199021
1205913119906 (119905) 119868
120577120574
1199022
V (119905)
+ 3119868120578120572
1199021
120591119906 (119905) 119868120577120574
1199022
1205882V (119905)
minus 3119868120578120572
1199021
1205912119906 (119905) 119868
120577120574
1199022
120588V (119905)
minus 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
1205883V (119905)]
(63)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
It is also noted that if we consider 119892(119905) as a unity and120573 = minus120572 and 120575 = minus120574 in Theorems 5 and 6 respectively andmake use of the relation (19) thenwe get the known fractionalintegral inequalities due to Brahim and Taf [15]
We conclude our present investigation by remarkingfurther that the results obtained here are general in characterand useful in deriving various inequalities involving frac-tional and 119902-fractional integral operators Moreover they areexpected to find some applications for establishing unique-ness of solutions in fractional boundary value problems andin fractional partial differential equations in the theory ofquantum calculus
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors should express their deep thanks for the editorialand reviewer boardmembers for accepting this paper in theirreputable journal
References
[1] R P Agrawal ldquoCertain fractional q-integrals and q-derivativesrdquoMathematical Proceedings of the Cambridge Philosophical Soci-ety vol 66 no 2 pp 365ndash370 1969
[2] R P Agarwal Difference Equations and Inequalities MarcelDekker 1992
[3] G A AnastassiouAdvances on Fractional Inequalities SpringerBriefs in Mathematics Springer New York NY USA 2011
[4] G E Andrews q-Series Their Development and Applications inAnalysis NumberTheory Combinatorics Physics and ComputerAlgebra American Mathematical Society 1986
[5] S Belarbi and Z Dahmani ldquoOn some new fractional integralinequalitiesrdquo Journal of Inequalities in Pure and Applied Mathe-matics vol 10 no 3 article 86 pp 1ndash5 2009
[6] Z Dahmani ldquoNew inequalities in fractional integralsrdquo Interna-tional Journal of Nonlinear Science vol 9 no 4 pp 493ndash4972010
[7] Z Dahmani O Mechouar and S Brahami ldquoCertain inequal-ities related to the Chebyshevrsquos functional involving a typeRiemann-Liouville operatorrdquo Bulletin of Mathematical Analysisand Applications vol 3 no 4 pp 38ndash44 2011
[8] S S Dragomir ldquoSome integral inequalities of Gruss typerdquoIndian Journal of Pure and Applied Mathematics vol 31 no 4pp 397ndash415 2000
[9] G Gasper and M Rahman Basic Hypergeometric Series Cam-bridge University Press Cambridge UK 2004
[10] S L Kalla and A Rao ldquoOn Gruss type inequality for hyperge-ometric fractional integralsrdquo Le Matematiche vol 66 no 1 pp57ndash64 2011
[11] H Ogunmez and U M Ozkan ldquoFractional quantum integralinequalitiesrdquo Journal of Inequalities and Applications vol 2011Article ID 787939 7 pages 2011
[12] H M Srivastava and J Choi Zeta and q-Zeta Functions andAssociated Series and Integrals Elsevier Science New York NYUSA 2012
[13] J Choi and P Agarwal ldquoCertain inequalities associated withSaigo fractional integrals and their q-analoguesrdquo Abstract andApplied Analysis vol 2014 Article ID 5792609 11 pages 2014
[14] D Baleanu S D Purohit and P Agarwal ldquoOn fractional inte-gral inequalities involving hypergeometric operatorsrdquo ChineseJournal of Mathematics vol 2014 Article ID 609476 5 pages2014
[15] K Brahim and S Taf ldquoSome fractional integral inequalities inquantum calculusrdquo Journal of Fractional Calculus and Applica-tions vol 4 no 2 pp 1ndash8 2013
[16] M Garg and L Chanchlani ldquoq-analogues of Saigorsquos fractionalcalculus operatorsrdquoBulletin ofMathematical Analysis andAppli-cations vol 3 no 4 pp 169ndash179 2011
[17] B Sroysang ldquoA study on a new fractional integral inequalities inquantum calculusrdquo Advanced Studies inTheoretical Physics vol7 no 14 pp 689ndash692 2013
[18] F H Jackson ldquoOn q-definite integralsrdquoTheQuarterly Journal ofPure and Applied Mathematics vol 41 pp 193ndash203 1910
Submit your manuscripts athttpwwwhindawicom
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Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Mathematical PhysicsAdvances in
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OptimizationJournal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Discrete Dynamics in Nature and Society
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Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 7
respectively Then the following inequality holds true For 119905 isin
(0infin)
10038161003816100381610038161003816119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905)
+ 119868120574120575120577
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
1199021
119906 (119905) ℎ (119905)
+ 119868120574120575120577
1199022
V (119905) ℎ (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
1199022
V (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574120575120577
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905)
minus 119868120574120575120577
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905)
minus 119868120574120575120577
1199022
V (119905) 119891 (119905) 119868120572120573120578
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574120575120577
1199022
V (119905) 119892 (119905) 1198681205721205731205781199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572120573120578
1199021
1205913119906 (119905) 119868
120574120575120577
1199022
V (119905)
+ 3119868120572120573120578
1199021
120591119906 (119905) 119868120574120575120577
1199022
1205882V (119905)
minus 3119868120572120573120578
1199021
1205912119906 (119905) 119868
120574120575120577
1199022
120588V (119905)
minus 119868120572120573120578
1199021
119906 (119905) 119868120574120575120577
1199022
1205883V (119905)]
(49)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof Multiplying both sides of (48) by F(120591 119905 119906(120591)
120572 120573 120578 1199021) and F(120588 119905 V(120588) 120574 120575 120577 119902
2) respectively in (29)
together with (30) taking the 119902-integration of the resultinginequality with respect to 120591 and 120588 from 0 to 119905 with the aid ofDefinition 1 and then applying (39) and (47) on the resultinginequality we get the desired result (49) This completes theproof of Theorem 12
Remark 13 It is easily seen that the special case 119906 = V of (49)inTheorem 12 reduces to that in Theorem 11
3 Special Cases and Concluding Remarks
In this section we consider some consequences of the mainresults derived in the preceding sections Following Garg andChanchlani [16] the operator (18) would reduce immediatelyto the extensively investigated 119902-analogue of Erdelyi-Koberand Riemann-Liouville type fractional integral operators in(19) and (20) respectively (see also [1])
For example if we consider120573 = minus120572 and 120575 = minus120574 (and V = 119906
additionally forTheorem 6) andmake use of the relation (20)Theorems 5 and 6 provide respectively the known fractionalintegral inequalities due to Sroysang [17]
Here we derive certain (presumably) new integralinequalities by setting 120573 = minus120572 and 120575 = minus120574 in Theorems 8to 12 respectively and applying the integral operator (19) tothe resulting inequalities we obtain four integral inequalities
involving 119902-Riemann-Liouville fractional integral operatorsstated in Corollaries 14 to 17 respectively below
Corollary 14 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(50)
Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120574
1199022
119906 (119905)
+ 119868120572
1199021
119906 (119905) ℎ (119905) 119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120572
1199021
119906 (119905) 119892 (119905) 119868120574
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120572
1199021
119906 (119905) 119891 (119905) 119868120574
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574
1199022
119906 (119905) 119891 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574
1199022
119906 (119905) 119892 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574
1199022
119906 (119905) ℎ (119905)
minus119868120572
1199021
119906 (119905) 119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572
1199021
119906 (119905) 119868120574
1199022
119906 (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(51)
for all 120572 gt 0 and 120574 gt 0
Corollary 15 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(52)
Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120572
1199021
119906 (119905) 119891 (119905) (119905) ℎ (119905) 119868120574
1199022
V (119905)
+ 119868120572
1199021
119906 (119905) ℎ (119905) 119868120574
1199022
V (119905) 119891 (119905) 119892 (119905)
+ 119868120572
1199021
119906 (119905) 119892 (119905) 119868120574
1199022
V (119905) 119891 (119905) ℎ (119905)
+ 119868120572
1199021
119906 (119905) 119891 (119905) 119868120574
1199022
119892 (119905) ℎ (119905) V (119905)
minus 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120574
1199022
V (119905) 119891 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120574
1199022
V (119905) 119892 (119905)
minus 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120574
1199022
V (119905) ℎ (119905)
minus119868120572
1199021
119906 (119905) 119868120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120572
1199021
119906 (119905) 119868120574
1199022
V (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(53)
for all 120572 gt 0 and 120574 gt 0
8 Abstract and Applied Analysis
Corollary 16 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572
1199021
119906 (119905)
+ 119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120572
1199021
119906 (119905) ℎ (119905)
+ 119868120574
1199022
119906 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574
1199022
119906 (119905) 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119891 (119905)
minus 119868120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119892 (119905)
minus 119868120574
1199022
119906 (119905) 119891 (119905) 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574
1199022
119906 (119905) 119892 (119905) 119868120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572
1199021
1205913119906 (119905) 119868
120574
1199022
119906 (119905)
+ 3119868120572
1199021
120591119906 (119905) 119868120574
1199022
1205882119906 (119905)
minus 3119868120572
1199021
1205912119906 (119905) 119868
120574
1199022
120588119906 (119905)
minus 119868120572
1199021
119906 (119905) 119868120574
1199022
1205883119906 (119905) ]
(54)
for all 120572 gt 0 and 120574 gt 0
Corollary 17 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572
1199021
119906 (119905)
+ 119868120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120572
1199021
119906 (119905) ℎ (119905)
+ 119868120574
1199022
V (119905) ℎ (119905) 1198681205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574
1199022
V (119905) 1198681205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205721199021
119906 (119905) 119891 (119905)
minus 119868120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119892 (119905)
minus 119868120574
1199022
V (119905) 119891 (119905) 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574
1199022
V (119905) 119892 (119905) 1198681205721199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572
1199021
1205913119906 (119905) 119868
120574
1199022
V (119905)
+ 3119868120572
1199021
120591119906 (119905) 119868120574
1199022
1205882V (119905)
minus 3119868120572
1199021
1205912119906 (119905) 119868
120574
1199022
120588V (119905)
minus119868120572
1199021
119906 (119905) 119868120574
1199022
1205883V (119905)]
(55)
for all 120572 gt 0 and 120574 gt 0
Furthermore if one sets 120573 = 0 = 120575 in Theorems 5 to12 and applying the integral operator (20) to the resulting
inequalities one obtains the following (presumably) newinequalities involving 119902-analogue of the Erdelyi-Kober typefractional integral operators asserted by Corollaries 18 to 23below
Corollary 18 Let 1199021 1199022
isin (0 119905) 119891 119892 and ℎ be threecontinuous and synchronous functions on [0infin) and 119906
[0infin) rarr [0infin) be a continuous functionThen the followinginequality holds true For 119905 isin (0infin)
119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
ge 119868120577120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)
(56)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 19 Let 1199021 1199022
isin (0 119905) 119891 119892 and ℎ be threecontinuous and synchronous functions on [0infin) and 119906 V
[0infin) rarr [0infin) be a continuous functionThen the followinginequality holds true For 119905 isin (0infin)
119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
ge 119868120577120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) 119892 (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) ℎ (119905)
(57)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 20 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(58)
Abstract and Applied Analysis 9
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905)
+ 119868120578120572
1199021
119906 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
1199021
119906 (119905) 119892 (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120578120572
1199021
119906 (119905) 119891 (119905) 119868120577120574
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119892 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120577120574
1199022
119906 (119905) ℎ (119905)
minus119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
119906 (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(59)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 21 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(60)
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
V (119905)
+ 119868120578120572
1199021
119906 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
1199021
119906 (119905) 119892 (119905) 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905)
+ 119868120578120572
1199021
119906 (119905) 119891 (119905) 119868120577120574
1199022
119892 (119905) ℎ (119905) V (119905)
minus 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119891 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119892 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120577120574
1199022
V (119905) ℎ (119905)
minus119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
V (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(61)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 22 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3
respectively Then the following inequality holds true For 119905 isin
(0infin)
10038161003816100381610038161003816119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120577120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905)
minus 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
minus 119868120577120574
1199022
119906 (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120577120574
1199022
119906 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120578120572
1199021
1205913119906 (119905) 119868
120574120575120577
1199022
119906 (119905)
+ 3119868120578120572
1199021
120591119906 (119905) 119868120577120574
1199022
1205882119906 (119905)
minus 3119868120578120572
1199021
1205912119906 (119905) 119868
120577120574
1199022
120588119906 (119905)
minus 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
1205883119906 (119905)]
(62)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 23 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120577120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905)
minus 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
minus 119868120577120574
1199022
V (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120577120574
1199022
V (119905) 119892 (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
10 Abstract and Applied Analysis
le 119871111987121198713[119868120578120572
1199021
1205913119906 (119905) 119868
120577120574
1199022
V (119905)
+ 3119868120578120572
1199021
120591119906 (119905) 119868120577120574
1199022
1205882V (119905)
minus 3119868120578120572
1199021
1205912119906 (119905) 119868
120577120574
1199022
120588V (119905)
minus 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
1205883V (119905)]
(63)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
It is also noted that if we consider 119892(119905) as a unity and120573 = minus120572 and 120575 = minus120574 in Theorems 5 and 6 respectively andmake use of the relation (19) thenwe get the known fractionalintegral inequalities due to Brahim and Taf [15]
We conclude our present investigation by remarkingfurther that the results obtained here are general in characterand useful in deriving various inequalities involving frac-tional and 119902-fractional integral operators Moreover they areexpected to find some applications for establishing unique-ness of solutions in fractional boundary value problems andin fractional partial differential equations in the theory ofquantum calculus
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors should express their deep thanks for the editorialand reviewer boardmembers for accepting this paper in theirreputable journal
References
[1] R P Agrawal ldquoCertain fractional q-integrals and q-derivativesrdquoMathematical Proceedings of the Cambridge Philosophical Soci-ety vol 66 no 2 pp 365ndash370 1969
[2] R P Agarwal Difference Equations and Inequalities MarcelDekker 1992
[3] G A AnastassiouAdvances on Fractional Inequalities SpringerBriefs in Mathematics Springer New York NY USA 2011
[4] G E Andrews q-Series Their Development and Applications inAnalysis NumberTheory Combinatorics Physics and ComputerAlgebra American Mathematical Society 1986
[5] S Belarbi and Z Dahmani ldquoOn some new fractional integralinequalitiesrdquo Journal of Inequalities in Pure and Applied Mathe-matics vol 10 no 3 article 86 pp 1ndash5 2009
[6] Z Dahmani ldquoNew inequalities in fractional integralsrdquo Interna-tional Journal of Nonlinear Science vol 9 no 4 pp 493ndash4972010
[7] Z Dahmani O Mechouar and S Brahami ldquoCertain inequal-ities related to the Chebyshevrsquos functional involving a typeRiemann-Liouville operatorrdquo Bulletin of Mathematical Analysisand Applications vol 3 no 4 pp 38ndash44 2011
[8] S S Dragomir ldquoSome integral inequalities of Gruss typerdquoIndian Journal of Pure and Applied Mathematics vol 31 no 4pp 397ndash415 2000
[9] G Gasper and M Rahman Basic Hypergeometric Series Cam-bridge University Press Cambridge UK 2004
[10] S L Kalla and A Rao ldquoOn Gruss type inequality for hyperge-ometric fractional integralsrdquo Le Matematiche vol 66 no 1 pp57ndash64 2011
[11] H Ogunmez and U M Ozkan ldquoFractional quantum integralinequalitiesrdquo Journal of Inequalities and Applications vol 2011Article ID 787939 7 pages 2011
[12] H M Srivastava and J Choi Zeta and q-Zeta Functions andAssociated Series and Integrals Elsevier Science New York NYUSA 2012
[13] J Choi and P Agarwal ldquoCertain inequalities associated withSaigo fractional integrals and their q-analoguesrdquo Abstract andApplied Analysis vol 2014 Article ID 5792609 11 pages 2014
[14] D Baleanu S D Purohit and P Agarwal ldquoOn fractional inte-gral inequalities involving hypergeometric operatorsrdquo ChineseJournal of Mathematics vol 2014 Article ID 609476 5 pages2014
[15] K Brahim and S Taf ldquoSome fractional integral inequalities inquantum calculusrdquo Journal of Fractional Calculus and Applica-tions vol 4 no 2 pp 1ndash8 2013
[16] M Garg and L Chanchlani ldquoq-analogues of Saigorsquos fractionalcalculus operatorsrdquoBulletin ofMathematical Analysis andAppli-cations vol 3 no 4 pp 169ndash179 2011
[17] B Sroysang ldquoA study on a new fractional integral inequalities inquantum calculusrdquo Advanced Studies inTheoretical Physics vol7 no 14 pp 689ndash692 2013
[18] F H Jackson ldquoOn q-definite integralsrdquoTheQuarterly Journal ofPure and Applied Mathematics vol 41 pp 193ndash203 1910
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Differential EquationsInternational Journal of
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Stochastic AnalysisInternational Journal of
8 Abstract and Applied Analysis
Corollary 16 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572
1199021
119906 (119905)
+ 119868120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120572
1199021
119906 (119905) ℎ (119905)
+ 119868120574
1199022
119906 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574
1199022
119906 (119905) 119868120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119891 (119905)
minus 119868120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119892 (119905)
minus 119868120574
1199022
119906 (119905) 119891 (119905) 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574
1199022
119906 (119905) 119892 (119905) 119868120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572
1199021
1205913119906 (119905) 119868
120574
1199022
119906 (119905)
+ 3119868120572
1199021
120591119906 (119905) 119868120574
1199022
1205882119906 (119905)
minus 3119868120572
1199021
1205912119906 (119905) 119868
120574
1199022
120588119906 (119905)
minus 119868120572
1199021
119906 (119905) 119868120574
1199022
1205883119906 (119905) ]
(54)
for all 120572 gt 0 and 120574 gt 0
Corollary 17 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)10038161003816100381610038161003816119868120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120572
1199021
119906 (119905)
+ 119868120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120572
1199021
119906 (119905) ℎ (119905)
+ 119868120574
1199022
V (119905) ℎ (119905) 1198681205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574
1199022
V (119905) 1198681205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205721199021
119906 (119905) 119891 (119905)
minus 119868120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120572
1199021
119906 (119905) 119892 (119905)
minus 119868120574
1199022
V (119905) 119891 (119905) 119868120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120574
1199022
V (119905) 119892 (119905) 1198681205721199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120572
1199021
1205913119906 (119905) 119868
120574
1199022
V (119905)
+ 3119868120572
1199021
120591119906 (119905) 119868120574
1199022
1205882V (119905)
minus 3119868120572
1199021
1205912119906 (119905) 119868
120574
1199022
120588V (119905)
minus119868120572
1199021
119906 (119905) 119868120574
1199022
1205883V (119905)]
(55)
for all 120572 gt 0 and 120574 gt 0
Furthermore if one sets 120573 = 0 = 120575 in Theorems 5 to12 and applying the integral operator (20) to the resulting
inequalities one obtains the following (presumably) newinequalities involving 119902-analogue of the Erdelyi-Kober typefractional integral operators asserted by Corollaries 18 to 23below
Corollary 18 Let 1199021 1199022
isin (0 119905) 119891 119892 and ℎ be threecontinuous and synchronous functions on [0infin) and 119906
[0infin) rarr [0infin) be a continuous functionThen the followinginequality holds true For 119905 isin (0infin)
119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
ge 119868120577120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)
(56)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 19 Let 1199021 1199022
isin (0 119905) 119891 119892 and ℎ be threecontinuous and synchronous functions on [0infin) and 119906 V
[0infin) rarr [0infin) be a continuous functionThen the followinginequality holds true For 119905 isin (0infin)
119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
ge 119868120577120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) 119892 (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) ℎ (119905)
(57)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 20 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(58)
Abstract and Applied Analysis 9
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905)
+ 119868120578120572
1199021
119906 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
1199021
119906 (119905) 119892 (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120578120572
1199021
119906 (119905) 119891 (119905) 119868120577120574
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119892 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120577120574
1199022
119906 (119905) ℎ (119905)
minus119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
119906 (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(59)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 21 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(60)
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
V (119905)
+ 119868120578120572
1199021
119906 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
1199021
119906 (119905) 119892 (119905) 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905)
+ 119868120578120572
1199021
119906 (119905) 119891 (119905) 119868120577120574
1199022
119892 (119905) ℎ (119905) V (119905)
minus 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119891 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119892 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120577120574
1199022
V (119905) ℎ (119905)
minus119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
V (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(61)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 22 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3
respectively Then the following inequality holds true For 119905 isin
(0infin)
10038161003816100381610038161003816119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120577120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905)
minus 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
minus 119868120577120574
1199022
119906 (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120577120574
1199022
119906 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120578120572
1199021
1205913119906 (119905) 119868
120574120575120577
1199022
119906 (119905)
+ 3119868120578120572
1199021
120591119906 (119905) 119868120577120574
1199022
1205882119906 (119905)
minus 3119868120578120572
1199021
1205912119906 (119905) 119868
120577120574
1199022
120588119906 (119905)
minus 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
1205883119906 (119905)]
(62)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 23 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120577120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905)
minus 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
minus 119868120577120574
1199022
V (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120577120574
1199022
V (119905) 119892 (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
10 Abstract and Applied Analysis
le 119871111987121198713[119868120578120572
1199021
1205913119906 (119905) 119868
120577120574
1199022
V (119905)
+ 3119868120578120572
1199021
120591119906 (119905) 119868120577120574
1199022
1205882V (119905)
minus 3119868120578120572
1199021
1205912119906 (119905) 119868
120577120574
1199022
120588V (119905)
minus 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
1205883V (119905)]
(63)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
It is also noted that if we consider 119892(119905) as a unity and120573 = minus120572 and 120575 = minus120574 in Theorems 5 and 6 respectively andmake use of the relation (19) thenwe get the known fractionalintegral inequalities due to Brahim and Taf [15]
We conclude our present investigation by remarkingfurther that the results obtained here are general in characterand useful in deriving various inequalities involving frac-tional and 119902-fractional integral operators Moreover they areexpected to find some applications for establishing unique-ness of solutions in fractional boundary value problems andin fractional partial differential equations in the theory ofquantum calculus
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors should express their deep thanks for the editorialand reviewer boardmembers for accepting this paper in theirreputable journal
References
[1] R P Agrawal ldquoCertain fractional q-integrals and q-derivativesrdquoMathematical Proceedings of the Cambridge Philosophical Soci-ety vol 66 no 2 pp 365ndash370 1969
[2] R P Agarwal Difference Equations and Inequalities MarcelDekker 1992
[3] G A AnastassiouAdvances on Fractional Inequalities SpringerBriefs in Mathematics Springer New York NY USA 2011
[4] G E Andrews q-Series Their Development and Applications inAnalysis NumberTheory Combinatorics Physics and ComputerAlgebra American Mathematical Society 1986
[5] S Belarbi and Z Dahmani ldquoOn some new fractional integralinequalitiesrdquo Journal of Inequalities in Pure and Applied Mathe-matics vol 10 no 3 article 86 pp 1ndash5 2009
[6] Z Dahmani ldquoNew inequalities in fractional integralsrdquo Interna-tional Journal of Nonlinear Science vol 9 no 4 pp 493ndash4972010
[7] Z Dahmani O Mechouar and S Brahami ldquoCertain inequal-ities related to the Chebyshevrsquos functional involving a typeRiemann-Liouville operatorrdquo Bulletin of Mathematical Analysisand Applications vol 3 no 4 pp 38ndash44 2011
[8] S S Dragomir ldquoSome integral inequalities of Gruss typerdquoIndian Journal of Pure and Applied Mathematics vol 31 no 4pp 397ndash415 2000
[9] G Gasper and M Rahman Basic Hypergeometric Series Cam-bridge University Press Cambridge UK 2004
[10] S L Kalla and A Rao ldquoOn Gruss type inequality for hyperge-ometric fractional integralsrdquo Le Matematiche vol 66 no 1 pp57ndash64 2011
[11] H Ogunmez and U M Ozkan ldquoFractional quantum integralinequalitiesrdquo Journal of Inequalities and Applications vol 2011Article ID 787939 7 pages 2011
[12] H M Srivastava and J Choi Zeta and q-Zeta Functions andAssociated Series and Integrals Elsevier Science New York NYUSA 2012
[13] J Choi and P Agarwal ldquoCertain inequalities associated withSaigo fractional integrals and their q-analoguesrdquo Abstract andApplied Analysis vol 2014 Article ID 5792609 11 pages 2014
[14] D Baleanu S D Purohit and P Agarwal ldquoOn fractional inte-gral inequalities involving hypergeometric operatorsrdquo ChineseJournal of Mathematics vol 2014 Article ID 609476 5 pages2014
[15] K Brahim and S Taf ldquoSome fractional integral inequalities inquantum calculusrdquo Journal of Fractional Calculus and Applica-tions vol 4 no 2 pp 1ndash8 2013
[16] M Garg and L Chanchlani ldquoq-analogues of Saigorsquos fractionalcalculus operatorsrdquoBulletin ofMathematical Analysis andAppli-cations vol 3 no 4 pp 169ndash179 2011
[17] B Sroysang ldquoA study on a new fractional integral inequalities inquantum calculusrdquo Advanced Studies inTheoretical Physics vol7 no 14 pp 689ndash692 2013
[18] F H Jackson ldquoOn q-definite integralsrdquoTheQuarterly Journal ofPure and Applied Mathematics vol 41 pp 193ndash203 1910
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Differential EquationsInternational Journal of
Volume 2014
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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Mathematical PhysicsAdvances in
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OptimizationJournal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Volume 2014
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Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 9
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905)
+ 119868120578120572
1199021
119906 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
1199021
119906 (119905) 119892 (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905)
+ 119868120578120572
1199021
119906 (119905) 119891 (119905) 119868120577120574
1199022
119892 (119905) ℎ (119905) 119906 (119905)
minus 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120577120574
1199022
119906 (119905) 119892 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120577120574
1199022
119906 (119905) ℎ (119905)
minus119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
119906 (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(59)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 21 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three continuousand synchronous functions on [0infin) satisfying the followingcondition
120595 le 119891 (119909) le Ψ 120601 le 119892 (119909) le Φ 120596 le ℎ (119909) le Ω
(120601 120595 120596ΦΨΩ isin R 119909 isin [0infin))
(60)
Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
V (119905)
+ 119868120578120572
1199021
119906 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
1199021
119906 (119905) 119892 (119905) 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905)
+ 119868120578120572
1199021
119906 (119905) 119891 (119905) 119868120577120574
1199022
119892 (119905) ℎ (119905) V (119905)
minus 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119891 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905) 119868120577120574
1199022
V (119905) 119892 (119905)
minus 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) 119868120577120574
1199022
V (119905) ℎ (119905)
minus119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905)10038161003816100381610038161003816
le 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
V (119905) (Ψ minus 120595) (Φ minus 120601) (Ω minus 120596)
(61)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 22 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3
respectively Then the following inequality holds true For 119905 isin
(0infin)
10038161003816100381610038161003816119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
119906 (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
119906 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
119906 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120577120574
1199022
119906 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905)
minus 119868120577120574
1199022
119906 (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
minus 119868120577120574
1199022
119906 (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120577120574
1199022
119906 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
le 119871111987121198713[119868120578120572
1199021
1205913119906 (119905) 119868
120574120575120577
1199022
119906 (119905)
+ 3119868120578120572
1199021
120591119906 (119905) 119868120577120574
1199022
1205882119906 (119905)
minus 3119868120578120572
1199021
1205912119906 (119905) 119868
120577120574
1199022
120588119906 (119905)
minus 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
1205883119906 (119905)]
(62)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Corollary 23 Let 1199021 1199022isin (0 119905) 119906 [0infin) rarr [0infin) be
a continuous function and 119891 119892 and ℎ be three Lipschitzianfunctions on [0infin) with the constants 119871
1 1198712 and 119871
3 respec-
tively Then the following inequality holds true For 119905 isin (0infin)
10038161003816100381610038161003816119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905)
+ 119868120577120574
1199022
V (119905) 119891 (119905) 119892 (119905) 119868120578120572
1199021
119906 (119905) ℎ (119905)
+ 119868120577120574
1199022
V (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120577120574
1199022
V (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) 119892 (119905) ℎ (119905)
minus 119868120577120574
1199022
V (119905) 119892 (119905) ℎ (119905) 1198681205781205721199021
119906 (119905) 119891 (119905)
minus 119868120577120574
1199022
V (119905) 119891 (119905) ℎ (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905)
minus 119868120577120574
1199022
V (119905) 119891 (119905) 119868120578120572
1199021
119906 (119905) 119892 (119905) ℎ (119905)
minus119868120577120574
1199022
V (119905) 119892 (119905) 1198681205781205721199021
119906 (119905) 119891 (119905) ℎ (119905)10038161003816100381610038161003816
10 Abstract and Applied Analysis
le 119871111987121198713[119868120578120572
1199021
1205913119906 (119905) 119868
120577120574
1199022
V (119905)
+ 3119868120578120572
1199021
120591119906 (119905) 119868120577120574
1199022
1205882V (119905)
minus 3119868120578120572
1199021
1205912119906 (119905) 119868
120577120574
1199022
120588V (119905)
minus 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
1205883V (119905)]
(63)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
It is also noted that if we consider 119892(119905) as a unity and120573 = minus120572 and 120575 = minus120574 in Theorems 5 and 6 respectively andmake use of the relation (19) thenwe get the known fractionalintegral inequalities due to Brahim and Taf [15]
We conclude our present investigation by remarkingfurther that the results obtained here are general in characterand useful in deriving various inequalities involving frac-tional and 119902-fractional integral operators Moreover they areexpected to find some applications for establishing unique-ness of solutions in fractional boundary value problems andin fractional partial differential equations in the theory ofquantum calculus
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors should express their deep thanks for the editorialand reviewer boardmembers for accepting this paper in theirreputable journal
References
[1] R P Agrawal ldquoCertain fractional q-integrals and q-derivativesrdquoMathematical Proceedings of the Cambridge Philosophical Soci-ety vol 66 no 2 pp 365ndash370 1969
[2] R P Agarwal Difference Equations and Inequalities MarcelDekker 1992
[3] G A AnastassiouAdvances on Fractional Inequalities SpringerBriefs in Mathematics Springer New York NY USA 2011
[4] G E Andrews q-Series Their Development and Applications inAnalysis NumberTheory Combinatorics Physics and ComputerAlgebra American Mathematical Society 1986
[5] S Belarbi and Z Dahmani ldquoOn some new fractional integralinequalitiesrdquo Journal of Inequalities in Pure and Applied Mathe-matics vol 10 no 3 article 86 pp 1ndash5 2009
[6] Z Dahmani ldquoNew inequalities in fractional integralsrdquo Interna-tional Journal of Nonlinear Science vol 9 no 4 pp 493ndash4972010
[7] Z Dahmani O Mechouar and S Brahami ldquoCertain inequal-ities related to the Chebyshevrsquos functional involving a typeRiemann-Liouville operatorrdquo Bulletin of Mathematical Analysisand Applications vol 3 no 4 pp 38ndash44 2011
[8] S S Dragomir ldquoSome integral inequalities of Gruss typerdquoIndian Journal of Pure and Applied Mathematics vol 31 no 4pp 397ndash415 2000
[9] G Gasper and M Rahman Basic Hypergeometric Series Cam-bridge University Press Cambridge UK 2004
[10] S L Kalla and A Rao ldquoOn Gruss type inequality for hyperge-ometric fractional integralsrdquo Le Matematiche vol 66 no 1 pp57ndash64 2011
[11] H Ogunmez and U M Ozkan ldquoFractional quantum integralinequalitiesrdquo Journal of Inequalities and Applications vol 2011Article ID 787939 7 pages 2011
[12] H M Srivastava and J Choi Zeta and q-Zeta Functions andAssociated Series and Integrals Elsevier Science New York NYUSA 2012
[13] J Choi and P Agarwal ldquoCertain inequalities associated withSaigo fractional integrals and their q-analoguesrdquo Abstract andApplied Analysis vol 2014 Article ID 5792609 11 pages 2014
[14] D Baleanu S D Purohit and P Agarwal ldquoOn fractional inte-gral inequalities involving hypergeometric operatorsrdquo ChineseJournal of Mathematics vol 2014 Article ID 609476 5 pages2014
[15] K Brahim and S Taf ldquoSome fractional integral inequalities inquantum calculusrdquo Journal of Fractional Calculus and Applica-tions vol 4 no 2 pp 1ndash8 2013
[16] M Garg and L Chanchlani ldquoq-analogues of Saigorsquos fractionalcalculus operatorsrdquoBulletin ofMathematical Analysis andAppli-cations vol 3 no 4 pp 169ndash179 2011
[17] B Sroysang ldquoA study on a new fractional integral inequalities inquantum calculusrdquo Advanced Studies inTheoretical Physics vol7 no 14 pp 689ndash692 2013
[18] F H Jackson ldquoOn q-definite integralsrdquoTheQuarterly Journal ofPure and Applied Mathematics vol 41 pp 193ndash203 1910
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
10 Abstract and Applied Analysis
le 119871111987121198713[119868120578120572
1199021
1205913119906 (119905) 119868
120577120574
1199022
V (119905)
+ 3119868120578120572
1199021
120591119906 (119905) 119868120577120574
1199022
1205882V (119905)
minus 3119868120578120572
1199021
1205912119906 (119905) 119868
120577120574
1199022
120588V (119905)
minus 119868120578120572
1199021
119906 (119905) 119868120577120574
1199022
1205883V (119905)]
(63)
for all 120572 gt 0 120574 gt 0 and 120578 120577 isin R
It is also noted that if we consider 119892(119905) as a unity and120573 = minus120572 and 120575 = minus120574 in Theorems 5 and 6 respectively andmake use of the relation (19) thenwe get the known fractionalintegral inequalities due to Brahim and Taf [15]
We conclude our present investigation by remarkingfurther that the results obtained here are general in characterand useful in deriving various inequalities involving frac-tional and 119902-fractional integral operators Moreover they areexpected to find some applications for establishing unique-ness of solutions in fractional boundary value problems andin fractional partial differential equations in the theory ofquantum calculus
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors should express their deep thanks for the editorialand reviewer boardmembers for accepting this paper in theirreputable journal
References
[1] R P Agrawal ldquoCertain fractional q-integrals and q-derivativesrdquoMathematical Proceedings of the Cambridge Philosophical Soci-ety vol 66 no 2 pp 365ndash370 1969
[2] R P Agarwal Difference Equations and Inequalities MarcelDekker 1992
[3] G A AnastassiouAdvances on Fractional Inequalities SpringerBriefs in Mathematics Springer New York NY USA 2011
[4] G E Andrews q-Series Their Development and Applications inAnalysis NumberTheory Combinatorics Physics and ComputerAlgebra American Mathematical Society 1986
[5] S Belarbi and Z Dahmani ldquoOn some new fractional integralinequalitiesrdquo Journal of Inequalities in Pure and Applied Mathe-matics vol 10 no 3 article 86 pp 1ndash5 2009
[6] Z Dahmani ldquoNew inequalities in fractional integralsrdquo Interna-tional Journal of Nonlinear Science vol 9 no 4 pp 493ndash4972010
[7] Z Dahmani O Mechouar and S Brahami ldquoCertain inequal-ities related to the Chebyshevrsquos functional involving a typeRiemann-Liouville operatorrdquo Bulletin of Mathematical Analysisand Applications vol 3 no 4 pp 38ndash44 2011
[8] S S Dragomir ldquoSome integral inequalities of Gruss typerdquoIndian Journal of Pure and Applied Mathematics vol 31 no 4pp 397ndash415 2000
[9] G Gasper and M Rahman Basic Hypergeometric Series Cam-bridge University Press Cambridge UK 2004
[10] S L Kalla and A Rao ldquoOn Gruss type inequality for hyperge-ometric fractional integralsrdquo Le Matematiche vol 66 no 1 pp57ndash64 2011
[11] H Ogunmez and U M Ozkan ldquoFractional quantum integralinequalitiesrdquo Journal of Inequalities and Applications vol 2011Article ID 787939 7 pages 2011
[12] H M Srivastava and J Choi Zeta and q-Zeta Functions andAssociated Series and Integrals Elsevier Science New York NYUSA 2012
[13] J Choi and P Agarwal ldquoCertain inequalities associated withSaigo fractional integrals and their q-analoguesrdquo Abstract andApplied Analysis vol 2014 Article ID 5792609 11 pages 2014
[14] D Baleanu S D Purohit and P Agarwal ldquoOn fractional inte-gral inequalities involving hypergeometric operatorsrdquo ChineseJournal of Mathematics vol 2014 Article ID 609476 5 pages2014
[15] K Brahim and S Taf ldquoSome fractional integral inequalities inquantum calculusrdquo Journal of Fractional Calculus and Applica-tions vol 4 no 2 pp 1ndash8 2013
[16] M Garg and L Chanchlani ldquoq-analogues of Saigorsquos fractionalcalculus operatorsrdquoBulletin ofMathematical Analysis andAppli-cations vol 3 no 4 pp 169ndash179 2011
[17] B Sroysang ldquoA study on a new fractional integral inequalities inquantum calculusrdquo Advanced Studies inTheoretical Physics vol7 no 14 pp 689ndash692 2013
[18] F H Jackson ldquoOn q-definite integralsrdquoTheQuarterly Journal ofPure and Applied Mathematics vol 41 pp 193ndash203 1910
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
