An Automatic Quadrature Schemes and Error Estimates for Semibounded Weighted Hadamard Type...

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Research Article An Automatic Quadrature Schemes and Error Estimates for Semibounded Weighted Hadamard Type Hypersingular Integrals Sirajo Lawan Bichi, 1,2 Z. K. Eshkuvatov, 1,3 and N. M. A. Nik Long 1,3 1 Department of Mathematics, Faculty of Science, Universiti Putra Malaysia, Selangor, Malaysia 2 Department of Mathematics, Faculty of Science, Bayero University Kano (B.U.K), Kano, Nigeria 3 Institute for Mathematical Research, Universiti Putra Malaysia, Selangor, Malaysia Correspondence should be addressed to Z. K. Eshkuvatov; [email protected] Received 4 July 2014; Revised 23 October 2014; Accepted 23 October 2014; Published 1 December 2014 Academic Editor: Gang Xu Copyright © 2014 Sirajo Lawan Bichi et al. 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. e approximate solutions for the semibounded Hadamard type hypersingular integrals (HSIs) for smooth density function are investigated. e automatic quadrature schemes (AQSs) are constructed by approximating the density function using the third and fourth kinds of Chebyshev polynomials. Error estimates for the semibounded solutions are obtained in the class of ℎ() ∈ , [−1, 1]. Numerical results for the obtained quadrature schemes revealed that the proposed methods are highly accurate when the density function ℎ () is any polynomial or rational functions. e results are in line with the theoretical findings. 1. Introduction In an attempt to solve the Cauchy’s problem for hyperbolic partial differential equations, Hadamard [1] introduced the concept of hypersingular integrals. He defined the hypersin- gular integrals to be the finite part of a divergent integral in which singularities are arranged at the endpoints of the interval for one-dimensional integrals and on the domain boundary for multidimensional integrals. ese forms of integrals are later identified as Hadamards finite part inte- grals or simply Hadamards integrals. Numerous applica- tions of the approximate solutions of Hadamard integrals have been found in mechanics, electrodynamics, aerody- namics, and acoustics with various numerical approaches [2–5]. Some authors applied various regularization meth- ods [2, 4, 5] for the transformation of the hypersingular integrals into singular or weakly singular integrals, while others applied the direct numerical computation of the finite part integrals by using various quadrature or cubature formulae [4, 6]. Hypersingular integral represents a natural extension of singular integrals in the Cauchy principal value: (ℎ, ) = = ∫ 1 −1 () ℎ () ( − ) 2 = (ℎ, ) , (1) where (ℎ, ) = − ∫ 1 −1 () ℎ () − . (2) Hui and Shia [7] developed Gaussian quadrature formula for hypersingular integrals with second-order singularities which have been extensively used for the solution of elasticity problems. Chen [8] used the hypersingular integral equation approach to study the plane elastic problem for crack prob- lem. Nik Long and Eshkuvatov [9] formulated the multiple curved crack problems into hypersingular integral equations via the complex variable function method. Obayis et al. [10] developed an automatic quadrature scheme (AQS) based Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2014, Article ID 387246, 13 pages http://dx.doi.org/10.1155/2014/387246

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Research ArticleAn Automatic Quadrature Schemes and ErrorEstimates for Semibounded Weighted Hadamard TypeHypersingular Integrals

Sirajo Lawan Bichi12 Z K Eshkuvatov13 and N M A Nik Long13

1 Department of Mathematics Faculty of Science Universiti Putra Malaysia Selangor Malaysia2 Department of Mathematics Faculty of Science Bayero University Kano (BUK) Kano Nigeria3 Institute for Mathematical Research Universiti Putra Malaysia Selangor Malaysia

Correspondence should be addressed to Z K Eshkuvatov zainidinupmedumy

Received 4 July 2014 Revised 23 October 2014 Accepted 23 October 2014 Published 1 December 2014

Academic Editor Gang Xu

Copyright copy 2014 Sirajo Lawan Bichi et al 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

The approximate solutions for the semibounded Hadamard type hypersingular integrals (HSIs) for smooth density function areinvestigated The automatic quadrature schemes (AQSs) are constructed by approximating the density function using the thirdand fourth kinds of Chebyshev polynomials Error estimates for the semibounded solutions are obtained in the class of ℎ(119905) isin119862119873120572[minus1 1] Numerical results for the obtained quadrature schemes revealed that the proposed methods are highly accurate when

the density function ℎ (119905) is any polynomial or rational functions The results are in line with the theoretical findings

1 Introduction

In an attempt to solve the Cauchyrsquos problem for hyperbolicpartial differential equations Hadamard [1] introduced theconcept of hypersingular integrals He defined the hypersin-gular integrals to be the finite part of a divergent integralin which singularities are arranged at the endpoints of theinterval for one-dimensional integrals and on the domainboundary for multidimensional integrals These forms ofintegrals are later identified as Hadamards finite part inte-grals or simply Hadamards integrals Numerous applica-tions of the approximate solutions of Hadamard integralshave been found in mechanics electrodynamics aerody-namics and acoustics with various numerical approaches[2ndash5] Some authors applied various regularization meth-ods [2 4 5] for the transformation of the hypersingularintegrals into singular or weakly singular integrals whileothers applied the direct numerical computation of thefinite part integrals by using various quadrature or cubatureformulae [4 6] Hypersingular integral represents a natural

extension of singular integrals in the Cauchy principalvalue

119867(ℎ 119909) = =int

1

minus1

119908 (119905) ℎ (119905)

(119905 minus 119909)2119889119905 =

119889

119889119909119862 (ℎ 119909) (1)

where

119862 (ℎ 119909) = minusint

1

minus1

119908 (119905) ℎ (119905)

119905 minus 119909119889119905 (2)

Hui and Shia [7] developed Gaussian quadrature formulafor hypersingular integrals with second-order singularitieswhich have been extensively used for the solution of elasticityproblems Chen [8] used the hypersingular integral equationapproach to study the plane elastic problem for crack prob-lem Nik Long and Eshkuvatov [9] formulated the multiplecurved crack problems into hypersingular integral equationsvia the complex variable function method Obayis et al [10]developed an automatic quadrature scheme (AQS) based

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 387246 13 pageshttpdxdoiorg1011552014387246

2 Abstract and Applied Analysis

on Clenshaw-Curtis Chebyshev quadrature methods for theevaluation of HSIs of the form

119876119894(119891 119909) = =int

1

minus1

119908119894 (119905) 119891 (119905)

(119905 minus 119909)2119889119905 119909 isin (minus1 1) 119894 = 0 1 2

(3)

where 1199080(119905) = 1 119908

1(119905) = radic1 minus 1199052 and 119908

2(119905) = 1radic1 minus 1199052

are the weight functions and 119891(119905) is a smooth or continuousError estimation for the developed AQS for 119894 = 0 wasobtained in the class of functions 119862119873+2120572[minus1 1]

In this note we have developed the automatic quadraturescheme based on Chebyshev Gauss quadrature formula forthe semibounded solutions of the hypersingular integrals ofthe form

119867119894 (ℎ 119909) =

119908119894 (119909)

120587=int

1

minus1

ℎ (119905)

119908119894 (119905) (119905 minus 119909)

2119889119905 119909 isin (minus1 1) (4)

where 1199081(119905) = radic(1 minus 119905)(1 + 119905) and 119908

2(119905) = radic(1 + 119905)(1 minus 119905)

are the weights and ℎ(119905) is a smooth function The pro-posed automatic quadrature schemes constructed cover thedifferent weights functions that have not been considered byObayis et al [10] Unlike the work of Obayis et al [10] whichtakes advantage of Clenshaw-Curtis Chebyshev quadraturemethods the construction of our AQSs is based on Cheby-shev Gauss quadrature methods The method proposed byHui and Shia [7] works only for even number 119873 of knotspoints whereas AQS for our problems works well for all 119873and converges very fast to the exact solution

Present in Section 2 are the mathematical concepts ofthe third and fourth kinds of Chebyshev polynomials andsingular integrals In Section 3 the construction of automaticquadrature schemes for semibounded solutions of hypersin-gular integrals is given Rate of convergence of the suggestedmethod is discussed in Section 4 Numerical examples andresults are provided in Section 5

2 Mathematical Concepts of Third andFourth Kinds of Chebyshev Polynomialsand Singular Integrals

The Chebyshev polynomials 119881119898(119905) and119882

119898(119905) are defined as

follows

Definition 1 (Mason and Handscomb [11]) The chebyshevpolynomial119881

119898(119905) of the third kind on [minus1 1] is a polynomial

of degree119898 in 119905 defined by

119881119898 (119905) =

cos (119898 + 12) 120579cos (12) 120579

119905 = cos 120579 0 le 120579 le 120587 (5)

Definition 2 (Mason and Handscomb [11]) The chebyshevpolynomial119882

119898(119905) of the fourth kind on [minus1 1] is a polyno-

mial of degree119898 in 119905 defined by

119882119898 (119905) =

sin (119898 + 12) 120579sin (12) 120579

119905 = cos 120579 0 le 120579 le 120587 (6)

Both polynomials

119875119894119898 (119905) =

119881119898 (119905) 119894 = 1

119882119898 (119905) 119894 = 2

(7)

satisfy the recurrence relation

119875119894119898 (119905) = 2119905119875119894(119898minus1) (119905) minus 119875119894(119898minus2) (119905)

119898 = 2 3 119894 = 1 2

(8)

with the same starting value

1198751198940 (119905) =

1198810 (119905) = 1 119894 = 1

1198820 (119905) = 1 119894 = 2

1198751198941 (119905) =

1198811 (119905) = 2119905 minus 1 119894 = 1

1198821 (119905) = 2119905 + 1 119894 = 2

(9)

The third and fourth kinds of Chebyshev polynomialsare orthogonal on [minus1 1] with the weights 119908

1(119905) and 119908

2(119905)

respectively They also have continuous orthogonality givenin [11] respectively as

⟨119881119898 119881119895⟩ = int

1

minus1

radic1 + 119905

1 minus 119905119881119898 (119905) 119881119895 (119905) 119889119905 =

0 119898 = 119895

120587 119898 = 119895

⟨119882119898119882119895⟩ = int

1

minus1

radic1 minus 119905

1 + 119905119882119898 (119905)119882119895 (119905) 119889119905

= 0 119898 = 119895 (119898 119895 le 119873)

120587 119898 = 119895

(10)

The third kind and fourth kind of Chebyshev polynomialshave discrete orthogonality respectively as

⟨119881119898 119881119895⟩ =

119873

sum

119896=0

(1 + 119905119896) 119881119898(119905119896) 119881119895(119905119896)

= 0 119898 = 119895 (119898 119895 le 119873)

119873 + 1 0 le 119895 = 119898 le 119873

⟨119882119898119882119895⟩ =

119873

sum

119896=0

(1 minus 119905119896)119882119898(119905119896)119882119895(119905119896)

= 0 119898 = 119895 (119898 119895 le 119873)

119873 + 1 119895 = 119898 (0 le 119898 le 119873)

(11)

The four kinds of Chebyshev polynomials are connectedvia the relations

(1 + 119905) 119881119896 (119905) = 119879119896+1 (119905) + 119879119896 (119905) (12)

119882119896 (119905) = 119880119896 (119905) + 119880119896minus1 (119905) (13)

(1 minus 119905)119882119896 (119905) = 119879119896+1 (119905) minus 119879119896 (119905) (14)

119881119896 (119905) = 119880119896 (119905) minus 119880119896minus1 (119905) (15)

Abstract and Applied Analysis 3

and the differentiation formulae119889

119889119909119879119896 (119905) = 119896119880119896minus1 (119905)

119889

119889119909119880119896 (119905) =

(119896 + 1) 119879119896+1 (119905) minus 119905119880119896

1199052 minus 1

(16)

The Hilbert type integral transform is as follows

Lemma 3 (Mason andHandscomb [11]) Consider the follow-ing

minusint

120587

0

cos119898120579cos 120579 minus cos120572

119889120579 = 120587sin119898120572sin120572

(17)

for any 120572 isin [0 120587]119898 = 1 2 3

Lemma4 (Mason andHandscomb [11]) Consider the follow-ing

minusint

120587

0

sin119898120579 sin 120579cos 120579 minus cos120572

119889120579 = minus120587 cos119898120572 (18)

for any 120572 isin [0 120587]119898 = 1 2 3

The estimations for some integrals are established

Lemma 5 (Israilov [12]) Let 0 lt 120590 le 1 and 1 minus 119909 ge 120575119873

0 lt 120575119873lt 12 and for all 119894 = 1 2

120574119894 (119909) = 119908119894 (119909) int

119909minus120575119873

minus1

119889119905

119908119894 (119905) (119905 minus 119909)

120574lowast

119894(119909) = 119908119894 (119909) int

1

119909+120575119873

119889119905

119908119894 (119905) (119905 minus 119909)

119879119894(119909 120575119873 120590) = 119908

119894 (119909) int

119909+120575119873

119909minus120575119873

|119905 minus 119909|120590minus1119889119905

119908119894 (119905)

(19)

where 1199081(119909) = radic(1 minus 119909)(1 + 119909) and 119908

2(119909) =

radic(1 + 119909)(1 minus 119909)The following estimations are true

(i) |120574119894(119909)| le ln(2120575

119873) + 120587radic2

(ii) 1205741(minus119909) = minus120574

lowast

2(119909) 1205742(minus119909) = minus120574

lowast

1(119909)

(iii) 119879119894(119909 120575119873 120590) le 120601

119894(120590)120575120590

119873 max120601

1(120590) 1206012(120590) le radic10 times

2minus120590+ (1 + radic10 times 2

minus1minus120590)120590minus1

3 Automatic Quadrature Schemes forSemibounded Hypersingular Integrals

To construct AQSs for the semiboundedHSIs (4) the densityfunction ℎ(119905) is approximated with the truncated series ofthird and forth kinds of Chebyshev polynomials

ℎ (119905) asymp 119878119894119873 (119905) =

119873

sum

119898=0

1198871119898119881119898 (119905) 119894 = 1

119873

sum

119898=0

1198872119898119882119898 (119905) 119894 = 2

(20)

Substituting (20) into (4) we have

119867119894 (ℎ 119909)

asymp 119867119894119873 (ℎ 119909)

=119908119894 (119905)

120587=int

1

minus1

119878119894119873 (119905)

119908119894 (119909) (119905 minus 119909)

2119889119905

=

radic1 minus 119909

1 + 119909

119873

sum

119898=0

1198871119898

119889

119889119909[1

120587minusint

1

minus1

(1 + 119905) 119881119898 (119905)

radic1 minus 1199052 (119905 minus 119909)119889119905] 119894 = 1

radic1 + 119909

1 minus 119909

119873

sum

119898=0

1198872119898

119889

119889119909[1

120587minusint

1

minus1

(1 minus 119905)119882119898 (119905)

radic1 minus 1199052 (119905 minus 119909)119889119905] 119894 = 2

=

radic1 minus 119909

1 + 1199091199021 (119909) 119894 = 1

radic1 + 119909

1 minus 1199091199022 (119909) 119894 = 2

(21)

where

119902119894 (119909) =

119889

119889119909

119873

sum

119898=0

1198871119898119882119898 (119909) 119894 = 1

119889

119889119909

119873

sum

119898=0

1198872119898119881119898 (119909) 119894 = 2

(22)

By applying the orthogonality relations (10) we obtain thecoefficients 119887

119894119898 119894 = 1 2 as

119887119894119898=

1

120587int

1

minus1

radic1 + 119905

1 minus 119905ℎ (119905) 119881119898 (119905) 119889119905 119894 = 1

1

120587int

1

minus1

radic1 minus 119905

1 + 119905ℎ (119905)119882119898 (119905) 119889119905 119894 = 2

(23)

For the discrete coefficients 119887119894119898 we choose the collocation

points 119905 = 119905119896 at the zeros of 119879

119873+1 namely

119905 = 119905119896= cos (2119896 + 1) 120587

2119873 + 2 119896 = 0 1 2 119873 (24)

and using the orthogonality conditions (11) and relations(12) and (14) respectively as well as applying compositetrapezium rule we arrive at

119887119894119898asymp

1

119873 + 32

119873

sum

119896=0

(1 + 119905119896) 119881119898(119905119896) ℎ (119905119896) 119894 = 1

1

119873 + 32

119873

sum

119896=0

(119905119896minus 1)119882

119898(119905119896) ℎ (119905119896) 119894 = 2

(25)

4 Error Estimation for the SemiboundedHypersingular Integrals

Let us introduce the classes of functions

(i) 119867120572([minus1 1] 119871) is a class of function satisfying Holdercondition on the interval with the index 120572 andconstant 119871

4 Abstract and Applied Analysis

(ii) 119862119873120572[minus1 1] = ℎ(119905) ℎ(119873)(119905) isin 119867120572([minus1 1] 119871119873)

(iii) 119862[minus1 1] is a class of continuous functions on theinterval [minus1 1] with the norm

ℎ = ℎ(119909)119862[minus11] = max119909isin[minus11]

|ℎ (119909)| (26)

(iv) 119862119903[minus1 1] = ℎ(119905) ℎ119903(119905) isin 119862([minus1 1])Let 119890119894119873(119905) = ℎ(119905) minus 119878

119894119873(119905) and let the maximum norm of

the error be defined as10038171003817100381710038171198901198941198731003817100381710038171003817 = max119905isin[minus11]

1003816100381610038161003816119890119894119873 (119905)1003816100381610038161003816 (27)

It is known in [11] that for any 119909 isin (minus1 1)

1198901119873 (119909) =

ℎ(119873)(120578119909)119882119873 (119909)

2119873119873 120578119909isin [minus1 1]

1198902119873 (119909) =

ℎ(119873)(120578119909) 119881119873 (119909)

2119873119873 120578119909isin [minus1 1]

(28)

1198901015840

1119873(119909) =

1

2119873119873[119882119873 (119909)

119889

119889119909ℎ(119873)(120578119909) + ℎ(119873)(120578119909)1198821015840

119873(119909)]

120578119909isin [minus1 1]

1198901015840

2119873(119909) =

1

2119873119873[119881119873 (119909)

119889

119889119909ℎ(119873)(120578119909) + ℎ(119873)(120578119909) 1198811015840

119873(119909)]

120578119909isin [minus1 1]

(29)

As a consequence of (29) we arrive at the followingestimations100381710038171003817100381711989011198731003817100381710038171003817 =100381710038171003817100381711989021198731003817100381710038171003817 le(2119873 + 1)1198721

2119873 sdot 119873

(30)100381710038171003817100381710038171198901015840

1119873

10038171003817100381710038171003817=100381710038171003817100381710038171198901015840

2119873

10038171003817100381710038171003817

le119872

12 sdot 2119873minus3 (119873 minus 3)[1 +

1198711119873

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)]

(31)

where119872 = max 119872

11198722 119872

1= max120578119909isin[minus11]

10038161003816100381610038161003816ℎ(119873)(120578119909)10038161003816100381610038161003816

1198722= max120578119909isin[minus11]

10038161003816100381610038161003816100381610038161003816

119889

119889119909ℎ(119873)(120578119909)

10038161003816100381610038161003816100381610038161003816lt infin

(32)

1198711119873= 1 +

3

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)(1 +

3

119873) (33)

Lemma 6 The Lipschitz constants 119871119895119894119873

119895 = 3 4 119894 = 1 2of 119881119873(119909) and119882

119873(119909) are

1003816100381610038161003816119881119873 (119909) minus 119881119873 (119910)1003816100381610038161003816 le 11987131119873

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

100381610038161003816100381610038161198811015840

119873(119909) minus 119881

1015840

119873(119910)10038161003816100381610038161003816le 11987132119873

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

1003816100381610038161003816119882119873 (119909) minus 119882119873 (119910)1003816100381610038161003816 le 11987141119873

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

100381610038161003816100381610038161198821015840

119873(119909) minus 119882

1015840

119873(119910)10038161003816100381610038161003816le 11987142119873

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

(34)

where

11987131119873= 11987141119873=119873

3(119873 + 1) (2119873 + 1)

11987132119873= 11987142119873=119873

15(119873 + 1) (119873 + 2) (2119873 + 1) (119873 minus 1)

(35)

Proof Let 11987131119873

be the Lipschitz constant of the third kindChebyshev polynomial 119881

119873(119905) that is

1003816100381610038161003816119881119873 (119905) minus 119881119873 (119909)1003816100381610038161003816 le 11987131119873 |119905 minus 119909| (36)

where

11987131119873= max120577119905isin[minus11]

100381610038161003816100381610038161198811015840

119873(120577119905)10038161003816100381610038161003816 (37)

which gives sequence of numbers

0 2 10 28 60 110 182 280 408 570 (38)

These sequences can be generated by

11987131(119873)

= 11987131(119873minus1)

+ 21198732 11987131(0)= 0 (39)

which leads to

11987131119873=119873

3(119873 + 1) (2119873 + 1) (40)

In a similar way we can get

11987141119873=119873

3(119873 + 1) (2119873 + 1) (41)

Let 11987132(119873)

be the Lipschitz constant of the derivative ofthird kind Chebyshev polynomial 1198811015840

119873(119905) that is

100381610038161003816100381610038161198811015840

119873(119905) minus 119881

1015840

119873(119909)10038161003816100381610038161003816le 11987132119873 |119905 minus 119909| (42)

where

11987132119873= max120577119905isin[minus11]

1003816100381610038161003816100381611988110158401015840

119873(120577119905)10038161003816100381610038161003816 (43)

This Lipschitz constants generate the following sequenceof numbers

0 0 8 56 216 616 1456 3024 5712 10032 (44)

and its recurrence relation is

11987132(119873+2)

= 11987132(119873+1)

+119872119873 (45)

where

11987132(0)= 0

119872(119873)= 8119863(119873+1)

119863(119873+1)

= 119863(119873)+ 119877(119873+1) 119863(1)= 1

119877(119873+1)

= 119877(119873)+ (119873 + 1)

2119877(1)= 1

(46)

Abstract and Applied Analysis 5

These recurrence relations lead to

11987132(119873)

=119873

5(119873 + 1) (119873 + 2) (2119873 + 1) (119873 minus 1) (47)

In a similar way we obtain

11987142(119873)

=119873

5(119873 + 1) (119873 + 2) (2119873 + 1) (119873 minus 1) (48)

Main results are given in the followingTheorem 7

Theorem 7 Let ℎ(119905) isin 119862119873120572[minus1 1] for 0 lt 120572 le 1 and let theseries of Chebyshev polynomials 119878

119894119873(119905) be defined by (20) then

the AQS (21) has an error bound

1003817100381710038171003817119864119894 (ℎ)1003817100381710038171003817119862 le

011119872

2119873minus4 (119873 minus 4)[1+238

1198731198713119873+05 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)]

119894 = 1 2

(49)

where119872 is defined by (32) and

1198713119873= 1 +

253

ln119873[1198711119873+ 007119871

2119873

+042 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)(1 +

111

ln119873)

+003 (23119873

3minus 69119873

2+ 100119873 minus 48)

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

1198711119873= 1 +

3

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)+

3 (7119873 minus 4)

2119873 (119873 minus 1) (119873 minus 2)

1198712119873= 1 +

3

119873+231198733minus 69119873

2+ 100119873 minus 48

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)(1 +

3

119873)

(50)

Proof of Theorem 7 In view of error norms (27) we cancompute the error bound of AQS (21) for 119894 = 1 2 as follows

119864119894 (ℎ 119909) = 119867119894 (ℎ 119909) minus 119867119894119873 (ℎ 119909)

=119908119894 (119909)

120587=int

1

minus1

119890119894119873 (119905) 119889119905

119908119894 (119905) (119905 minus 119909)

2

=119908119894 (119909)

120587=int

1

minus1

[119890119894119873 (119905) minus 119890119894119873 (119909) minus 119890

1015840

119894119873(119909) (119905 minus 119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

2

+119908119894 (119909)

120587=int

1

minus1

[119890119894119873 (119909) + 119890

1015840

119894119873(119909) (119905 minus 119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

2

(51)

Taylor series expansion of 119890119894119873(119905) around 119909 is given by

119890119894119873 (119905) = 119890119894119873 (119909) + 119890

1015840

119894119873(120577119905) (119905 minus 119909) (52)

where

120577119905isin (119909 119905) if 119905 isin (119909 1) (119905 119909) if 119905 isin (minus1 119909)

(53)

Due to (51)-(52) we have1003816100381610038161003816119864119894 (ℎ 119909)

1003816100381610038161003816 le 1198641198941 (ℎ 119909) + 1198641198942 (ℎ 119909) + 1198641198943 (ℎ 119909) (54)

where

1198641198941 (ℎ 119909) =

10038161003816100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909)

120587minusint

1

minus1

[1198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

10038161003816100381610038161003816100381610038161003816100381610038161003816

1198641198942 (ℎ 119909) =

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909)

120587119890119894119873 (119909) =int

1

minus1

119889119905

119908119894 (119905) (119905 minus 119909)

2

100381610038161003816100381610038161003816100381610038161003816

1198641198943 (ℎ 119909) =

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909)

1205871198901015840

119894119873(119909) minusint

1

minus1

119889119905

119908119894 (119905) (119905 minus 119909)

100381610038161003816100381610038161003816100381610038161003816

(55)

It is easy to check that

=int

1

minus1

radic1 + 119909

1 minus 119909

119889119905

(119905 minus 119909)2= =int

1

minus1

radic1 minus 119905

1 + 119905

119889119905

(119905 minus 119909)2= 0

minusint

1

minus1

radic1 + 119909

1 minus 119909

119889119905

119905 minus 119909= minusminusint

1

minus1

radic1 minus 119905

1 + 119905

119889119905

119905 minus 119909= 120587

(56)

For the estimation of 1198641198941(ℎ 119909) we divide the interval

(minus1 1) into three subintervals (minus1 119909 minus 120575119873) (119909 + 120575

119873 1) and

[119909 minus 120575119873 119909 + 120575

119873]

1198641198941 (ℎ 119909) le

1

120587[1198651198941 (119909) + 1198651198942 (119909) + 1198651198943 (119909)] (57)

where

1198651198941 (119909) =

10038161003816100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

119909minus120575119873

minus1

[1198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

10038161003816100381610038161003816100381610038161003816100381610038161003816

1198651198942 (119909) =

10038161003816100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

1

119909+120575119873

[1198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

10038161003816100381610038161003816100381610038161003816100381610038161003816

1198651198943 (119909) =

10038161003816100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

119909+120575119873

119909minus120575119873

[1198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

10038161003816100381610038161003816100381610038161003816100381610038161003816

(58)

For 1198651198941(119909) according to Lemma 5 we obtain

1198651198941 (119909) le 2

100381710038171003817100381710038171198901015840

119894119873

10038171003817100381710038171003817

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

119909minus120575119873

minus1

119889119905

119908119894 (119905) (119905 minus 119909)

100381610038161003816100381610038161003816100381610038161003816

le 2100381710038171003817100381710038171198901015840

119894119873

10038171003817100381710038171003817[ln 2120575119873

+120587

radic2]

(59)

In the similar way for 1198651198942(119909) we have

1198651198942 (119909) le 2

100381710038171003817100381710038171198901015840

119894119873

10038171003817100381710038171003817

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

1

119909+120575119873

119889119905

119908119894 (119905) (119905 minus 119909)

100381610038161003816100381610038161003816100381610038161003816

le 2100381710038171003817100381710038171198901015840

119894119873

10038171003817100381710038171003817[ln 2120575119873

+120587

radic2]

(60)

6 Abstract and Applied Analysis

Utilizing (30) and (31) into (59) and (60) we arrive at

1198651198941 (119909) = 1198651198942 (119909)

le(16)119872

2119873minus3 (119873 minus 3)[1 +

1198711119873

119873+

7119873 minus 4

(119873 minus 1) (119873 minus 2)]

times [ln 2120575119873

+120587

radic2]

(61)

For the error of 1198651198943(119909)

1198651198943 (119909) le 119908119894 (119909) int

119909+120590119873

119909minus120590119873

100381610038161003816100381610038161198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)10038161003816100381610038161003816

119908119894 (119905)1003816100381610038161003816120577119905 minus 119909

1003816100381610038161003816

119889119905 (62)

It follows from (29) and Lemma 6 that100381610038161003816100381610038161198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)10038161003816100381610038161003816

le1

2119873119873

times 1003816100381610038161003816119875119894119873 (120577119905) minus 119875119894119873 (119909)

1003816100381610038161003816

10038161003816100381610038161003816100381610038161003816

119889

119889119909ℎ(119873)(120577119905)

10038161003816100381610038161003816100381610038161003816

+10038161003816100381610038161003816ℎ(119873)(120577119905)10038161003816100381610038161003816

100381610038161003816100381610038161198751015840

119894119873(120577119905) minus 1198751015840

119894119873(119909)10038161003816100381610038161003816

le1

2119873119873

times 119871119895119894119873

1003816100381610038161003816120577119905 minus 1199091003816100381610038161003816

10038161003816100381610038161003816100381610038161003816

119889

119889119909ℎ(119873)(120577119905)

10038161003816100381610038161003816100381610038161003816+ 119871119895119894119873

10038161003816100381610038161003816ℎ(119873)(120577119905)10038161003816100381610038161003816

1003816100381610038161003816120577119905 minus 1199091003816100381610038161003816

le119872

480 sdot 2119873minus5 (119873 minus 5)

1198732(119873 + 1) (2119873 + 1)

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)

times [1 +1

119873+3

1198732]1003816100381610038161003816120577119905 minus 119909

1003816100381610038161003816

le119872

240 sdot 2119873minus5 (119873 minus 5)

times [1 +231198733minus 69119873

2+ 100119873 minus 48

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

times [1 +1

119873+3

1198732]1003816100381610038161003816120577119905 minus 119909

1003816100381610038161003816

le119872

240 sdot 2119873minus5 (119873 minus 5)

times [1 +1198712119873

119873+

231198733minus 69119873

2+ 100119873 minus 48

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

times1003816100381610038161003816120577119905 minus 119909

1003816100381610038161003816

(63)

where

1198712119873= 1 +

3

119873+231198733minus 69119873

2+ 100119873 minus 48

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)(1 +

3

119873)

(64)

Substituting (63) into (62) yields

1198651198943 (119909) le

119872

240 sdot 2119873minus5 (119873 minus 5)

times [1 +1198712119873

119873+

231198733minus 69119873

2+ 100119873 minus 48

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

times

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

119909+120575119873

119909minus120575119873

119889119905

119908119894 (119905)

100381610038161003816100381610038161003816100381610038161003816

(65)

Using Lemma 5 gives

1198651198943 (119909) le

337119872

240 sdot 2119873minus5 (119873 minus 5)

times [1+1198712119873

119873+231198733minus 69119873

2+ 100119873 minus 48

2 (119873minus1) (119873minus2) (119873minus3) (119873minus4)] 120575119873

(66)

By choosing 120575119873= 2119873

2 for119873 ge 5 we obtain

1198651198943 (119909) le

011119872

2119873minus3 (119873 minus 3)

times [1 +1198712119873

119873+

231198733minus 69119873

2+ 100119873 minus 48

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

(67)

By substituting (61) and (67) into (57) we arrive at

1198641198941 (ℎ 119909) le

064119872ln119873

120587 sdot 2119873minus3 (119873 minus 3)

times [1 +127

ln119873+211

ln119873(1198711119873+ 008119871

2119873)

+254 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)(1 +

111

ln119873)

+017 (23119873

3minus 69119873

2+ 100119873 minus 48)

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

(68)

1198641198941 (ℎ 119909) le

246119872ln119873

2119873minus3 (119873 minus 3)

times [1 +010

ln119873+1198711119873

119873+010119871

1119873

119873 ln119873

+2316 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)+005 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)

+1059 (23119873

3minus 69119873

2+ 100119873 minus 48)

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

(69)

Abstract and Applied Analysis 7

Table 1 Results of HSIs for 1199081(119905)

119873 119909 Exact solution Approximate solution Absolute error

2

minus099999 minus22360444988237596342 minus22360444988237596345 313 times 10minus16

minus091111 minus21535241487484869307 minus21535241487484869310 308 times 10minus18

minus071111 minus93563357403104129848 minus93563357403104129863 142 times 10minus18

minus031111 minus30963683860810919235 minus30963683860810919240 586 times 10minus19

000000 minus10000000000000000000 minus10000000000000000003 300 times 10minus19

031111 017718528456311131724 017718528456311131709 127 times 10minus19

071111 075786494952828070264 075786494952828070264 639 times 10minus21

091111 057031832778553277005 057031832778553277007 140 times 10minus20

099999 000670813126012938238 000670813126012938238 224 times 10minus22

minus1 minus05 0 05 1

Singular point values

minus500

minus1000

minus1500

minus2000

HSI

val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

30

20

10

times10minus16

Abso

lute

erro

r val

ues

(b)Figure 1 (a) Exact and AQS solutions when 119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5 and (b) absolute error between exact and AQS solution when119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5

Due to (31) and (56)1198641198942 (ℎ 119909) = 0

1198641198943 (ℎ 119909) le

2

radic1 minus 1199092

008119872

2119873minus3 (119873 minus 3)

times [1 +1198711119873

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)]

(70)

Since from the hypothesis in Lemma 5

1 minus 1199092ge 1 minus (1 minus 120575

119873)2

1

radic1 minus 1199092le

1

radic1 minus (1 minus 120575119873)2

leradic2

radic3120575119873

=119873

radic3

(71)

we have

1198641198943 (ℎ 119909) le

009119872

2119873minus3 (119873 minus 3)

times 119873[1 +1198711119873

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)]

(72)

Substituting (69)ndash(72) into (54) yields

1003816100381610038161003816119864119894 (ℎ 119909)1003816100381610038161003816 le

068119872 ln119873120587 sdot 2119873minus3 (119873 minus 3)

times [1 +127

ln119873+211

ln119873(1198711119873+ 008119871

2119873)

+254 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)(1 +

111

ln119873)

+017 (23119873

3minus69119873

2+100119873minus48)

(119873minus1) (119873 minus 2) (119873minus3) (119873minus4)+042119873

ln119873

+042

ln1198731198711119873+

042119873 (7119873minus4)

2 ln119873(119873minus1) (119873 minus 2)]

(73)

Choosing119873 ge 5 the statement ofTheorem 7 follows

8 Abstract and Applied Analysis

Table 2 Results of HSIs for 1199082(119905)

119873 119909 Exact solution Approximate solution Absolute error

2

minus099999 002012457266627356241 00201245726662735624 157 times 10minus21

minus091111 18643200698228626472 18643200698228626470 143 times 10minus19

minus071111 32232147018485970132 32232147018485970129 240 times 10minus19

minus031111 45263577087926478228 45263577087926478225 308 times 10minus19

000000 50000000000000000000 49999999999999999997 300 times 10minus19

031111 51810684429214884711 51810684429214884708 242 times 10minus19

071111 52460548398163357508 52460548398163357508 379 times 10minus20

091111 62854320328769430625 62854320328769430630 299 times 10minus19

099999 44723036596367022628 44723036596367022632 447 times 10minus17

minus1 minus05 0 05

400

300

200

100

1

Singular point values

HSI

val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

40

30

20

10

times10minus17

Abso

lute

erro

r val

ues

(b)

Figure 2 (a) Exact and AQS solutions when 119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5 and (b) absolute error between exact and AQS solution when119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5

5 Numerical Results and Discussion

In this session we evaluate the hypersingular integrals (4) fordifferent choices of ℎ(119905) and compare with the exact solution

Example 1 Consider the HSIs of the form

1198671 (ℎ 119909) =

1

120587radic1 minus 119909

1 + 119909=int

1

minus1

radic1 + 119905

1 minus 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (74)

where ℎ(119905) = 21199052 minus 3119905 + 5 The exact solution is 1198671(ℎ 119909) =

2radic(1 minus 119909)(1 + 119909)(minus1 + 4119909)

The numerical results in Table 1 show that the AQS givesa very good result for the quadratic function ℎ(119905) = 21199052 minus3119905 + 5 for the case 119894 = 1 and 119873 = 2 This is furtherillustrated in Figure 1(a) where the graphs of the exact andAQS solutions are superimposed for the same choices ofsingular point values Details of the accuracy of the proposedAQS is observed from the absolute error in Figure 1(b)

Example 2 Consider the HSIs of the form

1198672 (ℎ 119909) =

1

120587radic1 + 119909

1 minus 119909=int

1

minus1

radic1 minus 119905

1 + 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (75)

where ℎ(119905) = 21199052 minus 3119905 + 5 The exact solution is 1198672(ℎ 119909) =

radic(1 + 119909)(1 minus 119909)(5 minus 4119909)

The numerical results in Table 2 show that the AQS (21)gives a very good result for the quadratic function ℎ(119905) = 21199052minus3119905 + 5 for the case 119894 = 2 and 119873 = 2 Figure 2(a) indicatesthat the exact and AQS solutions are nearly the same Furtherdetails of the accuracy of the absolute error of the proposedAQS are demonstrated in Figure 2(b)

Example 3 Consider the HSIs of the form

1198671 (ℎ 119909) =

1

120587radic1 minus 119909

1 + 119909=int

1

minus1

radic1 + 119905

1 minus 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (76)

Abstract and Applied Analysis 9

Table 3 Results of HSIs for 1199081(119905)

119873 119909 Exact solution Approximate solution Absolute error

10

minus099999 minus25819308036170334212 minus25817083250027430060 223 times 10minus2

minus091111 minus22578118246552894121 minus22578564435672151572 446 times 10minus5

minus077111 minus10634637975870125656 minus10634476360338990858 162 times 10minus5

minus033111 minus02924243743021256602 minus02924196578457308097 472 times 10minus6

000000 minus01443375672974064411 minus01443397190188599320 215 times 10minus6

033111 minus00753154744680519215 minus00753153016422631300 173 times 10minus7

077111 minus00270285335549129334 minus00270280566695517436 477 times 10minus7

091111 minus00146928400579184367 minus00146941562302624467 132 times 10minus6

099999 minus00001434451425475821 minus00001431650832316927 280 times 10minus7

20

minus099999 minus25819308036170334212 minus25819308021411085378 148 times 10minus7

minus091111 minus22578118246552894121 minus22578118246107028644 446 times 10minus11

minus077111 minus10634637975870125656 minus10634637975433417458 437 times 10minus11

minus033111 minus02924243743021256602 minus02924243742855352783 166 times 10minus11

000000 minus01443375672974064411 minus01443375672897319217 768 times 10minus12

033111 minus00753154744680519215 minus00753154744695265617 148 times 10minus12

077111 minus00270285335549129334 minus00270285335485592653 635 times 10minus12

091111 minus00146928400579184367 minus00146928400450019102 129 times 10minus11

099999 minus00001434451425475821 minus00001434451390347442 351 times 10minus12

40

minus099999 minus25819308036170334212 minus25819308036170338346 404 times 10minus14

minus091111 minus22578118246552894121 minus22578118246552894116 450 times 10minus19

minus077111 minus10634637975870125656 minus10634637975870125637 185 times 10minus18

minus033111 minus02924243743021256602 minus02924243743021256612 967 times 10minus19

000000 minus01443375672974064411 minus01443375672974064412 113 times 10minus19

033111 minus00753154744680519215 minus00753154744680519211 400 times 10minus19

077111 minus00270285335549129334 minus00270285335549129336 172 times 10minus19

091111 minus00146928400579184367 minus00146928400579184373 573 times 10minus19

099999 minus00001434451425475821 minus00001434451425475826 486 times 10minus19

minus1 minus05 0 05 1

Singular point valuesminus50

minus100

minus150

minus200

minus250

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

0020

0015

0010

0005

Abso

lute

erro

r val

ues

(b)

Figure 3 (a) Exact and AQS solutions when 119873 = 10 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 10 and ℎ(119905) = 1(2 + 119905)

10 Abstract and Applied Analysis

minus1 minus05 0 05 1

Singular point valuesminus50

minus100

minus150

minus200

minus250

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

14

12

10

80

60

40

20

times10minus8

(b)

Figure 4 (a) Exact and AQS solutions when 119873 = 20 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 20 and ℎ(119905) = 1(2 + 119905)

minus1 minus05 0 05 1

Singular point values

minus50

minus100

minus150

minus250

HSI

s val

ues

minus200

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

40

30

20

10

times10minus14

(b)

Figure 5 (a) Exact and AQS solutions when 119873 = 40 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 40 and ℎ(119905) = 1(2 + 119905)

where ℎ(119905) = 1(2 + 119905) The exact solution is 1198671(ℎ 119909) =

minusradic(1 minus 119909)(1 + 119909)(radic33(2 + 119909)2)

The numerical results in Table 3 show that the AQS (21)is highly accurate for the rational function ℎ(119905) = 1(2 + 119905)for the case 119894 = 1 and119873 = 40 Figure 3(a) reveals that there jsgood agreement between the exact solution andAQS solutionwhich can be seen from absolute error in Figure 3(b) when119873 = 10 The absolute error is improved in Figure 4(b) when

119873 = 20 with the corresponding exact and AQS solution inFigure 4(a) Best result is obtained in Figures 5(a) and 5(b)when119873 = 40

Example 4 Consider the HSIs of the form

1198672 (ℎ 119909) =

1

120587radic1 + 119909

1 minus 119909=int

1

minus1

radic1 minus 119905

1 + 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (77)

Abstract and Applied Analysis 11

Table 4 Results of HSIs for 1199082(119905)

119873 119909 Exact solution Approximate solution Absolute error

10

minus099999 000387291557000340015 000387141593292902562 150 times 10minus6

minus091111 031504763162812503076 031505448317559772604 685 times 10minus6

minus077111 041231131092313516294 041230879229200872013 252 times 10minus6

minus033111 044083450592426358350 044083359884530457966 907 times 10minus7

000000 043301270189221932338 043301675286046974386 405 times 10minus6

033111 044963976686376800332 044963389568653105901 587 times 10minus6

077111 062742591722366873321 062741343576104182143 125 times 10minus5

091111 094767578680690190767 094770492350476572039 291 times 10minus5

099999 86066655193121619385 86053938870581623578 0127 times 10minus2

20

minus099999 000387291557000340015 000387291555159212453 184 times 10minus11

minus091111 031504763162812503076 031504763156789346359 602 times 10minus11

minus077111 041231131092313516294 041231131089736085386 258 times 10minus11

minus033111 044083450592426358350 044083450592691374976 265 times 10minus12

000000 043301270189221932338 043301270187830840205 139 times 10minus11

033111 044963976686376800332 044963976684337128944 204 times 10minus11

077111 062742591722366873321 062742591718912403402 346 times 10minus11

091111 094767578680690190767 094767578677558019222 313 times 10minus11

099999 86066655193121619385 86066655108152556362 850 times 10minus8

40

minus099999 000387291557000340015 000387291557000339911 104 times 10minus18

minus091111 031504763162812503076 031504763162812503215 140 times 10minus18

minus077111 041231131092313516294 041231131092313516515 228 times 10minus18

minus033111 044083450592426358350 044083450592426358289 614 times 10minus19

000000 043301270189221932338 043301270189221932269 675 times 10minus19

033111 044963976686376800332 044963976686376800195 131 times 10minus18

077111 062742591722366873321 062742591722366873124 191 times 10minus18

091111 094767578680690190767 094767578680690190858 131 times 10minus18

099999 86066655193121619385 86066655193121618680 705 times 10minus16

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

0012

0010

0008

0006

0004

0002

Abso

lute

erro

r val

ues

(b)

Figure 6 (a) Exact and AQS solutions when 119873 = 10 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 10 and ℎ(119905) = 1(2 + 119905)

12 Abstract and Applied Analysis

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

times10minus8

Abso

lute

erro

r val

ues

(b)

Figure 7 (a) Exact and AQS solutions when 119873 = 20 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 20 and ℎ(119905) = 1(2 + 119905)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

Exact solution plotAQS solution plot

80

70

60

50

40

30

20

10

(a)

minus1 minus05 0 05 1

Singular point values

70

60

50

40

30

20

10

times10minus16

Abso

lute

erro

r val

ues

(b)

Figure 8 (a) Exact and AQS solutions when 119873 = 40 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 40 and ℎ(119905) = 1(2 + 119905)

where ℎ(119905) = 1(2 + 119905) The exact solution is 1198672(ℎ 119909) =

radic(1 + 119909)(1 minus 119909)(radic3(2 + 119909)2)

The numerical results in Table 4 show that the AQS (21)is highly accurate for the rational function ℎ(119905) = 1(2 + 119905)for the case 119894 = 2 and119873 = 40 Figure 6(a) shows that there is

good agreement between the exact solution andAQS solutionwhich can be seen from absolute error in Figure 6(b) when119873 = 10 The absolute error is reduced in Figure 7(b) when119873 = 20 with the corresponding exact and AQS solutionin Figure 7(a) The best result is obtained and illustrated inFigures 8(a) and 8(b) when119873 = 40

Abstract and Applied Analysis 13

6 Conclusion

In this paper we have constructed the automatic quadratureschemes for the semibounded weighted Hadamard typehypersingular integrals using the truncated series of the thirdand fourth kinds of Chebyshev polynomials The unknowncoefficients are found by applying the discrete orthogonalitycondition and the choice of collocation points on (minus1 1)The error estimations for the semibounded solutions of thehypersingular integrals are obtained Numerical exampleshave shown the accuracy of the developed methods andagreed with theoretical findings

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Hadamard Le Probleme de Cauchy et les Equations auxDerivees Particelles Lineaires Hyperboliques Hermann ParisFrance 1932

[2] E Lutz L Gray and A Ingraffea ldquoAn overview of integrationmethods for hypersingular integralsrdquo in Boundary Elements CA Brebbia and G S Gipson Eds vol 8 pp 913ndash925 CMPSouthampton UK 1991

[3] S G Samko Hypersingular Integrals and Their ApplicationsTaylor amp Francis London UK 2002

[4] I V Boikov ldquoNumerical methods of computation of singularand hypersingular integralsrdquo International Journal of Mathe-matics and Mathematical Sciences vol 28 no 3 pp 127ndash1792001

[5] I K Lifanov L N Poltavskii and G M Vainikko Hypersin-gular Integral Equations and Their Applications Chapman ampHallCRC A CRC Press Company London UK 2004

[6] H R Kutt ldquoQuadrature formulae for finite part integralsrdquoReport WISK 178 The National Research Institute for Mathe-matical Sciences Pretoria South Africa 1975

[7] C-Y Hui and D Shia ldquoEvaluations of hypersingular integralsusingGaussian quadraturerdquo International Journal for NumericalMethods in Engineering vol 44 no 2 pp 205ndash214 1999

[8] Y Z Chen ldquoNumerical solution of a curved crack problem byusing hypersingular integral equation approachrdquo EngineeringFracture Mechanics vol 46 no 2 pp 275ndash283 1993

[9] N M A Nik Long and Z K Eshkuvatov ldquoHypersingularintegral equation for multiple curved cracks problem in planeelasticityrdquo International Journal of Solids and Structures vol 46no 13 pp 2611ndash2617 2009

[10] S J Obaiys Z K Eshkuvatov and N M Nik Long ldquoOn errorestimation of automatic quadrature scheme for the evaluationof Hadamard integral of second order singularityrdquo University ofBucharest Scientific Bulletin Series A Applied Mathematics andPhysics vol 75 no 1 pp 85ndash98 2013

[11] J CMason andDCHandscombChebyshev Polynomials CRCPress 2003

[12] M I Israilov Semi-Analytic Solution of Singular Integral Equa-tions of First Kind by Quadrature Formula Numerical Integra-tion and Related Problems FAN Press 1990 (Russian)

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Journal 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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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 2: An Automatic Quadrature Schemes and Error Estimates for Semibounded Weighted Hadamard Type Hypersingular Integrals

2 Abstract and Applied Analysis

on Clenshaw-Curtis Chebyshev quadrature methods for theevaluation of HSIs of the form

119876119894(119891 119909) = =int

1

minus1

119908119894 (119905) 119891 (119905)

(119905 minus 119909)2119889119905 119909 isin (minus1 1) 119894 = 0 1 2

(3)

where 1199080(119905) = 1 119908

1(119905) = radic1 minus 1199052 and 119908

2(119905) = 1radic1 minus 1199052

are the weight functions and 119891(119905) is a smooth or continuousError estimation for the developed AQS for 119894 = 0 wasobtained in the class of functions 119862119873+2120572[minus1 1]

In this note we have developed the automatic quadraturescheme based on Chebyshev Gauss quadrature formula forthe semibounded solutions of the hypersingular integrals ofthe form

119867119894 (ℎ 119909) =

119908119894 (119909)

120587=int

1

minus1

ℎ (119905)

119908119894 (119905) (119905 minus 119909)

2119889119905 119909 isin (minus1 1) (4)

where 1199081(119905) = radic(1 minus 119905)(1 + 119905) and 119908

2(119905) = radic(1 + 119905)(1 minus 119905)

are the weights and ℎ(119905) is a smooth function The pro-posed automatic quadrature schemes constructed cover thedifferent weights functions that have not been considered byObayis et al [10] Unlike the work of Obayis et al [10] whichtakes advantage of Clenshaw-Curtis Chebyshev quadraturemethods the construction of our AQSs is based on Cheby-shev Gauss quadrature methods The method proposed byHui and Shia [7] works only for even number 119873 of knotspoints whereas AQS for our problems works well for all 119873and converges very fast to the exact solution

Present in Section 2 are the mathematical concepts ofthe third and fourth kinds of Chebyshev polynomials andsingular integrals In Section 3 the construction of automaticquadrature schemes for semibounded solutions of hypersin-gular integrals is given Rate of convergence of the suggestedmethod is discussed in Section 4 Numerical examples andresults are provided in Section 5

2 Mathematical Concepts of Third andFourth Kinds of Chebyshev Polynomialsand Singular Integrals

The Chebyshev polynomials 119881119898(119905) and119882

119898(119905) are defined as

follows

Definition 1 (Mason and Handscomb [11]) The chebyshevpolynomial119881

119898(119905) of the third kind on [minus1 1] is a polynomial

of degree119898 in 119905 defined by

119881119898 (119905) =

cos (119898 + 12) 120579cos (12) 120579

119905 = cos 120579 0 le 120579 le 120587 (5)

Definition 2 (Mason and Handscomb [11]) The chebyshevpolynomial119882

119898(119905) of the fourth kind on [minus1 1] is a polyno-

mial of degree119898 in 119905 defined by

119882119898 (119905) =

sin (119898 + 12) 120579sin (12) 120579

119905 = cos 120579 0 le 120579 le 120587 (6)

Both polynomials

119875119894119898 (119905) =

119881119898 (119905) 119894 = 1

119882119898 (119905) 119894 = 2

(7)

satisfy the recurrence relation

119875119894119898 (119905) = 2119905119875119894(119898minus1) (119905) minus 119875119894(119898minus2) (119905)

119898 = 2 3 119894 = 1 2

(8)

with the same starting value

1198751198940 (119905) =

1198810 (119905) = 1 119894 = 1

1198820 (119905) = 1 119894 = 2

1198751198941 (119905) =

1198811 (119905) = 2119905 minus 1 119894 = 1

1198821 (119905) = 2119905 + 1 119894 = 2

(9)

The third and fourth kinds of Chebyshev polynomialsare orthogonal on [minus1 1] with the weights 119908

1(119905) and 119908

2(119905)

respectively They also have continuous orthogonality givenin [11] respectively as

⟨119881119898 119881119895⟩ = int

1

minus1

radic1 + 119905

1 minus 119905119881119898 (119905) 119881119895 (119905) 119889119905 =

0 119898 = 119895

120587 119898 = 119895

⟨119882119898119882119895⟩ = int

1

minus1

radic1 minus 119905

1 + 119905119882119898 (119905)119882119895 (119905) 119889119905

= 0 119898 = 119895 (119898 119895 le 119873)

120587 119898 = 119895

(10)

The third kind and fourth kind of Chebyshev polynomialshave discrete orthogonality respectively as

⟨119881119898 119881119895⟩ =

119873

sum

119896=0

(1 + 119905119896) 119881119898(119905119896) 119881119895(119905119896)

= 0 119898 = 119895 (119898 119895 le 119873)

119873 + 1 0 le 119895 = 119898 le 119873

⟨119882119898119882119895⟩ =

119873

sum

119896=0

(1 minus 119905119896)119882119898(119905119896)119882119895(119905119896)

= 0 119898 = 119895 (119898 119895 le 119873)

119873 + 1 119895 = 119898 (0 le 119898 le 119873)

(11)

The four kinds of Chebyshev polynomials are connectedvia the relations

(1 + 119905) 119881119896 (119905) = 119879119896+1 (119905) + 119879119896 (119905) (12)

119882119896 (119905) = 119880119896 (119905) + 119880119896minus1 (119905) (13)

(1 minus 119905)119882119896 (119905) = 119879119896+1 (119905) minus 119879119896 (119905) (14)

119881119896 (119905) = 119880119896 (119905) minus 119880119896minus1 (119905) (15)

Abstract and Applied Analysis 3

and the differentiation formulae119889

119889119909119879119896 (119905) = 119896119880119896minus1 (119905)

119889

119889119909119880119896 (119905) =

(119896 + 1) 119879119896+1 (119905) minus 119905119880119896

1199052 minus 1

(16)

The Hilbert type integral transform is as follows

Lemma 3 (Mason andHandscomb [11]) Consider the follow-ing

minusint

120587

0

cos119898120579cos 120579 minus cos120572

119889120579 = 120587sin119898120572sin120572

(17)

for any 120572 isin [0 120587]119898 = 1 2 3

Lemma4 (Mason andHandscomb [11]) Consider the follow-ing

minusint

120587

0

sin119898120579 sin 120579cos 120579 minus cos120572

119889120579 = minus120587 cos119898120572 (18)

for any 120572 isin [0 120587]119898 = 1 2 3

The estimations for some integrals are established

Lemma 5 (Israilov [12]) Let 0 lt 120590 le 1 and 1 minus 119909 ge 120575119873

0 lt 120575119873lt 12 and for all 119894 = 1 2

120574119894 (119909) = 119908119894 (119909) int

119909minus120575119873

minus1

119889119905

119908119894 (119905) (119905 minus 119909)

120574lowast

119894(119909) = 119908119894 (119909) int

1

119909+120575119873

119889119905

119908119894 (119905) (119905 minus 119909)

119879119894(119909 120575119873 120590) = 119908

119894 (119909) int

119909+120575119873

119909minus120575119873

|119905 minus 119909|120590minus1119889119905

119908119894 (119905)

(19)

where 1199081(119909) = radic(1 minus 119909)(1 + 119909) and 119908

2(119909) =

radic(1 + 119909)(1 minus 119909)The following estimations are true

(i) |120574119894(119909)| le ln(2120575

119873) + 120587radic2

(ii) 1205741(minus119909) = minus120574

lowast

2(119909) 1205742(minus119909) = minus120574

lowast

1(119909)

(iii) 119879119894(119909 120575119873 120590) le 120601

119894(120590)120575120590

119873 max120601

1(120590) 1206012(120590) le radic10 times

2minus120590+ (1 + radic10 times 2

minus1minus120590)120590minus1

3 Automatic Quadrature Schemes forSemibounded Hypersingular Integrals

To construct AQSs for the semiboundedHSIs (4) the densityfunction ℎ(119905) is approximated with the truncated series ofthird and forth kinds of Chebyshev polynomials

ℎ (119905) asymp 119878119894119873 (119905) =

119873

sum

119898=0

1198871119898119881119898 (119905) 119894 = 1

119873

sum

119898=0

1198872119898119882119898 (119905) 119894 = 2

(20)

Substituting (20) into (4) we have

119867119894 (ℎ 119909)

asymp 119867119894119873 (ℎ 119909)

=119908119894 (119905)

120587=int

1

minus1

119878119894119873 (119905)

119908119894 (119909) (119905 minus 119909)

2119889119905

=

radic1 minus 119909

1 + 119909

119873

sum

119898=0

1198871119898

119889

119889119909[1

120587minusint

1

minus1

(1 + 119905) 119881119898 (119905)

radic1 minus 1199052 (119905 minus 119909)119889119905] 119894 = 1

radic1 + 119909

1 minus 119909

119873

sum

119898=0

1198872119898

119889

119889119909[1

120587minusint

1

minus1

(1 minus 119905)119882119898 (119905)

radic1 minus 1199052 (119905 minus 119909)119889119905] 119894 = 2

=

radic1 minus 119909

1 + 1199091199021 (119909) 119894 = 1

radic1 + 119909

1 minus 1199091199022 (119909) 119894 = 2

(21)

where

119902119894 (119909) =

119889

119889119909

119873

sum

119898=0

1198871119898119882119898 (119909) 119894 = 1

119889

119889119909

119873

sum

119898=0

1198872119898119881119898 (119909) 119894 = 2

(22)

By applying the orthogonality relations (10) we obtain thecoefficients 119887

119894119898 119894 = 1 2 as

119887119894119898=

1

120587int

1

minus1

radic1 + 119905

1 minus 119905ℎ (119905) 119881119898 (119905) 119889119905 119894 = 1

1

120587int

1

minus1

radic1 minus 119905

1 + 119905ℎ (119905)119882119898 (119905) 119889119905 119894 = 2

(23)

For the discrete coefficients 119887119894119898 we choose the collocation

points 119905 = 119905119896 at the zeros of 119879

119873+1 namely

119905 = 119905119896= cos (2119896 + 1) 120587

2119873 + 2 119896 = 0 1 2 119873 (24)

and using the orthogonality conditions (11) and relations(12) and (14) respectively as well as applying compositetrapezium rule we arrive at

119887119894119898asymp

1

119873 + 32

119873

sum

119896=0

(1 + 119905119896) 119881119898(119905119896) ℎ (119905119896) 119894 = 1

1

119873 + 32

119873

sum

119896=0

(119905119896minus 1)119882

119898(119905119896) ℎ (119905119896) 119894 = 2

(25)

4 Error Estimation for the SemiboundedHypersingular Integrals

Let us introduce the classes of functions

(i) 119867120572([minus1 1] 119871) is a class of function satisfying Holdercondition on the interval with the index 120572 andconstant 119871

4 Abstract and Applied Analysis

(ii) 119862119873120572[minus1 1] = ℎ(119905) ℎ(119873)(119905) isin 119867120572([minus1 1] 119871119873)

(iii) 119862[minus1 1] is a class of continuous functions on theinterval [minus1 1] with the norm

ℎ = ℎ(119909)119862[minus11] = max119909isin[minus11]

|ℎ (119909)| (26)

(iv) 119862119903[minus1 1] = ℎ(119905) ℎ119903(119905) isin 119862([minus1 1])Let 119890119894119873(119905) = ℎ(119905) minus 119878

119894119873(119905) and let the maximum norm of

the error be defined as10038171003817100381710038171198901198941198731003817100381710038171003817 = max119905isin[minus11]

1003816100381610038161003816119890119894119873 (119905)1003816100381610038161003816 (27)

It is known in [11] that for any 119909 isin (minus1 1)

1198901119873 (119909) =

ℎ(119873)(120578119909)119882119873 (119909)

2119873119873 120578119909isin [minus1 1]

1198902119873 (119909) =

ℎ(119873)(120578119909) 119881119873 (119909)

2119873119873 120578119909isin [minus1 1]

(28)

1198901015840

1119873(119909) =

1

2119873119873[119882119873 (119909)

119889

119889119909ℎ(119873)(120578119909) + ℎ(119873)(120578119909)1198821015840

119873(119909)]

120578119909isin [minus1 1]

1198901015840

2119873(119909) =

1

2119873119873[119881119873 (119909)

119889

119889119909ℎ(119873)(120578119909) + ℎ(119873)(120578119909) 1198811015840

119873(119909)]

120578119909isin [minus1 1]

(29)

As a consequence of (29) we arrive at the followingestimations100381710038171003817100381711989011198731003817100381710038171003817 =100381710038171003817100381711989021198731003817100381710038171003817 le(2119873 + 1)1198721

2119873 sdot 119873

(30)100381710038171003817100381710038171198901015840

1119873

10038171003817100381710038171003817=100381710038171003817100381710038171198901015840

2119873

10038171003817100381710038171003817

le119872

12 sdot 2119873minus3 (119873 minus 3)[1 +

1198711119873

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)]

(31)

where119872 = max 119872

11198722 119872

1= max120578119909isin[minus11]

10038161003816100381610038161003816ℎ(119873)(120578119909)10038161003816100381610038161003816

1198722= max120578119909isin[minus11]

10038161003816100381610038161003816100381610038161003816

119889

119889119909ℎ(119873)(120578119909)

10038161003816100381610038161003816100381610038161003816lt infin

(32)

1198711119873= 1 +

3

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)(1 +

3

119873) (33)

Lemma 6 The Lipschitz constants 119871119895119894119873

119895 = 3 4 119894 = 1 2of 119881119873(119909) and119882

119873(119909) are

1003816100381610038161003816119881119873 (119909) minus 119881119873 (119910)1003816100381610038161003816 le 11987131119873

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

100381610038161003816100381610038161198811015840

119873(119909) minus 119881

1015840

119873(119910)10038161003816100381610038161003816le 11987132119873

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

1003816100381610038161003816119882119873 (119909) minus 119882119873 (119910)1003816100381610038161003816 le 11987141119873

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

100381610038161003816100381610038161198821015840

119873(119909) minus 119882

1015840

119873(119910)10038161003816100381610038161003816le 11987142119873

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

(34)

where

11987131119873= 11987141119873=119873

3(119873 + 1) (2119873 + 1)

11987132119873= 11987142119873=119873

15(119873 + 1) (119873 + 2) (2119873 + 1) (119873 minus 1)

(35)

Proof Let 11987131119873

be the Lipschitz constant of the third kindChebyshev polynomial 119881

119873(119905) that is

1003816100381610038161003816119881119873 (119905) minus 119881119873 (119909)1003816100381610038161003816 le 11987131119873 |119905 minus 119909| (36)

where

11987131119873= max120577119905isin[minus11]

100381610038161003816100381610038161198811015840

119873(120577119905)10038161003816100381610038161003816 (37)

which gives sequence of numbers

0 2 10 28 60 110 182 280 408 570 (38)

These sequences can be generated by

11987131(119873)

= 11987131(119873minus1)

+ 21198732 11987131(0)= 0 (39)

which leads to

11987131119873=119873

3(119873 + 1) (2119873 + 1) (40)

In a similar way we can get

11987141119873=119873

3(119873 + 1) (2119873 + 1) (41)

Let 11987132(119873)

be the Lipschitz constant of the derivative ofthird kind Chebyshev polynomial 1198811015840

119873(119905) that is

100381610038161003816100381610038161198811015840

119873(119905) minus 119881

1015840

119873(119909)10038161003816100381610038161003816le 11987132119873 |119905 minus 119909| (42)

where

11987132119873= max120577119905isin[minus11]

1003816100381610038161003816100381611988110158401015840

119873(120577119905)10038161003816100381610038161003816 (43)

This Lipschitz constants generate the following sequenceof numbers

0 0 8 56 216 616 1456 3024 5712 10032 (44)

and its recurrence relation is

11987132(119873+2)

= 11987132(119873+1)

+119872119873 (45)

where

11987132(0)= 0

119872(119873)= 8119863(119873+1)

119863(119873+1)

= 119863(119873)+ 119877(119873+1) 119863(1)= 1

119877(119873+1)

= 119877(119873)+ (119873 + 1)

2119877(1)= 1

(46)

Abstract and Applied Analysis 5

These recurrence relations lead to

11987132(119873)

=119873

5(119873 + 1) (119873 + 2) (2119873 + 1) (119873 minus 1) (47)

In a similar way we obtain

11987142(119873)

=119873

5(119873 + 1) (119873 + 2) (2119873 + 1) (119873 minus 1) (48)

Main results are given in the followingTheorem 7

Theorem 7 Let ℎ(119905) isin 119862119873120572[minus1 1] for 0 lt 120572 le 1 and let theseries of Chebyshev polynomials 119878

119894119873(119905) be defined by (20) then

the AQS (21) has an error bound

1003817100381710038171003817119864119894 (ℎ)1003817100381710038171003817119862 le

011119872

2119873minus4 (119873 minus 4)[1+238

1198731198713119873+05 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)]

119894 = 1 2

(49)

where119872 is defined by (32) and

1198713119873= 1 +

253

ln119873[1198711119873+ 007119871

2119873

+042 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)(1 +

111

ln119873)

+003 (23119873

3minus 69119873

2+ 100119873 minus 48)

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

1198711119873= 1 +

3

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)+

3 (7119873 minus 4)

2119873 (119873 minus 1) (119873 minus 2)

1198712119873= 1 +

3

119873+231198733minus 69119873

2+ 100119873 minus 48

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)(1 +

3

119873)

(50)

Proof of Theorem 7 In view of error norms (27) we cancompute the error bound of AQS (21) for 119894 = 1 2 as follows

119864119894 (ℎ 119909) = 119867119894 (ℎ 119909) minus 119867119894119873 (ℎ 119909)

=119908119894 (119909)

120587=int

1

minus1

119890119894119873 (119905) 119889119905

119908119894 (119905) (119905 minus 119909)

2

=119908119894 (119909)

120587=int

1

minus1

[119890119894119873 (119905) minus 119890119894119873 (119909) minus 119890

1015840

119894119873(119909) (119905 minus 119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

2

+119908119894 (119909)

120587=int

1

minus1

[119890119894119873 (119909) + 119890

1015840

119894119873(119909) (119905 minus 119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

2

(51)

Taylor series expansion of 119890119894119873(119905) around 119909 is given by

119890119894119873 (119905) = 119890119894119873 (119909) + 119890

1015840

119894119873(120577119905) (119905 minus 119909) (52)

where

120577119905isin (119909 119905) if 119905 isin (119909 1) (119905 119909) if 119905 isin (minus1 119909)

(53)

Due to (51)-(52) we have1003816100381610038161003816119864119894 (ℎ 119909)

1003816100381610038161003816 le 1198641198941 (ℎ 119909) + 1198641198942 (ℎ 119909) + 1198641198943 (ℎ 119909) (54)

where

1198641198941 (ℎ 119909) =

10038161003816100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909)

120587minusint

1

minus1

[1198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

10038161003816100381610038161003816100381610038161003816100381610038161003816

1198641198942 (ℎ 119909) =

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909)

120587119890119894119873 (119909) =int

1

minus1

119889119905

119908119894 (119905) (119905 minus 119909)

2

100381610038161003816100381610038161003816100381610038161003816

1198641198943 (ℎ 119909) =

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909)

1205871198901015840

119894119873(119909) minusint

1

minus1

119889119905

119908119894 (119905) (119905 minus 119909)

100381610038161003816100381610038161003816100381610038161003816

(55)

It is easy to check that

=int

1

minus1

radic1 + 119909

1 minus 119909

119889119905

(119905 minus 119909)2= =int

1

minus1

radic1 minus 119905

1 + 119905

119889119905

(119905 minus 119909)2= 0

minusint

1

minus1

radic1 + 119909

1 minus 119909

119889119905

119905 minus 119909= minusminusint

1

minus1

radic1 minus 119905

1 + 119905

119889119905

119905 minus 119909= 120587

(56)

For the estimation of 1198641198941(ℎ 119909) we divide the interval

(minus1 1) into three subintervals (minus1 119909 minus 120575119873) (119909 + 120575

119873 1) and

[119909 minus 120575119873 119909 + 120575

119873]

1198641198941 (ℎ 119909) le

1

120587[1198651198941 (119909) + 1198651198942 (119909) + 1198651198943 (119909)] (57)

where

1198651198941 (119909) =

10038161003816100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

119909minus120575119873

minus1

[1198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

10038161003816100381610038161003816100381610038161003816100381610038161003816

1198651198942 (119909) =

10038161003816100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

1

119909+120575119873

[1198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

10038161003816100381610038161003816100381610038161003816100381610038161003816

1198651198943 (119909) =

10038161003816100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

119909+120575119873

119909minus120575119873

[1198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

10038161003816100381610038161003816100381610038161003816100381610038161003816

(58)

For 1198651198941(119909) according to Lemma 5 we obtain

1198651198941 (119909) le 2

100381710038171003817100381710038171198901015840

119894119873

10038171003817100381710038171003817

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

119909minus120575119873

minus1

119889119905

119908119894 (119905) (119905 minus 119909)

100381610038161003816100381610038161003816100381610038161003816

le 2100381710038171003817100381710038171198901015840

119894119873

10038171003817100381710038171003817[ln 2120575119873

+120587

radic2]

(59)

In the similar way for 1198651198942(119909) we have

1198651198942 (119909) le 2

100381710038171003817100381710038171198901015840

119894119873

10038171003817100381710038171003817

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

1

119909+120575119873

119889119905

119908119894 (119905) (119905 minus 119909)

100381610038161003816100381610038161003816100381610038161003816

le 2100381710038171003817100381710038171198901015840

119894119873

10038171003817100381710038171003817[ln 2120575119873

+120587

radic2]

(60)

6 Abstract and Applied Analysis

Utilizing (30) and (31) into (59) and (60) we arrive at

1198651198941 (119909) = 1198651198942 (119909)

le(16)119872

2119873minus3 (119873 minus 3)[1 +

1198711119873

119873+

7119873 minus 4

(119873 minus 1) (119873 minus 2)]

times [ln 2120575119873

+120587

radic2]

(61)

For the error of 1198651198943(119909)

1198651198943 (119909) le 119908119894 (119909) int

119909+120590119873

119909minus120590119873

100381610038161003816100381610038161198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)10038161003816100381610038161003816

119908119894 (119905)1003816100381610038161003816120577119905 minus 119909

1003816100381610038161003816

119889119905 (62)

It follows from (29) and Lemma 6 that100381610038161003816100381610038161198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)10038161003816100381610038161003816

le1

2119873119873

times 1003816100381610038161003816119875119894119873 (120577119905) minus 119875119894119873 (119909)

1003816100381610038161003816

10038161003816100381610038161003816100381610038161003816

119889

119889119909ℎ(119873)(120577119905)

10038161003816100381610038161003816100381610038161003816

+10038161003816100381610038161003816ℎ(119873)(120577119905)10038161003816100381610038161003816

100381610038161003816100381610038161198751015840

119894119873(120577119905) minus 1198751015840

119894119873(119909)10038161003816100381610038161003816

le1

2119873119873

times 119871119895119894119873

1003816100381610038161003816120577119905 minus 1199091003816100381610038161003816

10038161003816100381610038161003816100381610038161003816

119889

119889119909ℎ(119873)(120577119905)

10038161003816100381610038161003816100381610038161003816+ 119871119895119894119873

10038161003816100381610038161003816ℎ(119873)(120577119905)10038161003816100381610038161003816

1003816100381610038161003816120577119905 minus 1199091003816100381610038161003816

le119872

480 sdot 2119873minus5 (119873 minus 5)

1198732(119873 + 1) (2119873 + 1)

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)

times [1 +1

119873+3

1198732]1003816100381610038161003816120577119905 minus 119909

1003816100381610038161003816

le119872

240 sdot 2119873minus5 (119873 minus 5)

times [1 +231198733minus 69119873

2+ 100119873 minus 48

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

times [1 +1

119873+3

1198732]1003816100381610038161003816120577119905 minus 119909

1003816100381610038161003816

le119872

240 sdot 2119873minus5 (119873 minus 5)

times [1 +1198712119873

119873+

231198733minus 69119873

2+ 100119873 minus 48

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

times1003816100381610038161003816120577119905 minus 119909

1003816100381610038161003816

(63)

where

1198712119873= 1 +

3

119873+231198733minus 69119873

2+ 100119873 minus 48

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)(1 +

3

119873)

(64)

Substituting (63) into (62) yields

1198651198943 (119909) le

119872

240 sdot 2119873minus5 (119873 minus 5)

times [1 +1198712119873

119873+

231198733minus 69119873

2+ 100119873 minus 48

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

times

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

119909+120575119873

119909minus120575119873

119889119905

119908119894 (119905)

100381610038161003816100381610038161003816100381610038161003816

(65)

Using Lemma 5 gives

1198651198943 (119909) le

337119872

240 sdot 2119873minus5 (119873 minus 5)

times [1+1198712119873

119873+231198733minus 69119873

2+ 100119873 minus 48

2 (119873minus1) (119873minus2) (119873minus3) (119873minus4)] 120575119873

(66)

By choosing 120575119873= 2119873

2 for119873 ge 5 we obtain

1198651198943 (119909) le

011119872

2119873minus3 (119873 minus 3)

times [1 +1198712119873

119873+

231198733minus 69119873

2+ 100119873 minus 48

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

(67)

By substituting (61) and (67) into (57) we arrive at

1198641198941 (ℎ 119909) le

064119872ln119873

120587 sdot 2119873minus3 (119873 minus 3)

times [1 +127

ln119873+211

ln119873(1198711119873+ 008119871

2119873)

+254 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)(1 +

111

ln119873)

+017 (23119873

3minus 69119873

2+ 100119873 minus 48)

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

(68)

1198641198941 (ℎ 119909) le

246119872ln119873

2119873minus3 (119873 minus 3)

times [1 +010

ln119873+1198711119873

119873+010119871

1119873

119873 ln119873

+2316 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)+005 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)

+1059 (23119873

3minus 69119873

2+ 100119873 minus 48)

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

(69)

Abstract and Applied Analysis 7

Table 1 Results of HSIs for 1199081(119905)

119873 119909 Exact solution Approximate solution Absolute error

2

minus099999 minus22360444988237596342 minus22360444988237596345 313 times 10minus16

minus091111 minus21535241487484869307 minus21535241487484869310 308 times 10minus18

minus071111 minus93563357403104129848 minus93563357403104129863 142 times 10minus18

minus031111 minus30963683860810919235 minus30963683860810919240 586 times 10minus19

000000 minus10000000000000000000 minus10000000000000000003 300 times 10minus19

031111 017718528456311131724 017718528456311131709 127 times 10minus19

071111 075786494952828070264 075786494952828070264 639 times 10minus21

091111 057031832778553277005 057031832778553277007 140 times 10minus20

099999 000670813126012938238 000670813126012938238 224 times 10minus22

minus1 minus05 0 05 1

Singular point values

minus500

minus1000

minus1500

minus2000

HSI

val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

30

20

10

times10minus16

Abso

lute

erro

r val

ues

(b)Figure 1 (a) Exact and AQS solutions when 119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5 and (b) absolute error between exact and AQS solution when119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5

Due to (31) and (56)1198641198942 (ℎ 119909) = 0

1198641198943 (ℎ 119909) le

2

radic1 minus 1199092

008119872

2119873minus3 (119873 minus 3)

times [1 +1198711119873

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)]

(70)

Since from the hypothesis in Lemma 5

1 minus 1199092ge 1 minus (1 minus 120575

119873)2

1

radic1 minus 1199092le

1

radic1 minus (1 minus 120575119873)2

leradic2

radic3120575119873

=119873

radic3

(71)

we have

1198641198943 (ℎ 119909) le

009119872

2119873minus3 (119873 minus 3)

times 119873[1 +1198711119873

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)]

(72)

Substituting (69)ndash(72) into (54) yields

1003816100381610038161003816119864119894 (ℎ 119909)1003816100381610038161003816 le

068119872 ln119873120587 sdot 2119873minus3 (119873 minus 3)

times [1 +127

ln119873+211

ln119873(1198711119873+ 008119871

2119873)

+254 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)(1 +

111

ln119873)

+017 (23119873

3minus69119873

2+100119873minus48)

(119873minus1) (119873 minus 2) (119873minus3) (119873minus4)+042119873

ln119873

+042

ln1198731198711119873+

042119873 (7119873minus4)

2 ln119873(119873minus1) (119873 minus 2)]

(73)

Choosing119873 ge 5 the statement ofTheorem 7 follows

8 Abstract and Applied Analysis

Table 2 Results of HSIs for 1199082(119905)

119873 119909 Exact solution Approximate solution Absolute error

2

minus099999 002012457266627356241 00201245726662735624 157 times 10minus21

minus091111 18643200698228626472 18643200698228626470 143 times 10minus19

minus071111 32232147018485970132 32232147018485970129 240 times 10minus19

minus031111 45263577087926478228 45263577087926478225 308 times 10minus19

000000 50000000000000000000 49999999999999999997 300 times 10minus19

031111 51810684429214884711 51810684429214884708 242 times 10minus19

071111 52460548398163357508 52460548398163357508 379 times 10minus20

091111 62854320328769430625 62854320328769430630 299 times 10minus19

099999 44723036596367022628 44723036596367022632 447 times 10minus17

minus1 minus05 0 05

400

300

200

100

1

Singular point values

HSI

val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

40

30

20

10

times10minus17

Abso

lute

erro

r val

ues

(b)

Figure 2 (a) Exact and AQS solutions when 119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5 and (b) absolute error between exact and AQS solution when119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5

5 Numerical Results and Discussion

In this session we evaluate the hypersingular integrals (4) fordifferent choices of ℎ(119905) and compare with the exact solution

Example 1 Consider the HSIs of the form

1198671 (ℎ 119909) =

1

120587radic1 minus 119909

1 + 119909=int

1

minus1

radic1 + 119905

1 minus 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (74)

where ℎ(119905) = 21199052 minus 3119905 + 5 The exact solution is 1198671(ℎ 119909) =

2radic(1 minus 119909)(1 + 119909)(minus1 + 4119909)

The numerical results in Table 1 show that the AQS givesa very good result for the quadratic function ℎ(119905) = 21199052 minus3119905 + 5 for the case 119894 = 1 and 119873 = 2 This is furtherillustrated in Figure 1(a) where the graphs of the exact andAQS solutions are superimposed for the same choices ofsingular point values Details of the accuracy of the proposedAQS is observed from the absolute error in Figure 1(b)

Example 2 Consider the HSIs of the form

1198672 (ℎ 119909) =

1

120587radic1 + 119909

1 minus 119909=int

1

minus1

radic1 minus 119905

1 + 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (75)

where ℎ(119905) = 21199052 minus 3119905 + 5 The exact solution is 1198672(ℎ 119909) =

radic(1 + 119909)(1 minus 119909)(5 minus 4119909)

The numerical results in Table 2 show that the AQS (21)gives a very good result for the quadratic function ℎ(119905) = 21199052minus3119905 + 5 for the case 119894 = 2 and 119873 = 2 Figure 2(a) indicatesthat the exact and AQS solutions are nearly the same Furtherdetails of the accuracy of the absolute error of the proposedAQS are demonstrated in Figure 2(b)

Example 3 Consider the HSIs of the form

1198671 (ℎ 119909) =

1

120587radic1 minus 119909

1 + 119909=int

1

minus1

radic1 + 119905

1 minus 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (76)

Abstract and Applied Analysis 9

Table 3 Results of HSIs for 1199081(119905)

119873 119909 Exact solution Approximate solution Absolute error

10

minus099999 minus25819308036170334212 minus25817083250027430060 223 times 10minus2

minus091111 minus22578118246552894121 minus22578564435672151572 446 times 10minus5

minus077111 minus10634637975870125656 minus10634476360338990858 162 times 10minus5

minus033111 minus02924243743021256602 minus02924196578457308097 472 times 10minus6

000000 minus01443375672974064411 minus01443397190188599320 215 times 10minus6

033111 minus00753154744680519215 minus00753153016422631300 173 times 10minus7

077111 minus00270285335549129334 minus00270280566695517436 477 times 10minus7

091111 minus00146928400579184367 minus00146941562302624467 132 times 10minus6

099999 minus00001434451425475821 minus00001431650832316927 280 times 10minus7

20

minus099999 minus25819308036170334212 minus25819308021411085378 148 times 10minus7

minus091111 minus22578118246552894121 minus22578118246107028644 446 times 10minus11

minus077111 minus10634637975870125656 minus10634637975433417458 437 times 10minus11

minus033111 minus02924243743021256602 minus02924243742855352783 166 times 10minus11

000000 minus01443375672974064411 minus01443375672897319217 768 times 10minus12

033111 minus00753154744680519215 minus00753154744695265617 148 times 10minus12

077111 minus00270285335549129334 minus00270285335485592653 635 times 10minus12

091111 minus00146928400579184367 minus00146928400450019102 129 times 10minus11

099999 minus00001434451425475821 minus00001434451390347442 351 times 10minus12

40

minus099999 minus25819308036170334212 minus25819308036170338346 404 times 10minus14

minus091111 minus22578118246552894121 minus22578118246552894116 450 times 10minus19

minus077111 minus10634637975870125656 minus10634637975870125637 185 times 10minus18

minus033111 minus02924243743021256602 minus02924243743021256612 967 times 10minus19

000000 minus01443375672974064411 minus01443375672974064412 113 times 10minus19

033111 minus00753154744680519215 minus00753154744680519211 400 times 10minus19

077111 minus00270285335549129334 minus00270285335549129336 172 times 10minus19

091111 minus00146928400579184367 minus00146928400579184373 573 times 10minus19

099999 minus00001434451425475821 minus00001434451425475826 486 times 10minus19

minus1 minus05 0 05 1

Singular point valuesminus50

minus100

minus150

minus200

minus250

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

0020

0015

0010

0005

Abso

lute

erro

r val

ues

(b)

Figure 3 (a) Exact and AQS solutions when 119873 = 10 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 10 and ℎ(119905) = 1(2 + 119905)

10 Abstract and Applied Analysis

minus1 minus05 0 05 1

Singular point valuesminus50

minus100

minus150

minus200

minus250

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

14

12

10

80

60

40

20

times10minus8

(b)

Figure 4 (a) Exact and AQS solutions when 119873 = 20 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 20 and ℎ(119905) = 1(2 + 119905)

minus1 minus05 0 05 1

Singular point values

minus50

minus100

minus150

minus250

HSI

s val

ues

minus200

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

40

30

20

10

times10minus14

(b)

Figure 5 (a) Exact and AQS solutions when 119873 = 40 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 40 and ℎ(119905) = 1(2 + 119905)

where ℎ(119905) = 1(2 + 119905) The exact solution is 1198671(ℎ 119909) =

minusradic(1 minus 119909)(1 + 119909)(radic33(2 + 119909)2)

The numerical results in Table 3 show that the AQS (21)is highly accurate for the rational function ℎ(119905) = 1(2 + 119905)for the case 119894 = 1 and119873 = 40 Figure 3(a) reveals that there jsgood agreement between the exact solution andAQS solutionwhich can be seen from absolute error in Figure 3(b) when119873 = 10 The absolute error is improved in Figure 4(b) when

119873 = 20 with the corresponding exact and AQS solution inFigure 4(a) Best result is obtained in Figures 5(a) and 5(b)when119873 = 40

Example 4 Consider the HSIs of the form

1198672 (ℎ 119909) =

1

120587radic1 + 119909

1 minus 119909=int

1

minus1

radic1 minus 119905

1 + 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (77)

Abstract and Applied Analysis 11

Table 4 Results of HSIs for 1199082(119905)

119873 119909 Exact solution Approximate solution Absolute error

10

minus099999 000387291557000340015 000387141593292902562 150 times 10minus6

minus091111 031504763162812503076 031505448317559772604 685 times 10minus6

minus077111 041231131092313516294 041230879229200872013 252 times 10minus6

minus033111 044083450592426358350 044083359884530457966 907 times 10minus7

000000 043301270189221932338 043301675286046974386 405 times 10minus6

033111 044963976686376800332 044963389568653105901 587 times 10minus6

077111 062742591722366873321 062741343576104182143 125 times 10minus5

091111 094767578680690190767 094770492350476572039 291 times 10minus5

099999 86066655193121619385 86053938870581623578 0127 times 10minus2

20

minus099999 000387291557000340015 000387291555159212453 184 times 10minus11

minus091111 031504763162812503076 031504763156789346359 602 times 10minus11

minus077111 041231131092313516294 041231131089736085386 258 times 10minus11

minus033111 044083450592426358350 044083450592691374976 265 times 10minus12

000000 043301270189221932338 043301270187830840205 139 times 10minus11

033111 044963976686376800332 044963976684337128944 204 times 10minus11

077111 062742591722366873321 062742591718912403402 346 times 10minus11

091111 094767578680690190767 094767578677558019222 313 times 10minus11

099999 86066655193121619385 86066655108152556362 850 times 10minus8

40

minus099999 000387291557000340015 000387291557000339911 104 times 10minus18

minus091111 031504763162812503076 031504763162812503215 140 times 10minus18

minus077111 041231131092313516294 041231131092313516515 228 times 10minus18

minus033111 044083450592426358350 044083450592426358289 614 times 10minus19

000000 043301270189221932338 043301270189221932269 675 times 10minus19

033111 044963976686376800332 044963976686376800195 131 times 10minus18

077111 062742591722366873321 062742591722366873124 191 times 10minus18

091111 094767578680690190767 094767578680690190858 131 times 10minus18

099999 86066655193121619385 86066655193121618680 705 times 10minus16

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

0012

0010

0008

0006

0004

0002

Abso

lute

erro

r val

ues

(b)

Figure 6 (a) Exact and AQS solutions when 119873 = 10 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 10 and ℎ(119905) = 1(2 + 119905)

12 Abstract and Applied Analysis

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

times10minus8

Abso

lute

erro

r val

ues

(b)

Figure 7 (a) Exact and AQS solutions when 119873 = 20 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 20 and ℎ(119905) = 1(2 + 119905)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

Exact solution plotAQS solution plot

80

70

60

50

40

30

20

10

(a)

minus1 minus05 0 05 1

Singular point values

70

60

50

40

30

20

10

times10minus16

Abso

lute

erro

r val

ues

(b)

Figure 8 (a) Exact and AQS solutions when 119873 = 40 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 40 and ℎ(119905) = 1(2 + 119905)

where ℎ(119905) = 1(2 + 119905) The exact solution is 1198672(ℎ 119909) =

radic(1 + 119909)(1 minus 119909)(radic3(2 + 119909)2)

The numerical results in Table 4 show that the AQS (21)is highly accurate for the rational function ℎ(119905) = 1(2 + 119905)for the case 119894 = 2 and119873 = 40 Figure 6(a) shows that there is

good agreement between the exact solution andAQS solutionwhich can be seen from absolute error in Figure 6(b) when119873 = 10 The absolute error is reduced in Figure 7(b) when119873 = 20 with the corresponding exact and AQS solutionin Figure 7(a) The best result is obtained and illustrated inFigures 8(a) and 8(b) when119873 = 40

Abstract and Applied Analysis 13

6 Conclusion

In this paper we have constructed the automatic quadratureschemes for the semibounded weighted Hadamard typehypersingular integrals using the truncated series of the thirdand fourth kinds of Chebyshev polynomials The unknowncoefficients are found by applying the discrete orthogonalitycondition and the choice of collocation points on (minus1 1)The error estimations for the semibounded solutions of thehypersingular integrals are obtained Numerical exampleshave shown the accuracy of the developed methods andagreed with theoretical findings

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Hadamard Le Probleme de Cauchy et les Equations auxDerivees Particelles Lineaires Hyperboliques Hermann ParisFrance 1932

[2] E Lutz L Gray and A Ingraffea ldquoAn overview of integrationmethods for hypersingular integralsrdquo in Boundary Elements CA Brebbia and G S Gipson Eds vol 8 pp 913ndash925 CMPSouthampton UK 1991

[3] S G Samko Hypersingular Integrals and Their ApplicationsTaylor amp Francis London UK 2002

[4] I V Boikov ldquoNumerical methods of computation of singularand hypersingular integralsrdquo International Journal of Mathe-matics and Mathematical Sciences vol 28 no 3 pp 127ndash1792001

[5] I K Lifanov L N Poltavskii and G M Vainikko Hypersin-gular Integral Equations and Their Applications Chapman ampHallCRC A CRC Press Company London UK 2004

[6] H R Kutt ldquoQuadrature formulae for finite part integralsrdquoReport WISK 178 The National Research Institute for Mathe-matical Sciences Pretoria South Africa 1975

[7] C-Y Hui and D Shia ldquoEvaluations of hypersingular integralsusingGaussian quadraturerdquo International Journal for NumericalMethods in Engineering vol 44 no 2 pp 205ndash214 1999

[8] Y Z Chen ldquoNumerical solution of a curved crack problem byusing hypersingular integral equation approachrdquo EngineeringFracture Mechanics vol 46 no 2 pp 275ndash283 1993

[9] N M A Nik Long and Z K Eshkuvatov ldquoHypersingularintegral equation for multiple curved cracks problem in planeelasticityrdquo International Journal of Solids and Structures vol 46no 13 pp 2611ndash2617 2009

[10] S J Obaiys Z K Eshkuvatov and N M Nik Long ldquoOn errorestimation of automatic quadrature scheme for the evaluationof Hadamard integral of second order singularityrdquo University ofBucharest Scientific Bulletin Series A Applied Mathematics andPhysics vol 75 no 1 pp 85ndash98 2013

[11] J CMason andDCHandscombChebyshev Polynomials CRCPress 2003

[12] M I Israilov Semi-Analytic Solution of Singular Integral Equa-tions of First Kind by Quadrature Formula Numerical Integra-tion and Related Problems FAN Press 1990 (Russian)

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Stochastic AnalysisInternational Journal of

Page 3: An Automatic Quadrature Schemes and Error Estimates for Semibounded Weighted Hadamard Type Hypersingular Integrals

Abstract and Applied Analysis 3

and the differentiation formulae119889

119889119909119879119896 (119905) = 119896119880119896minus1 (119905)

119889

119889119909119880119896 (119905) =

(119896 + 1) 119879119896+1 (119905) minus 119905119880119896

1199052 minus 1

(16)

The Hilbert type integral transform is as follows

Lemma 3 (Mason andHandscomb [11]) Consider the follow-ing

minusint

120587

0

cos119898120579cos 120579 minus cos120572

119889120579 = 120587sin119898120572sin120572

(17)

for any 120572 isin [0 120587]119898 = 1 2 3

Lemma4 (Mason andHandscomb [11]) Consider the follow-ing

minusint

120587

0

sin119898120579 sin 120579cos 120579 minus cos120572

119889120579 = minus120587 cos119898120572 (18)

for any 120572 isin [0 120587]119898 = 1 2 3

The estimations for some integrals are established

Lemma 5 (Israilov [12]) Let 0 lt 120590 le 1 and 1 minus 119909 ge 120575119873

0 lt 120575119873lt 12 and for all 119894 = 1 2

120574119894 (119909) = 119908119894 (119909) int

119909minus120575119873

minus1

119889119905

119908119894 (119905) (119905 minus 119909)

120574lowast

119894(119909) = 119908119894 (119909) int

1

119909+120575119873

119889119905

119908119894 (119905) (119905 minus 119909)

119879119894(119909 120575119873 120590) = 119908

119894 (119909) int

119909+120575119873

119909minus120575119873

|119905 minus 119909|120590minus1119889119905

119908119894 (119905)

(19)

where 1199081(119909) = radic(1 minus 119909)(1 + 119909) and 119908

2(119909) =

radic(1 + 119909)(1 minus 119909)The following estimations are true

(i) |120574119894(119909)| le ln(2120575

119873) + 120587radic2

(ii) 1205741(minus119909) = minus120574

lowast

2(119909) 1205742(minus119909) = minus120574

lowast

1(119909)

(iii) 119879119894(119909 120575119873 120590) le 120601

119894(120590)120575120590

119873 max120601

1(120590) 1206012(120590) le radic10 times

2minus120590+ (1 + radic10 times 2

minus1minus120590)120590minus1

3 Automatic Quadrature Schemes forSemibounded Hypersingular Integrals

To construct AQSs for the semiboundedHSIs (4) the densityfunction ℎ(119905) is approximated with the truncated series ofthird and forth kinds of Chebyshev polynomials

ℎ (119905) asymp 119878119894119873 (119905) =

119873

sum

119898=0

1198871119898119881119898 (119905) 119894 = 1

119873

sum

119898=0

1198872119898119882119898 (119905) 119894 = 2

(20)

Substituting (20) into (4) we have

119867119894 (ℎ 119909)

asymp 119867119894119873 (ℎ 119909)

=119908119894 (119905)

120587=int

1

minus1

119878119894119873 (119905)

119908119894 (119909) (119905 minus 119909)

2119889119905

=

radic1 minus 119909

1 + 119909

119873

sum

119898=0

1198871119898

119889

119889119909[1

120587minusint

1

minus1

(1 + 119905) 119881119898 (119905)

radic1 minus 1199052 (119905 minus 119909)119889119905] 119894 = 1

radic1 + 119909

1 minus 119909

119873

sum

119898=0

1198872119898

119889

119889119909[1

120587minusint

1

minus1

(1 minus 119905)119882119898 (119905)

radic1 minus 1199052 (119905 minus 119909)119889119905] 119894 = 2

=

radic1 minus 119909

1 + 1199091199021 (119909) 119894 = 1

radic1 + 119909

1 minus 1199091199022 (119909) 119894 = 2

(21)

where

119902119894 (119909) =

119889

119889119909

119873

sum

119898=0

1198871119898119882119898 (119909) 119894 = 1

119889

119889119909

119873

sum

119898=0

1198872119898119881119898 (119909) 119894 = 2

(22)

By applying the orthogonality relations (10) we obtain thecoefficients 119887

119894119898 119894 = 1 2 as

119887119894119898=

1

120587int

1

minus1

radic1 + 119905

1 minus 119905ℎ (119905) 119881119898 (119905) 119889119905 119894 = 1

1

120587int

1

minus1

radic1 minus 119905

1 + 119905ℎ (119905)119882119898 (119905) 119889119905 119894 = 2

(23)

For the discrete coefficients 119887119894119898 we choose the collocation

points 119905 = 119905119896 at the zeros of 119879

119873+1 namely

119905 = 119905119896= cos (2119896 + 1) 120587

2119873 + 2 119896 = 0 1 2 119873 (24)

and using the orthogonality conditions (11) and relations(12) and (14) respectively as well as applying compositetrapezium rule we arrive at

119887119894119898asymp

1

119873 + 32

119873

sum

119896=0

(1 + 119905119896) 119881119898(119905119896) ℎ (119905119896) 119894 = 1

1

119873 + 32

119873

sum

119896=0

(119905119896minus 1)119882

119898(119905119896) ℎ (119905119896) 119894 = 2

(25)

4 Error Estimation for the SemiboundedHypersingular Integrals

Let us introduce the classes of functions

(i) 119867120572([minus1 1] 119871) is a class of function satisfying Holdercondition on the interval with the index 120572 andconstant 119871

4 Abstract and Applied Analysis

(ii) 119862119873120572[minus1 1] = ℎ(119905) ℎ(119873)(119905) isin 119867120572([minus1 1] 119871119873)

(iii) 119862[minus1 1] is a class of continuous functions on theinterval [minus1 1] with the norm

ℎ = ℎ(119909)119862[minus11] = max119909isin[minus11]

|ℎ (119909)| (26)

(iv) 119862119903[minus1 1] = ℎ(119905) ℎ119903(119905) isin 119862([minus1 1])Let 119890119894119873(119905) = ℎ(119905) minus 119878

119894119873(119905) and let the maximum norm of

the error be defined as10038171003817100381710038171198901198941198731003817100381710038171003817 = max119905isin[minus11]

1003816100381610038161003816119890119894119873 (119905)1003816100381610038161003816 (27)

It is known in [11] that for any 119909 isin (minus1 1)

1198901119873 (119909) =

ℎ(119873)(120578119909)119882119873 (119909)

2119873119873 120578119909isin [minus1 1]

1198902119873 (119909) =

ℎ(119873)(120578119909) 119881119873 (119909)

2119873119873 120578119909isin [minus1 1]

(28)

1198901015840

1119873(119909) =

1

2119873119873[119882119873 (119909)

119889

119889119909ℎ(119873)(120578119909) + ℎ(119873)(120578119909)1198821015840

119873(119909)]

120578119909isin [minus1 1]

1198901015840

2119873(119909) =

1

2119873119873[119881119873 (119909)

119889

119889119909ℎ(119873)(120578119909) + ℎ(119873)(120578119909) 1198811015840

119873(119909)]

120578119909isin [minus1 1]

(29)

As a consequence of (29) we arrive at the followingestimations100381710038171003817100381711989011198731003817100381710038171003817 =100381710038171003817100381711989021198731003817100381710038171003817 le(2119873 + 1)1198721

2119873 sdot 119873

(30)100381710038171003817100381710038171198901015840

1119873

10038171003817100381710038171003817=100381710038171003817100381710038171198901015840

2119873

10038171003817100381710038171003817

le119872

12 sdot 2119873minus3 (119873 minus 3)[1 +

1198711119873

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)]

(31)

where119872 = max 119872

11198722 119872

1= max120578119909isin[minus11]

10038161003816100381610038161003816ℎ(119873)(120578119909)10038161003816100381610038161003816

1198722= max120578119909isin[minus11]

10038161003816100381610038161003816100381610038161003816

119889

119889119909ℎ(119873)(120578119909)

10038161003816100381610038161003816100381610038161003816lt infin

(32)

1198711119873= 1 +

3

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)(1 +

3

119873) (33)

Lemma 6 The Lipschitz constants 119871119895119894119873

119895 = 3 4 119894 = 1 2of 119881119873(119909) and119882

119873(119909) are

1003816100381610038161003816119881119873 (119909) minus 119881119873 (119910)1003816100381610038161003816 le 11987131119873

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

100381610038161003816100381610038161198811015840

119873(119909) minus 119881

1015840

119873(119910)10038161003816100381610038161003816le 11987132119873

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

1003816100381610038161003816119882119873 (119909) minus 119882119873 (119910)1003816100381610038161003816 le 11987141119873

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

100381610038161003816100381610038161198821015840

119873(119909) minus 119882

1015840

119873(119910)10038161003816100381610038161003816le 11987142119873

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

(34)

where

11987131119873= 11987141119873=119873

3(119873 + 1) (2119873 + 1)

11987132119873= 11987142119873=119873

15(119873 + 1) (119873 + 2) (2119873 + 1) (119873 minus 1)

(35)

Proof Let 11987131119873

be the Lipschitz constant of the third kindChebyshev polynomial 119881

119873(119905) that is

1003816100381610038161003816119881119873 (119905) minus 119881119873 (119909)1003816100381610038161003816 le 11987131119873 |119905 minus 119909| (36)

where

11987131119873= max120577119905isin[minus11]

100381610038161003816100381610038161198811015840

119873(120577119905)10038161003816100381610038161003816 (37)

which gives sequence of numbers

0 2 10 28 60 110 182 280 408 570 (38)

These sequences can be generated by

11987131(119873)

= 11987131(119873minus1)

+ 21198732 11987131(0)= 0 (39)

which leads to

11987131119873=119873

3(119873 + 1) (2119873 + 1) (40)

In a similar way we can get

11987141119873=119873

3(119873 + 1) (2119873 + 1) (41)

Let 11987132(119873)

be the Lipschitz constant of the derivative ofthird kind Chebyshev polynomial 1198811015840

119873(119905) that is

100381610038161003816100381610038161198811015840

119873(119905) minus 119881

1015840

119873(119909)10038161003816100381610038161003816le 11987132119873 |119905 minus 119909| (42)

where

11987132119873= max120577119905isin[minus11]

1003816100381610038161003816100381611988110158401015840

119873(120577119905)10038161003816100381610038161003816 (43)

This Lipschitz constants generate the following sequenceof numbers

0 0 8 56 216 616 1456 3024 5712 10032 (44)

and its recurrence relation is

11987132(119873+2)

= 11987132(119873+1)

+119872119873 (45)

where

11987132(0)= 0

119872(119873)= 8119863(119873+1)

119863(119873+1)

= 119863(119873)+ 119877(119873+1) 119863(1)= 1

119877(119873+1)

= 119877(119873)+ (119873 + 1)

2119877(1)= 1

(46)

Abstract and Applied Analysis 5

These recurrence relations lead to

11987132(119873)

=119873

5(119873 + 1) (119873 + 2) (2119873 + 1) (119873 minus 1) (47)

In a similar way we obtain

11987142(119873)

=119873

5(119873 + 1) (119873 + 2) (2119873 + 1) (119873 minus 1) (48)

Main results are given in the followingTheorem 7

Theorem 7 Let ℎ(119905) isin 119862119873120572[minus1 1] for 0 lt 120572 le 1 and let theseries of Chebyshev polynomials 119878

119894119873(119905) be defined by (20) then

the AQS (21) has an error bound

1003817100381710038171003817119864119894 (ℎ)1003817100381710038171003817119862 le

011119872

2119873minus4 (119873 minus 4)[1+238

1198731198713119873+05 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)]

119894 = 1 2

(49)

where119872 is defined by (32) and

1198713119873= 1 +

253

ln119873[1198711119873+ 007119871

2119873

+042 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)(1 +

111

ln119873)

+003 (23119873

3minus 69119873

2+ 100119873 minus 48)

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

1198711119873= 1 +

3

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)+

3 (7119873 minus 4)

2119873 (119873 minus 1) (119873 minus 2)

1198712119873= 1 +

3

119873+231198733minus 69119873

2+ 100119873 minus 48

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)(1 +

3

119873)

(50)

Proof of Theorem 7 In view of error norms (27) we cancompute the error bound of AQS (21) for 119894 = 1 2 as follows

119864119894 (ℎ 119909) = 119867119894 (ℎ 119909) minus 119867119894119873 (ℎ 119909)

=119908119894 (119909)

120587=int

1

minus1

119890119894119873 (119905) 119889119905

119908119894 (119905) (119905 minus 119909)

2

=119908119894 (119909)

120587=int

1

minus1

[119890119894119873 (119905) minus 119890119894119873 (119909) minus 119890

1015840

119894119873(119909) (119905 minus 119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

2

+119908119894 (119909)

120587=int

1

minus1

[119890119894119873 (119909) + 119890

1015840

119894119873(119909) (119905 minus 119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

2

(51)

Taylor series expansion of 119890119894119873(119905) around 119909 is given by

119890119894119873 (119905) = 119890119894119873 (119909) + 119890

1015840

119894119873(120577119905) (119905 minus 119909) (52)

where

120577119905isin (119909 119905) if 119905 isin (119909 1) (119905 119909) if 119905 isin (minus1 119909)

(53)

Due to (51)-(52) we have1003816100381610038161003816119864119894 (ℎ 119909)

1003816100381610038161003816 le 1198641198941 (ℎ 119909) + 1198641198942 (ℎ 119909) + 1198641198943 (ℎ 119909) (54)

where

1198641198941 (ℎ 119909) =

10038161003816100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909)

120587minusint

1

minus1

[1198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

10038161003816100381610038161003816100381610038161003816100381610038161003816

1198641198942 (ℎ 119909) =

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909)

120587119890119894119873 (119909) =int

1

minus1

119889119905

119908119894 (119905) (119905 minus 119909)

2

100381610038161003816100381610038161003816100381610038161003816

1198641198943 (ℎ 119909) =

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909)

1205871198901015840

119894119873(119909) minusint

1

minus1

119889119905

119908119894 (119905) (119905 minus 119909)

100381610038161003816100381610038161003816100381610038161003816

(55)

It is easy to check that

=int

1

minus1

radic1 + 119909

1 minus 119909

119889119905

(119905 minus 119909)2= =int

1

minus1

radic1 minus 119905

1 + 119905

119889119905

(119905 minus 119909)2= 0

minusint

1

minus1

radic1 + 119909

1 minus 119909

119889119905

119905 minus 119909= minusminusint

1

minus1

radic1 minus 119905

1 + 119905

119889119905

119905 minus 119909= 120587

(56)

For the estimation of 1198641198941(ℎ 119909) we divide the interval

(minus1 1) into three subintervals (minus1 119909 minus 120575119873) (119909 + 120575

119873 1) and

[119909 minus 120575119873 119909 + 120575

119873]

1198641198941 (ℎ 119909) le

1

120587[1198651198941 (119909) + 1198651198942 (119909) + 1198651198943 (119909)] (57)

where

1198651198941 (119909) =

10038161003816100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

119909minus120575119873

minus1

[1198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

10038161003816100381610038161003816100381610038161003816100381610038161003816

1198651198942 (119909) =

10038161003816100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

1

119909+120575119873

[1198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

10038161003816100381610038161003816100381610038161003816100381610038161003816

1198651198943 (119909) =

10038161003816100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

119909+120575119873

119909minus120575119873

[1198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

10038161003816100381610038161003816100381610038161003816100381610038161003816

(58)

For 1198651198941(119909) according to Lemma 5 we obtain

1198651198941 (119909) le 2

100381710038171003817100381710038171198901015840

119894119873

10038171003817100381710038171003817

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

119909minus120575119873

minus1

119889119905

119908119894 (119905) (119905 minus 119909)

100381610038161003816100381610038161003816100381610038161003816

le 2100381710038171003817100381710038171198901015840

119894119873

10038171003817100381710038171003817[ln 2120575119873

+120587

radic2]

(59)

In the similar way for 1198651198942(119909) we have

1198651198942 (119909) le 2

100381710038171003817100381710038171198901015840

119894119873

10038171003817100381710038171003817

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

1

119909+120575119873

119889119905

119908119894 (119905) (119905 minus 119909)

100381610038161003816100381610038161003816100381610038161003816

le 2100381710038171003817100381710038171198901015840

119894119873

10038171003817100381710038171003817[ln 2120575119873

+120587

radic2]

(60)

6 Abstract and Applied Analysis

Utilizing (30) and (31) into (59) and (60) we arrive at

1198651198941 (119909) = 1198651198942 (119909)

le(16)119872

2119873minus3 (119873 minus 3)[1 +

1198711119873

119873+

7119873 minus 4

(119873 minus 1) (119873 minus 2)]

times [ln 2120575119873

+120587

radic2]

(61)

For the error of 1198651198943(119909)

1198651198943 (119909) le 119908119894 (119909) int

119909+120590119873

119909minus120590119873

100381610038161003816100381610038161198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)10038161003816100381610038161003816

119908119894 (119905)1003816100381610038161003816120577119905 minus 119909

1003816100381610038161003816

119889119905 (62)

It follows from (29) and Lemma 6 that100381610038161003816100381610038161198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)10038161003816100381610038161003816

le1

2119873119873

times 1003816100381610038161003816119875119894119873 (120577119905) minus 119875119894119873 (119909)

1003816100381610038161003816

10038161003816100381610038161003816100381610038161003816

119889

119889119909ℎ(119873)(120577119905)

10038161003816100381610038161003816100381610038161003816

+10038161003816100381610038161003816ℎ(119873)(120577119905)10038161003816100381610038161003816

100381610038161003816100381610038161198751015840

119894119873(120577119905) minus 1198751015840

119894119873(119909)10038161003816100381610038161003816

le1

2119873119873

times 119871119895119894119873

1003816100381610038161003816120577119905 minus 1199091003816100381610038161003816

10038161003816100381610038161003816100381610038161003816

119889

119889119909ℎ(119873)(120577119905)

10038161003816100381610038161003816100381610038161003816+ 119871119895119894119873

10038161003816100381610038161003816ℎ(119873)(120577119905)10038161003816100381610038161003816

1003816100381610038161003816120577119905 minus 1199091003816100381610038161003816

le119872

480 sdot 2119873minus5 (119873 minus 5)

1198732(119873 + 1) (2119873 + 1)

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)

times [1 +1

119873+3

1198732]1003816100381610038161003816120577119905 minus 119909

1003816100381610038161003816

le119872

240 sdot 2119873minus5 (119873 minus 5)

times [1 +231198733minus 69119873

2+ 100119873 minus 48

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

times [1 +1

119873+3

1198732]1003816100381610038161003816120577119905 minus 119909

1003816100381610038161003816

le119872

240 sdot 2119873minus5 (119873 minus 5)

times [1 +1198712119873

119873+

231198733minus 69119873

2+ 100119873 minus 48

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

times1003816100381610038161003816120577119905 minus 119909

1003816100381610038161003816

(63)

where

1198712119873= 1 +

3

119873+231198733minus 69119873

2+ 100119873 minus 48

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)(1 +

3

119873)

(64)

Substituting (63) into (62) yields

1198651198943 (119909) le

119872

240 sdot 2119873minus5 (119873 minus 5)

times [1 +1198712119873

119873+

231198733minus 69119873

2+ 100119873 minus 48

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

times

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

119909+120575119873

119909minus120575119873

119889119905

119908119894 (119905)

100381610038161003816100381610038161003816100381610038161003816

(65)

Using Lemma 5 gives

1198651198943 (119909) le

337119872

240 sdot 2119873minus5 (119873 minus 5)

times [1+1198712119873

119873+231198733minus 69119873

2+ 100119873 minus 48

2 (119873minus1) (119873minus2) (119873minus3) (119873minus4)] 120575119873

(66)

By choosing 120575119873= 2119873

2 for119873 ge 5 we obtain

1198651198943 (119909) le

011119872

2119873minus3 (119873 minus 3)

times [1 +1198712119873

119873+

231198733minus 69119873

2+ 100119873 minus 48

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

(67)

By substituting (61) and (67) into (57) we arrive at

1198641198941 (ℎ 119909) le

064119872ln119873

120587 sdot 2119873minus3 (119873 minus 3)

times [1 +127

ln119873+211

ln119873(1198711119873+ 008119871

2119873)

+254 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)(1 +

111

ln119873)

+017 (23119873

3minus 69119873

2+ 100119873 minus 48)

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

(68)

1198641198941 (ℎ 119909) le

246119872ln119873

2119873minus3 (119873 minus 3)

times [1 +010

ln119873+1198711119873

119873+010119871

1119873

119873 ln119873

+2316 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)+005 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)

+1059 (23119873

3minus 69119873

2+ 100119873 minus 48)

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

(69)

Abstract and Applied Analysis 7

Table 1 Results of HSIs for 1199081(119905)

119873 119909 Exact solution Approximate solution Absolute error

2

minus099999 minus22360444988237596342 minus22360444988237596345 313 times 10minus16

minus091111 minus21535241487484869307 minus21535241487484869310 308 times 10minus18

minus071111 minus93563357403104129848 minus93563357403104129863 142 times 10minus18

minus031111 minus30963683860810919235 minus30963683860810919240 586 times 10minus19

000000 minus10000000000000000000 minus10000000000000000003 300 times 10minus19

031111 017718528456311131724 017718528456311131709 127 times 10minus19

071111 075786494952828070264 075786494952828070264 639 times 10minus21

091111 057031832778553277005 057031832778553277007 140 times 10minus20

099999 000670813126012938238 000670813126012938238 224 times 10minus22

minus1 minus05 0 05 1

Singular point values

minus500

minus1000

minus1500

minus2000

HSI

val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

30

20

10

times10minus16

Abso

lute

erro

r val

ues

(b)Figure 1 (a) Exact and AQS solutions when 119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5 and (b) absolute error between exact and AQS solution when119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5

Due to (31) and (56)1198641198942 (ℎ 119909) = 0

1198641198943 (ℎ 119909) le

2

radic1 minus 1199092

008119872

2119873minus3 (119873 minus 3)

times [1 +1198711119873

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)]

(70)

Since from the hypothesis in Lemma 5

1 minus 1199092ge 1 minus (1 minus 120575

119873)2

1

radic1 minus 1199092le

1

radic1 minus (1 minus 120575119873)2

leradic2

radic3120575119873

=119873

radic3

(71)

we have

1198641198943 (ℎ 119909) le

009119872

2119873minus3 (119873 minus 3)

times 119873[1 +1198711119873

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)]

(72)

Substituting (69)ndash(72) into (54) yields

1003816100381610038161003816119864119894 (ℎ 119909)1003816100381610038161003816 le

068119872 ln119873120587 sdot 2119873minus3 (119873 minus 3)

times [1 +127

ln119873+211

ln119873(1198711119873+ 008119871

2119873)

+254 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)(1 +

111

ln119873)

+017 (23119873

3minus69119873

2+100119873minus48)

(119873minus1) (119873 minus 2) (119873minus3) (119873minus4)+042119873

ln119873

+042

ln1198731198711119873+

042119873 (7119873minus4)

2 ln119873(119873minus1) (119873 minus 2)]

(73)

Choosing119873 ge 5 the statement ofTheorem 7 follows

8 Abstract and Applied Analysis

Table 2 Results of HSIs for 1199082(119905)

119873 119909 Exact solution Approximate solution Absolute error

2

minus099999 002012457266627356241 00201245726662735624 157 times 10minus21

minus091111 18643200698228626472 18643200698228626470 143 times 10minus19

minus071111 32232147018485970132 32232147018485970129 240 times 10minus19

minus031111 45263577087926478228 45263577087926478225 308 times 10minus19

000000 50000000000000000000 49999999999999999997 300 times 10minus19

031111 51810684429214884711 51810684429214884708 242 times 10minus19

071111 52460548398163357508 52460548398163357508 379 times 10minus20

091111 62854320328769430625 62854320328769430630 299 times 10minus19

099999 44723036596367022628 44723036596367022632 447 times 10minus17

minus1 minus05 0 05

400

300

200

100

1

Singular point values

HSI

val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

40

30

20

10

times10minus17

Abso

lute

erro

r val

ues

(b)

Figure 2 (a) Exact and AQS solutions when 119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5 and (b) absolute error between exact and AQS solution when119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5

5 Numerical Results and Discussion

In this session we evaluate the hypersingular integrals (4) fordifferent choices of ℎ(119905) and compare with the exact solution

Example 1 Consider the HSIs of the form

1198671 (ℎ 119909) =

1

120587radic1 minus 119909

1 + 119909=int

1

minus1

radic1 + 119905

1 minus 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (74)

where ℎ(119905) = 21199052 minus 3119905 + 5 The exact solution is 1198671(ℎ 119909) =

2radic(1 minus 119909)(1 + 119909)(minus1 + 4119909)

The numerical results in Table 1 show that the AQS givesa very good result for the quadratic function ℎ(119905) = 21199052 minus3119905 + 5 for the case 119894 = 1 and 119873 = 2 This is furtherillustrated in Figure 1(a) where the graphs of the exact andAQS solutions are superimposed for the same choices ofsingular point values Details of the accuracy of the proposedAQS is observed from the absolute error in Figure 1(b)

Example 2 Consider the HSIs of the form

1198672 (ℎ 119909) =

1

120587radic1 + 119909

1 minus 119909=int

1

minus1

radic1 minus 119905

1 + 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (75)

where ℎ(119905) = 21199052 minus 3119905 + 5 The exact solution is 1198672(ℎ 119909) =

radic(1 + 119909)(1 minus 119909)(5 minus 4119909)

The numerical results in Table 2 show that the AQS (21)gives a very good result for the quadratic function ℎ(119905) = 21199052minus3119905 + 5 for the case 119894 = 2 and 119873 = 2 Figure 2(a) indicatesthat the exact and AQS solutions are nearly the same Furtherdetails of the accuracy of the absolute error of the proposedAQS are demonstrated in Figure 2(b)

Example 3 Consider the HSIs of the form

1198671 (ℎ 119909) =

1

120587radic1 minus 119909

1 + 119909=int

1

minus1

radic1 + 119905

1 minus 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (76)

Abstract and Applied Analysis 9

Table 3 Results of HSIs for 1199081(119905)

119873 119909 Exact solution Approximate solution Absolute error

10

minus099999 minus25819308036170334212 minus25817083250027430060 223 times 10minus2

minus091111 minus22578118246552894121 minus22578564435672151572 446 times 10minus5

minus077111 minus10634637975870125656 minus10634476360338990858 162 times 10minus5

minus033111 minus02924243743021256602 minus02924196578457308097 472 times 10minus6

000000 minus01443375672974064411 minus01443397190188599320 215 times 10minus6

033111 minus00753154744680519215 minus00753153016422631300 173 times 10minus7

077111 minus00270285335549129334 minus00270280566695517436 477 times 10minus7

091111 minus00146928400579184367 minus00146941562302624467 132 times 10minus6

099999 minus00001434451425475821 minus00001431650832316927 280 times 10minus7

20

minus099999 minus25819308036170334212 minus25819308021411085378 148 times 10minus7

minus091111 minus22578118246552894121 minus22578118246107028644 446 times 10minus11

minus077111 minus10634637975870125656 minus10634637975433417458 437 times 10minus11

minus033111 minus02924243743021256602 minus02924243742855352783 166 times 10minus11

000000 minus01443375672974064411 minus01443375672897319217 768 times 10minus12

033111 minus00753154744680519215 minus00753154744695265617 148 times 10minus12

077111 minus00270285335549129334 minus00270285335485592653 635 times 10minus12

091111 minus00146928400579184367 minus00146928400450019102 129 times 10minus11

099999 minus00001434451425475821 minus00001434451390347442 351 times 10minus12

40

minus099999 minus25819308036170334212 minus25819308036170338346 404 times 10minus14

minus091111 minus22578118246552894121 minus22578118246552894116 450 times 10minus19

minus077111 minus10634637975870125656 minus10634637975870125637 185 times 10minus18

minus033111 minus02924243743021256602 minus02924243743021256612 967 times 10minus19

000000 minus01443375672974064411 minus01443375672974064412 113 times 10minus19

033111 minus00753154744680519215 minus00753154744680519211 400 times 10minus19

077111 minus00270285335549129334 minus00270285335549129336 172 times 10minus19

091111 minus00146928400579184367 minus00146928400579184373 573 times 10minus19

099999 minus00001434451425475821 minus00001434451425475826 486 times 10minus19

minus1 minus05 0 05 1

Singular point valuesminus50

minus100

minus150

minus200

minus250

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

0020

0015

0010

0005

Abso

lute

erro

r val

ues

(b)

Figure 3 (a) Exact and AQS solutions when 119873 = 10 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 10 and ℎ(119905) = 1(2 + 119905)

10 Abstract and Applied Analysis

minus1 minus05 0 05 1

Singular point valuesminus50

minus100

minus150

minus200

minus250

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

14

12

10

80

60

40

20

times10minus8

(b)

Figure 4 (a) Exact and AQS solutions when 119873 = 20 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 20 and ℎ(119905) = 1(2 + 119905)

minus1 minus05 0 05 1

Singular point values

minus50

minus100

minus150

minus250

HSI

s val

ues

minus200

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

40

30

20

10

times10minus14

(b)

Figure 5 (a) Exact and AQS solutions when 119873 = 40 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 40 and ℎ(119905) = 1(2 + 119905)

where ℎ(119905) = 1(2 + 119905) The exact solution is 1198671(ℎ 119909) =

minusradic(1 minus 119909)(1 + 119909)(radic33(2 + 119909)2)

The numerical results in Table 3 show that the AQS (21)is highly accurate for the rational function ℎ(119905) = 1(2 + 119905)for the case 119894 = 1 and119873 = 40 Figure 3(a) reveals that there jsgood agreement between the exact solution andAQS solutionwhich can be seen from absolute error in Figure 3(b) when119873 = 10 The absolute error is improved in Figure 4(b) when

119873 = 20 with the corresponding exact and AQS solution inFigure 4(a) Best result is obtained in Figures 5(a) and 5(b)when119873 = 40

Example 4 Consider the HSIs of the form

1198672 (ℎ 119909) =

1

120587radic1 + 119909

1 minus 119909=int

1

minus1

radic1 minus 119905

1 + 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (77)

Abstract and Applied Analysis 11

Table 4 Results of HSIs for 1199082(119905)

119873 119909 Exact solution Approximate solution Absolute error

10

minus099999 000387291557000340015 000387141593292902562 150 times 10minus6

minus091111 031504763162812503076 031505448317559772604 685 times 10minus6

minus077111 041231131092313516294 041230879229200872013 252 times 10minus6

minus033111 044083450592426358350 044083359884530457966 907 times 10minus7

000000 043301270189221932338 043301675286046974386 405 times 10minus6

033111 044963976686376800332 044963389568653105901 587 times 10minus6

077111 062742591722366873321 062741343576104182143 125 times 10minus5

091111 094767578680690190767 094770492350476572039 291 times 10minus5

099999 86066655193121619385 86053938870581623578 0127 times 10minus2

20

minus099999 000387291557000340015 000387291555159212453 184 times 10minus11

minus091111 031504763162812503076 031504763156789346359 602 times 10minus11

minus077111 041231131092313516294 041231131089736085386 258 times 10minus11

minus033111 044083450592426358350 044083450592691374976 265 times 10minus12

000000 043301270189221932338 043301270187830840205 139 times 10minus11

033111 044963976686376800332 044963976684337128944 204 times 10minus11

077111 062742591722366873321 062742591718912403402 346 times 10minus11

091111 094767578680690190767 094767578677558019222 313 times 10minus11

099999 86066655193121619385 86066655108152556362 850 times 10minus8

40

minus099999 000387291557000340015 000387291557000339911 104 times 10minus18

minus091111 031504763162812503076 031504763162812503215 140 times 10minus18

minus077111 041231131092313516294 041231131092313516515 228 times 10minus18

minus033111 044083450592426358350 044083450592426358289 614 times 10minus19

000000 043301270189221932338 043301270189221932269 675 times 10minus19

033111 044963976686376800332 044963976686376800195 131 times 10minus18

077111 062742591722366873321 062742591722366873124 191 times 10minus18

091111 094767578680690190767 094767578680690190858 131 times 10minus18

099999 86066655193121619385 86066655193121618680 705 times 10minus16

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

0012

0010

0008

0006

0004

0002

Abso

lute

erro

r val

ues

(b)

Figure 6 (a) Exact and AQS solutions when 119873 = 10 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 10 and ℎ(119905) = 1(2 + 119905)

12 Abstract and Applied Analysis

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

times10minus8

Abso

lute

erro

r val

ues

(b)

Figure 7 (a) Exact and AQS solutions when 119873 = 20 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 20 and ℎ(119905) = 1(2 + 119905)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

Exact solution plotAQS solution plot

80

70

60

50

40

30

20

10

(a)

minus1 minus05 0 05 1

Singular point values

70

60

50

40

30

20

10

times10minus16

Abso

lute

erro

r val

ues

(b)

Figure 8 (a) Exact and AQS solutions when 119873 = 40 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 40 and ℎ(119905) = 1(2 + 119905)

where ℎ(119905) = 1(2 + 119905) The exact solution is 1198672(ℎ 119909) =

radic(1 + 119909)(1 minus 119909)(radic3(2 + 119909)2)

The numerical results in Table 4 show that the AQS (21)is highly accurate for the rational function ℎ(119905) = 1(2 + 119905)for the case 119894 = 2 and119873 = 40 Figure 6(a) shows that there is

good agreement between the exact solution andAQS solutionwhich can be seen from absolute error in Figure 6(b) when119873 = 10 The absolute error is reduced in Figure 7(b) when119873 = 20 with the corresponding exact and AQS solutionin Figure 7(a) The best result is obtained and illustrated inFigures 8(a) and 8(b) when119873 = 40

Abstract and Applied Analysis 13

6 Conclusion

In this paper we have constructed the automatic quadratureschemes for the semibounded weighted Hadamard typehypersingular integrals using the truncated series of the thirdand fourth kinds of Chebyshev polynomials The unknowncoefficients are found by applying the discrete orthogonalitycondition and the choice of collocation points on (minus1 1)The error estimations for the semibounded solutions of thehypersingular integrals are obtained Numerical exampleshave shown the accuracy of the developed methods andagreed with theoretical findings

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Hadamard Le Probleme de Cauchy et les Equations auxDerivees Particelles Lineaires Hyperboliques Hermann ParisFrance 1932

[2] E Lutz L Gray and A Ingraffea ldquoAn overview of integrationmethods for hypersingular integralsrdquo in Boundary Elements CA Brebbia and G S Gipson Eds vol 8 pp 913ndash925 CMPSouthampton UK 1991

[3] S G Samko Hypersingular Integrals and Their ApplicationsTaylor amp Francis London UK 2002

[4] I V Boikov ldquoNumerical methods of computation of singularand hypersingular integralsrdquo International Journal of Mathe-matics and Mathematical Sciences vol 28 no 3 pp 127ndash1792001

[5] I K Lifanov L N Poltavskii and G M Vainikko Hypersin-gular Integral Equations and Their Applications Chapman ampHallCRC A CRC Press Company London UK 2004

[6] H R Kutt ldquoQuadrature formulae for finite part integralsrdquoReport WISK 178 The National Research Institute for Mathe-matical Sciences Pretoria South Africa 1975

[7] C-Y Hui and D Shia ldquoEvaluations of hypersingular integralsusingGaussian quadraturerdquo International Journal for NumericalMethods in Engineering vol 44 no 2 pp 205ndash214 1999

[8] Y Z Chen ldquoNumerical solution of a curved crack problem byusing hypersingular integral equation approachrdquo EngineeringFracture Mechanics vol 46 no 2 pp 275ndash283 1993

[9] N M A Nik Long and Z K Eshkuvatov ldquoHypersingularintegral equation for multiple curved cracks problem in planeelasticityrdquo International Journal of Solids and Structures vol 46no 13 pp 2611ndash2617 2009

[10] S J Obaiys Z K Eshkuvatov and N M Nik Long ldquoOn errorestimation of automatic quadrature scheme for the evaluationof Hadamard integral of second order singularityrdquo University ofBucharest Scientific Bulletin Series A Applied Mathematics andPhysics vol 75 no 1 pp 85ndash98 2013

[11] J CMason andDCHandscombChebyshev Polynomials CRCPress 2003

[12] M I Israilov Semi-Analytic Solution of Singular Integral Equa-tions of First Kind by Quadrature Formula Numerical Integra-tion and Related Problems FAN Press 1990 (Russian)

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Stochastic AnalysisInternational Journal of

Page 4: An Automatic Quadrature Schemes and Error Estimates for Semibounded Weighted Hadamard Type Hypersingular Integrals

4 Abstract and Applied Analysis

(ii) 119862119873120572[minus1 1] = ℎ(119905) ℎ(119873)(119905) isin 119867120572([minus1 1] 119871119873)

(iii) 119862[minus1 1] is a class of continuous functions on theinterval [minus1 1] with the norm

ℎ = ℎ(119909)119862[minus11] = max119909isin[minus11]

|ℎ (119909)| (26)

(iv) 119862119903[minus1 1] = ℎ(119905) ℎ119903(119905) isin 119862([minus1 1])Let 119890119894119873(119905) = ℎ(119905) minus 119878

119894119873(119905) and let the maximum norm of

the error be defined as10038171003817100381710038171198901198941198731003817100381710038171003817 = max119905isin[minus11]

1003816100381610038161003816119890119894119873 (119905)1003816100381610038161003816 (27)

It is known in [11] that for any 119909 isin (minus1 1)

1198901119873 (119909) =

ℎ(119873)(120578119909)119882119873 (119909)

2119873119873 120578119909isin [minus1 1]

1198902119873 (119909) =

ℎ(119873)(120578119909) 119881119873 (119909)

2119873119873 120578119909isin [minus1 1]

(28)

1198901015840

1119873(119909) =

1

2119873119873[119882119873 (119909)

119889

119889119909ℎ(119873)(120578119909) + ℎ(119873)(120578119909)1198821015840

119873(119909)]

120578119909isin [minus1 1]

1198901015840

2119873(119909) =

1

2119873119873[119881119873 (119909)

119889

119889119909ℎ(119873)(120578119909) + ℎ(119873)(120578119909) 1198811015840

119873(119909)]

120578119909isin [minus1 1]

(29)

As a consequence of (29) we arrive at the followingestimations100381710038171003817100381711989011198731003817100381710038171003817 =100381710038171003817100381711989021198731003817100381710038171003817 le(2119873 + 1)1198721

2119873 sdot 119873

(30)100381710038171003817100381710038171198901015840

1119873

10038171003817100381710038171003817=100381710038171003817100381710038171198901015840

2119873

10038171003817100381710038171003817

le119872

12 sdot 2119873minus3 (119873 minus 3)[1 +

1198711119873

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)]

(31)

where119872 = max 119872

11198722 119872

1= max120578119909isin[minus11]

10038161003816100381610038161003816ℎ(119873)(120578119909)10038161003816100381610038161003816

1198722= max120578119909isin[minus11]

10038161003816100381610038161003816100381610038161003816

119889

119889119909ℎ(119873)(120578119909)

10038161003816100381610038161003816100381610038161003816lt infin

(32)

1198711119873= 1 +

3

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)(1 +

3

119873) (33)

Lemma 6 The Lipschitz constants 119871119895119894119873

119895 = 3 4 119894 = 1 2of 119881119873(119909) and119882

119873(119909) are

1003816100381610038161003816119881119873 (119909) minus 119881119873 (119910)1003816100381610038161003816 le 11987131119873

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

100381610038161003816100381610038161198811015840

119873(119909) minus 119881

1015840

119873(119910)10038161003816100381610038161003816le 11987132119873

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

1003816100381610038161003816119882119873 (119909) minus 119882119873 (119910)1003816100381610038161003816 le 11987141119873

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

100381610038161003816100381610038161198821015840

119873(119909) minus 119882

1015840

119873(119910)10038161003816100381610038161003816le 11987142119873

1003816100381610038161003816119909 minus 1199101003816100381610038161003816

(34)

where

11987131119873= 11987141119873=119873

3(119873 + 1) (2119873 + 1)

11987132119873= 11987142119873=119873

15(119873 + 1) (119873 + 2) (2119873 + 1) (119873 minus 1)

(35)

Proof Let 11987131119873

be the Lipschitz constant of the third kindChebyshev polynomial 119881

119873(119905) that is

1003816100381610038161003816119881119873 (119905) minus 119881119873 (119909)1003816100381610038161003816 le 11987131119873 |119905 minus 119909| (36)

where

11987131119873= max120577119905isin[minus11]

100381610038161003816100381610038161198811015840

119873(120577119905)10038161003816100381610038161003816 (37)

which gives sequence of numbers

0 2 10 28 60 110 182 280 408 570 (38)

These sequences can be generated by

11987131(119873)

= 11987131(119873minus1)

+ 21198732 11987131(0)= 0 (39)

which leads to

11987131119873=119873

3(119873 + 1) (2119873 + 1) (40)

In a similar way we can get

11987141119873=119873

3(119873 + 1) (2119873 + 1) (41)

Let 11987132(119873)

be the Lipschitz constant of the derivative ofthird kind Chebyshev polynomial 1198811015840

119873(119905) that is

100381610038161003816100381610038161198811015840

119873(119905) minus 119881

1015840

119873(119909)10038161003816100381610038161003816le 11987132119873 |119905 minus 119909| (42)

where

11987132119873= max120577119905isin[minus11]

1003816100381610038161003816100381611988110158401015840

119873(120577119905)10038161003816100381610038161003816 (43)

This Lipschitz constants generate the following sequenceof numbers

0 0 8 56 216 616 1456 3024 5712 10032 (44)

and its recurrence relation is

11987132(119873+2)

= 11987132(119873+1)

+119872119873 (45)

where

11987132(0)= 0

119872(119873)= 8119863(119873+1)

119863(119873+1)

= 119863(119873)+ 119877(119873+1) 119863(1)= 1

119877(119873+1)

= 119877(119873)+ (119873 + 1)

2119877(1)= 1

(46)

Abstract and Applied Analysis 5

These recurrence relations lead to

11987132(119873)

=119873

5(119873 + 1) (119873 + 2) (2119873 + 1) (119873 minus 1) (47)

In a similar way we obtain

11987142(119873)

=119873

5(119873 + 1) (119873 + 2) (2119873 + 1) (119873 minus 1) (48)

Main results are given in the followingTheorem 7

Theorem 7 Let ℎ(119905) isin 119862119873120572[minus1 1] for 0 lt 120572 le 1 and let theseries of Chebyshev polynomials 119878

119894119873(119905) be defined by (20) then

the AQS (21) has an error bound

1003817100381710038171003817119864119894 (ℎ)1003817100381710038171003817119862 le

011119872

2119873minus4 (119873 minus 4)[1+238

1198731198713119873+05 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)]

119894 = 1 2

(49)

where119872 is defined by (32) and

1198713119873= 1 +

253

ln119873[1198711119873+ 007119871

2119873

+042 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)(1 +

111

ln119873)

+003 (23119873

3minus 69119873

2+ 100119873 minus 48)

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

1198711119873= 1 +

3

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)+

3 (7119873 minus 4)

2119873 (119873 minus 1) (119873 minus 2)

1198712119873= 1 +

3

119873+231198733minus 69119873

2+ 100119873 minus 48

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)(1 +

3

119873)

(50)

Proof of Theorem 7 In view of error norms (27) we cancompute the error bound of AQS (21) for 119894 = 1 2 as follows

119864119894 (ℎ 119909) = 119867119894 (ℎ 119909) minus 119867119894119873 (ℎ 119909)

=119908119894 (119909)

120587=int

1

minus1

119890119894119873 (119905) 119889119905

119908119894 (119905) (119905 minus 119909)

2

=119908119894 (119909)

120587=int

1

minus1

[119890119894119873 (119905) minus 119890119894119873 (119909) minus 119890

1015840

119894119873(119909) (119905 minus 119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

2

+119908119894 (119909)

120587=int

1

minus1

[119890119894119873 (119909) + 119890

1015840

119894119873(119909) (119905 minus 119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

2

(51)

Taylor series expansion of 119890119894119873(119905) around 119909 is given by

119890119894119873 (119905) = 119890119894119873 (119909) + 119890

1015840

119894119873(120577119905) (119905 minus 119909) (52)

where

120577119905isin (119909 119905) if 119905 isin (119909 1) (119905 119909) if 119905 isin (minus1 119909)

(53)

Due to (51)-(52) we have1003816100381610038161003816119864119894 (ℎ 119909)

1003816100381610038161003816 le 1198641198941 (ℎ 119909) + 1198641198942 (ℎ 119909) + 1198641198943 (ℎ 119909) (54)

where

1198641198941 (ℎ 119909) =

10038161003816100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909)

120587minusint

1

minus1

[1198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

10038161003816100381610038161003816100381610038161003816100381610038161003816

1198641198942 (ℎ 119909) =

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909)

120587119890119894119873 (119909) =int

1

minus1

119889119905

119908119894 (119905) (119905 minus 119909)

2

100381610038161003816100381610038161003816100381610038161003816

1198641198943 (ℎ 119909) =

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909)

1205871198901015840

119894119873(119909) minusint

1

minus1

119889119905

119908119894 (119905) (119905 minus 119909)

100381610038161003816100381610038161003816100381610038161003816

(55)

It is easy to check that

=int

1

minus1

radic1 + 119909

1 minus 119909

119889119905

(119905 minus 119909)2= =int

1

minus1

radic1 minus 119905

1 + 119905

119889119905

(119905 minus 119909)2= 0

minusint

1

minus1

radic1 + 119909

1 minus 119909

119889119905

119905 minus 119909= minusminusint

1

minus1

radic1 minus 119905

1 + 119905

119889119905

119905 minus 119909= 120587

(56)

For the estimation of 1198641198941(ℎ 119909) we divide the interval

(minus1 1) into three subintervals (minus1 119909 minus 120575119873) (119909 + 120575

119873 1) and

[119909 minus 120575119873 119909 + 120575

119873]

1198641198941 (ℎ 119909) le

1

120587[1198651198941 (119909) + 1198651198942 (119909) + 1198651198943 (119909)] (57)

where

1198651198941 (119909) =

10038161003816100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

119909minus120575119873

minus1

[1198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

10038161003816100381610038161003816100381610038161003816100381610038161003816

1198651198942 (119909) =

10038161003816100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

1

119909+120575119873

[1198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

10038161003816100381610038161003816100381610038161003816100381610038161003816

1198651198943 (119909) =

10038161003816100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

119909+120575119873

119909minus120575119873

[1198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

10038161003816100381610038161003816100381610038161003816100381610038161003816

(58)

For 1198651198941(119909) according to Lemma 5 we obtain

1198651198941 (119909) le 2

100381710038171003817100381710038171198901015840

119894119873

10038171003817100381710038171003817

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

119909minus120575119873

minus1

119889119905

119908119894 (119905) (119905 minus 119909)

100381610038161003816100381610038161003816100381610038161003816

le 2100381710038171003817100381710038171198901015840

119894119873

10038171003817100381710038171003817[ln 2120575119873

+120587

radic2]

(59)

In the similar way for 1198651198942(119909) we have

1198651198942 (119909) le 2

100381710038171003817100381710038171198901015840

119894119873

10038171003817100381710038171003817

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

1

119909+120575119873

119889119905

119908119894 (119905) (119905 minus 119909)

100381610038161003816100381610038161003816100381610038161003816

le 2100381710038171003817100381710038171198901015840

119894119873

10038171003817100381710038171003817[ln 2120575119873

+120587

radic2]

(60)

6 Abstract and Applied Analysis

Utilizing (30) and (31) into (59) and (60) we arrive at

1198651198941 (119909) = 1198651198942 (119909)

le(16)119872

2119873minus3 (119873 minus 3)[1 +

1198711119873

119873+

7119873 minus 4

(119873 minus 1) (119873 minus 2)]

times [ln 2120575119873

+120587

radic2]

(61)

For the error of 1198651198943(119909)

1198651198943 (119909) le 119908119894 (119909) int

119909+120590119873

119909minus120590119873

100381610038161003816100381610038161198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)10038161003816100381610038161003816

119908119894 (119905)1003816100381610038161003816120577119905 minus 119909

1003816100381610038161003816

119889119905 (62)

It follows from (29) and Lemma 6 that100381610038161003816100381610038161198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)10038161003816100381610038161003816

le1

2119873119873

times 1003816100381610038161003816119875119894119873 (120577119905) minus 119875119894119873 (119909)

1003816100381610038161003816

10038161003816100381610038161003816100381610038161003816

119889

119889119909ℎ(119873)(120577119905)

10038161003816100381610038161003816100381610038161003816

+10038161003816100381610038161003816ℎ(119873)(120577119905)10038161003816100381610038161003816

100381610038161003816100381610038161198751015840

119894119873(120577119905) minus 1198751015840

119894119873(119909)10038161003816100381610038161003816

le1

2119873119873

times 119871119895119894119873

1003816100381610038161003816120577119905 minus 1199091003816100381610038161003816

10038161003816100381610038161003816100381610038161003816

119889

119889119909ℎ(119873)(120577119905)

10038161003816100381610038161003816100381610038161003816+ 119871119895119894119873

10038161003816100381610038161003816ℎ(119873)(120577119905)10038161003816100381610038161003816

1003816100381610038161003816120577119905 minus 1199091003816100381610038161003816

le119872

480 sdot 2119873minus5 (119873 minus 5)

1198732(119873 + 1) (2119873 + 1)

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)

times [1 +1

119873+3

1198732]1003816100381610038161003816120577119905 minus 119909

1003816100381610038161003816

le119872

240 sdot 2119873minus5 (119873 minus 5)

times [1 +231198733minus 69119873

2+ 100119873 minus 48

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

times [1 +1

119873+3

1198732]1003816100381610038161003816120577119905 minus 119909

1003816100381610038161003816

le119872

240 sdot 2119873minus5 (119873 minus 5)

times [1 +1198712119873

119873+

231198733minus 69119873

2+ 100119873 minus 48

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

times1003816100381610038161003816120577119905 minus 119909

1003816100381610038161003816

(63)

where

1198712119873= 1 +

3

119873+231198733minus 69119873

2+ 100119873 minus 48

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)(1 +

3

119873)

(64)

Substituting (63) into (62) yields

1198651198943 (119909) le

119872

240 sdot 2119873minus5 (119873 minus 5)

times [1 +1198712119873

119873+

231198733minus 69119873

2+ 100119873 minus 48

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

times

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

119909+120575119873

119909minus120575119873

119889119905

119908119894 (119905)

100381610038161003816100381610038161003816100381610038161003816

(65)

Using Lemma 5 gives

1198651198943 (119909) le

337119872

240 sdot 2119873minus5 (119873 minus 5)

times [1+1198712119873

119873+231198733minus 69119873

2+ 100119873 minus 48

2 (119873minus1) (119873minus2) (119873minus3) (119873minus4)] 120575119873

(66)

By choosing 120575119873= 2119873

2 for119873 ge 5 we obtain

1198651198943 (119909) le

011119872

2119873minus3 (119873 minus 3)

times [1 +1198712119873

119873+

231198733minus 69119873

2+ 100119873 minus 48

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

(67)

By substituting (61) and (67) into (57) we arrive at

1198641198941 (ℎ 119909) le

064119872ln119873

120587 sdot 2119873minus3 (119873 minus 3)

times [1 +127

ln119873+211

ln119873(1198711119873+ 008119871

2119873)

+254 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)(1 +

111

ln119873)

+017 (23119873

3minus 69119873

2+ 100119873 minus 48)

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

(68)

1198641198941 (ℎ 119909) le

246119872ln119873

2119873minus3 (119873 minus 3)

times [1 +010

ln119873+1198711119873

119873+010119871

1119873

119873 ln119873

+2316 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)+005 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)

+1059 (23119873

3minus 69119873

2+ 100119873 minus 48)

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

(69)

Abstract and Applied Analysis 7

Table 1 Results of HSIs for 1199081(119905)

119873 119909 Exact solution Approximate solution Absolute error

2

minus099999 minus22360444988237596342 minus22360444988237596345 313 times 10minus16

minus091111 minus21535241487484869307 minus21535241487484869310 308 times 10minus18

minus071111 minus93563357403104129848 minus93563357403104129863 142 times 10minus18

minus031111 minus30963683860810919235 minus30963683860810919240 586 times 10minus19

000000 minus10000000000000000000 minus10000000000000000003 300 times 10minus19

031111 017718528456311131724 017718528456311131709 127 times 10minus19

071111 075786494952828070264 075786494952828070264 639 times 10minus21

091111 057031832778553277005 057031832778553277007 140 times 10minus20

099999 000670813126012938238 000670813126012938238 224 times 10minus22

minus1 minus05 0 05 1

Singular point values

minus500

minus1000

minus1500

minus2000

HSI

val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

30

20

10

times10minus16

Abso

lute

erro

r val

ues

(b)Figure 1 (a) Exact and AQS solutions when 119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5 and (b) absolute error between exact and AQS solution when119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5

Due to (31) and (56)1198641198942 (ℎ 119909) = 0

1198641198943 (ℎ 119909) le

2

radic1 minus 1199092

008119872

2119873minus3 (119873 minus 3)

times [1 +1198711119873

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)]

(70)

Since from the hypothesis in Lemma 5

1 minus 1199092ge 1 minus (1 minus 120575

119873)2

1

radic1 minus 1199092le

1

radic1 minus (1 minus 120575119873)2

leradic2

radic3120575119873

=119873

radic3

(71)

we have

1198641198943 (ℎ 119909) le

009119872

2119873minus3 (119873 minus 3)

times 119873[1 +1198711119873

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)]

(72)

Substituting (69)ndash(72) into (54) yields

1003816100381610038161003816119864119894 (ℎ 119909)1003816100381610038161003816 le

068119872 ln119873120587 sdot 2119873minus3 (119873 minus 3)

times [1 +127

ln119873+211

ln119873(1198711119873+ 008119871

2119873)

+254 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)(1 +

111

ln119873)

+017 (23119873

3minus69119873

2+100119873minus48)

(119873minus1) (119873 minus 2) (119873minus3) (119873minus4)+042119873

ln119873

+042

ln1198731198711119873+

042119873 (7119873minus4)

2 ln119873(119873minus1) (119873 minus 2)]

(73)

Choosing119873 ge 5 the statement ofTheorem 7 follows

8 Abstract and Applied Analysis

Table 2 Results of HSIs for 1199082(119905)

119873 119909 Exact solution Approximate solution Absolute error

2

minus099999 002012457266627356241 00201245726662735624 157 times 10minus21

minus091111 18643200698228626472 18643200698228626470 143 times 10minus19

minus071111 32232147018485970132 32232147018485970129 240 times 10minus19

minus031111 45263577087926478228 45263577087926478225 308 times 10minus19

000000 50000000000000000000 49999999999999999997 300 times 10minus19

031111 51810684429214884711 51810684429214884708 242 times 10minus19

071111 52460548398163357508 52460548398163357508 379 times 10minus20

091111 62854320328769430625 62854320328769430630 299 times 10minus19

099999 44723036596367022628 44723036596367022632 447 times 10minus17

minus1 minus05 0 05

400

300

200

100

1

Singular point values

HSI

val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

40

30

20

10

times10minus17

Abso

lute

erro

r val

ues

(b)

Figure 2 (a) Exact and AQS solutions when 119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5 and (b) absolute error between exact and AQS solution when119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5

5 Numerical Results and Discussion

In this session we evaluate the hypersingular integrals (4) fordifferent choices of ℎ(119905) and compare with the exact solution

Example 1 Consider the HSIs of the form

1198671 (ℎ 119909) =

1

120587radic1 minus 119909

1 + 119909=int

1

minus1

radic1 + 119905

1 minus 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (74)

where ℎ(119905) = 21199052 minus 3119905 + 5 The exact solution is 1198671(ℎ 119909) =

2radic(1 minus 119909)(1 + 119909)(minus1 + 4119909)

The numerical results in Table 1 show that the AQS givesa very good result for the quadratic function ℎ(119905) = 21199052 minus3119905 + 5 for the case 119894 = 1 and 119873 = 2 This is furtherillustrated in Figure 1(a) where the graphs of the exact andAQS solutions are superimposed for the same choices ofsingular point values Details of the accuracy of the proposedAQS is observed from the absolute error in Figure 1(b)

Example 2 Consider the HSIs of the form

1198672 (ℎ 119909) =

1

120587radic1 + 119909

1 minus 119909=int

1

minus1

radic1 minus 119905

1 + 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (75)

where ℎ(119905) = 21199052 minus 3119905 + 5 The exact solution is 1198672(ℎ 119909) =

radic(1 + 119909)(1 minus 119909)(5 minus 4119909)

The numerical results in Table 2 show that the AQS (21)gives a very good result for the quadratic function ℎ(119905) = 21199052minus3119905 + 5 for the case 119894 = 2 and 119873 = 2 Figure 2(a) indicatesthat the exact and AQS solutions are nearly the same Furtherdetails of the accuracy of the absolute error of the proposedAQS are demonstrated in Figure 2(b)

Example 3 Consider the HSIs of the form

1198671 (ℎ 119909) =

1

120587radic1 minus 119909

1 + 119909=int

1

minus1

radic1 + 119905

1 minus 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (76)

Abstract and Applied Analysis 9

Table 3 Results of HSIs for 1199081(119905)

119873 119909 Exact solution Approximate solution Absolute error

10

minus099999 minus25819308036170334212 minus25817083250027430060 223 times 10minus2

minus091111 minus22578118246552894121 minus22578564435672151572 446 times 10minus5

minus077111 minus10634637975870125656 minus10634476360338990858 162 times 10minus5

minus033111 minus02924243743021256602 minus02924196578457308097 472 times 10minus6

000000 minus01443375672974064411 minus01443397190188599320 215 times 10minus6

033111 minus00753154744680519215 minus00753153016422631300 173 times 10minus7

077111 minus00270285335549129334 minus00270280566695517436 477 times 10minus7

091111 minus00146928400579184367 minus00146941562302624467 132 times 10minus6

099999 minus00001434451425475821 minus00001431650832316927 280 times 10minus7

20

minus099999 minus25819308036170334212 minus25819308021411085378 148 times 10minus7

minus091111 minus22578118246552894121 minus22578118246107028644 446 times 10minus11

minus077111 minus10634637975870125656 minus10634637975433417458 437 times 10minus11

minus033111 minus02924243743021256602 minus02924243742855352783 166 times 10minus11

000000 minus01443375672974064411 minus01443375672897319217 768 times 10minus12

033111 minus00753154744680519215 minus00753154744695265617 148 times 10minus12

077111 minus00270285335549129334 minus00270285335485592653 635 times 10minus12

091111 minus00146928400579184367 minus00146928400450019102 129 times 10minus11

099999 minus00001434451425475821 minus00001434451390347442 351 times 10minus12

40

minus099999 minus25819308036170334212 minus25819308036170338346 404 times 10minus14

minus091111 minus22578118246552894121 minus22578118246552894116 450 times 10minus19

minus077111 minus10634637975870125656 minus10634637975870125637 185 times 10minus18

minus033111 minus02924243743021256602 minus02924243743021256612 967 times 10minus19

000000 minus01443375672974064411 minus01443375672974064412 113 times 10minus19

033111 minus00753154744680519215 minus00753154744680519211 400 times 10minus19

077111 minus00270285335549129334 minus00270285335549129336 172 times 10minus19

091111 minus00146928400579184367 minus00146928400579184373 573 times 10minus19

099999 minus00001434451425475821 minus00001434451425475826 486 times 10minus19

minus1 minus05 0 05 1

Singular point valuesminus50

minus100

minus150

minus200

minus250

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

0020

0015

0010

0005

Abso

lute

erro

r val

ues

(b)

Figure 3 (a) Exact and AQS solutions when 119873 = 10 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 10 and ℎ(119905) = 1(2 + 119905)

10 Abstract and Applied Analysis

minus1 minus05 0 05 1

Singular point valuesminus50

minus100

minus150

minus200

minus250

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

14

12

10

80

60

40

20

times10minus8

(b)

Figure 4 (a) Exact and AQS solutions when 119873 = 20 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 20 and ℎ(119905) = 1(2 + 119905)

minus1 minus05 0 05 1

Singular point values

minus50

minus100

minus150

minus250

HSI

s val

ues

minus200

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

40

30

20

10

times10minus14

(b)

Figure 5 (a) Exact and AQS solutions when 119873 = 40 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 40 and ℎ(119905) = 1(2 + 119905)

where ℎ(119905) = 1(2 + 119905) The exact solution is 1198671(ℎ 119909) =

minusradic(1 minus 119909)(1 + 119909)(radic33(2 + 119909)2)

The numerical results in Table 3 show that the AQS (21)is highly accurate for the rational function ℎ(119905) = 1(2 + 119905)for the case 119894 = 1 and119873 = 40 Figure 3(a) reveals that there jsgood agreement between the exact solution andAQS solutionwhich can be seen from absolute error in Figure 3(b) when119873 = 10 The absolute error is improved in Figure 4(b) when

119873 = 20 with the corresponding exact and AQS solution inFigure 4(a) Best result is obtained in Figures 5(a) and 5(b)when119873 = 40

Example 4 Consider the HSIs of the form

1198672 (ℎ 119909) =

1

120587radic1 + 119909

1 minus 119909=int

1

minus1

radic1 minus 119905

1 + 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (77)

Abstract and Applied Analysis 11

Table 4 Results of HSIs for 1199082(119905)

119873 119909 Exact solution Approximate solution Absolute error

10

minus099999 000387291557000340015 000387141593292902562 150 times 10minus6

minus091111 031504763162812503076 031505448317559772604 685 times 10minus6

minus077111 041231131092313516294 041230879229200872013 252 times 10minus6

minus033111 044083450592426358350 044083359884530457966 907 times 10minus7

000000 043301270189221932338 043301675286046974386 405 times 10minus6

033111 044963976686376800332 044963389568653105901 587 times 10minus6

077111 062742591722366873321 062741343576104182143 125 times 10minus5

091111 094767578680690190767 094770492350476572039 291 times 10minus5

099999 86066655193121619385 86053938870581623578 0127 times 10minus2

20

minus099999 000387291557000340015 000387291555159212453 184 times 10minus11

minus091111 031504763162812503076 031504763156789346359 602 times 10minus11

minus077111 041231131092313516294 041231131089736085386 258 times 10minus11

minus033111 044083450592426358350 044083450592691374976 265 times 10minus12

000000 043301270189221932338 043301270187830840205 139 times 10minus11

033111 044963976686376800332 044963976684337128944 204 times 10minus11

077111 062742591722366873321 062742591718912403402 346 times 10minus11

091111 094767578680690190767 094767578677558019222 313 times 10minus11

099999 86066655193121619385 86066655108152556362 850 times 10minus8

40

minus099999 000387291557000340015 000387291557000339911 104 times 10minus18

minus091111 031504763162812503076 031504763162812503215 140 times 10minus18

minus077111 041231131092313516294 041231131092313516515 228 times 10minus18

minus033111 044083450592426358350 044083450592426358289 614 times 10minus19

000000 043301270189221932338 043301270189221932269 675 times 10minus19

033111 044963976686376800332 044963976686376800195 131 times 10minus18

077111 062742591722366873321 062742591722366873124 191 times 10minus18

091111 094767578680690190767 094767578680690190858 131 times 10minus18

099999 86066655193121619385 86066655193121618680 705 times 10minus16

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

0012

0010

0008

0006

0004

0002

Abso

lute

erro

r val

ues

(b)

Figure 6 (a) Exact and AQS solutions when 119873 = 10 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 10 and ℎ(119905) = 1(2 + 119905)

12 Abstract and Applied Analysis

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

times10minus8

Abso

lute

erro

r val

ues

(b)

Figure 7 (a) Exact and AQS solutions when 119873 = 20 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 20 and ℎ(119905) = 1(2 + 119905)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

Exact solution plotAQS solution plot

80

70

60

50

40

30

20

10

(a)

minus1 minus05 0 05 1

Singular point values

70

60

50

40

30

20

10

times10minus16

Abso

lute

erro

r val

ues

(b)

Figure 8 (a) Exact and AQS solutions when 119873 = 40 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 40 and ℎ(119905) = 1(2 + 119905)

where ℎ(119905) = 1(2 + 119905) The exact solution is 1198672(ℎ 119909) =

radic(1 + 119909)(1 minus 119909)(radic3(2 + 119909)2)

The numerical results in Table 4 show that the AQS (21)is highly accurate for the rational function ℎ(119905) = 1(2 + 119905)for the case 119894 = 2 and119873 = 40 Figure 6(a) shows that there is

good agreement between the exact solution andAQS solutionwhich can be seen from absolute error in Figure 6(b) when119873 = 10 The absolute error is reduced in Figure 7(b) when119873 = 20 with the corresponding exact and AQS solutionin Figure 7(a) The best result is obtained and illustrated inFigures 8(a) and 8(b) when119873 = 40

Abstract and Applied Analysis 13

6 Conclusion

In this paper we have constructed the automatic quadratureschemes for the semibounded weighted Hadamard typehypersingular integrals using the truncated series of the thirdand fourth kinds of Chebyshev polynomials The unknowncoefficients are found by applying the discrete orthogonalitycondition and the choice of collocation points on (minus1 1)The error estimations for the semibounded solutions of thehypersingular integrals are obtained Numerical exampleshave shown the accuracy of the developed methods andagreed with theoretical findings

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Hadamard Le Probleme de Cauchy et les Equations auxDerivees Particelles Lineaires Hyperboliques Hermann ParisFrance 1932

[2] E Lutz L Gray and A Ingraffea ldquoAn overview of integrationmethods for hypersingular integralsrdquo in Boundary Elements CA Brebbia and G S Gipson Eds vol 8 pp 913ndash925 CMPSouthampton UK 1991

[3] S G Samko Hypersingular Integrals and Their ApplicationsTaylor amp Francis London UK 2002

[4] I V Boikov ldquoNumerical methods of computation of singularand hypersingular integralsrdquo International Journal of Mathe-matics and Mathematical Sciences vol 28 no 3 pp 127ndash1792001

[5] I K Lifanov L N Poltavskii and G M Vainikko Hypersin-gular Integral Equations and Their Applications Chapman ampHallCRC A CRC Press Company London UK 2004

[6] H R Kutt ldquoQuadrature formulae for finite part integralsrdquoReport WISK 178 The National Research Institute for Mathe-matical Sciences Pretoria South Africa 1975

[7] C-Y Hui and D Shia ldquoEvaluations of hypersingular integralsusingGaussian quadraturerdquo International Journal for NumericalMethods in Engineering vol 44 no 2 pp 205ndash214 1999

[8] Y Z Chen ldquoNumerical solution of a curved crack problem byusing hypersingular integral equation approachrdquo EngineeringFracture Mechanics vol 46 no 2 pp 275ndash283 1993

[9] N M A Nik Long and Z K Eshkuvatov ldquoHypersingularintegral equation for multiple curved cracks problem in planeelasticityrdquo International Journal of Solids and Structures vol 46no 13 pp 2611ndash2617 2009

[10] S J Obaiys Z K Eshkuvatov and N M Nik Long ldquoOn errorestimation of automatic quadrature scheme for the evaluationof Hadamard integral of second order singularityrdquo University ofBucharest Scientific Bulletin Series A Applied Mathematics andPhysics vol 75 no 1 pp 85ndash98 2013

[11] J CMason andDCHandscombChebyshev Polynomials CRCPress 2003

[12] M I Israilov Semi-Analytic Solution of Singular Integral Equa-tions of First Kind by Quadrature Formula Numerical Integra-tion and Related Problems FAN Press 1990 (Russian)

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Stochastic AnalysisInternational Journal of

Page 5: An Automatic Quadrature Schemes and Error Estimates for Semibounded Weighted Hadamard Type Hypersingular Integrals

Abstract and Applied Analysis 5

These recurrence relations lead to

11987132(119873)

=119873

5(119873 + 1) (119873 + 2) (2119873 + 1) (119873 minus 1) (47)

In a similar way we obtain

11987142(119873)

=119873

5(119873 + 1) (119873 + 2) (2119873 + 1) (119873 minus 1) (48)

Main results are given in the followingTheorem 7

Theorem 7 Let ℎ(119905) isin 119862119873120572[minus1 1] for 0 lt 120572 le 1 and let theseries of Chebyshev polynomials 119878

119894119873(119905) be defined by (20) then

the AQS (21) has an error bound

1003817100381710038171003817119864119894 (ℎ)1003817100381710038171003817119862 le

011119872

2119873minus4 (119873 minus 4)[1+238

1198731198713119873+05 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)]

119894 = 1 2

(49)

where119872 is defined by (32) and

1198713119873= 1 +

253

ln119873[1198711119873+ 007119871

2119873

+042 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)(1 +

111

ln119873)

+003 (23119873

3minus 69119873

2+ 100119873 minus 48)

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

1198711119873= 1 +

3

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)+

3 (7119873 minus 4)

2119873 (119873 minus 1) (119873 minus 2)

1198712119873= 1 +

3

119873+231198733minus 69119873

2+ 100119873 minus 48

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)(1 +

3

119873)

(50)

Proof of Theorem 7 In view of error norms (27) we cancompute the error bound of AQS (21) for 119894 = 1 2 as follows

119864119894 (ℎ 119909) = 119867119894 (ℎ 119909) minus 119867119894119873 (ℎ 119909)

=119908119894 (119909)

120587=int

1

minus1

119890119894119873 (119905) 119889119905

119908119894 (119905) (119905 minus 119909)

2

=119908119894 (119909)

120587=int

1

minus1

[119890119894119873 (119905) minus 119890119894119873 (119909) minus 119890

1015840

119894119873(119909) (119905 minus 119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

2

+119908119894 (119909)

120587=int

1

minus1

[119890119894119873 (119909) + 119890

1015840

119894119873(119909) (119905 minus 119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

2

(51)

Taylor series expansion of 119890119894119873(119905) around 119909 is given by

119890119894119873 (119905) = 119890119894119873 (119909) + 119890

1015840

119894119873(120577119905) (119905 minus 119909) (52)

where

120577119905isin (119909 119905) if 119905 isin (119909 1) (119905 119909) if 119905 isin (minus1 119909)

(53)

Due to (51)-(52) we have1003816100381610038161003816119864119894 (ℎ 119909)

1003816100381610038161003816 le 1198641198941 (ℎ 119909) + 1198641198942 (ℎ 119909) + 1198641198943 (ℎ 119909) (54)

where

1198641198941 (ℎ 119909) =

10038161003816100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909)

120587minusint

1

minus1

[1198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

10038161003816100381610038161003816100381610038161003816100381610038161003816

1198641198942 (ℎ 119909) =

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909)

120587119890119894119873 (119909) =int

1

minus1

119889119905

119908119894 (119905) (119905 minus 119909)

2

100381610038161003816100381610038161003816100381610038161003816

1198641198943 (ℎ 119909) =

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909)

1205871198901015840

119894119873(119909) minusint

1

minus1

119889119905

119908119894 (119905) (119905 minus 119909)

100381610038161003816100381610038161003816100381610038161003816

(55)

It is easy to check that

=int

1

minus1

radic1 + 119909

1 minus 119909

119889119905

(119905 minus 119909)2= =int

1

minus1

radic1 minus 119905

1 + 119905

119889119905

(119905 minus 119909)2= 0

minusint

1

minus1

radic1 + 119909

1 minus 119909

119889119905

119905 minus 119909= minusminusint

1

minus1

radic1 minus 119905

1 + 119905

119889119905

119905 minus 119909= 120587

(56)

For the estimation of 1198641198941(ℎ 119909) we divide the interval

(minus1 1) into three subintervals (minus1 119909 minus 120575119873) (119909 + 120575

119873 1) and

[119909 minus 120575119873 119909 + 120575

119873]

1198641198941 (ℎ 119909) le

1

120587[1198651198941 (119909) + 1198651198942 (119909) + 1198651198943 (119909)] (57)

where

1198651198941 (119909) =

10038161003816100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

119909minus120575119873

minus1

[1198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

10038161003816100381610038161003816100381610038161003816100381610038161003816

1198651198942 (119909) =

10038161003816100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

1

119909+120575119873

[1198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

10038161003816100381610038161003816100381610038161003816100381610038161003816

1198651198943 (119909) =

10038161003816100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

119909+120575119873

119909minus120575119873

[1198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)] 119889119905

119908119894 (119905) (119905 minus 119909)

10038161003816100381610038161003816100381610038161003816100381610038161003816

(58)

For 1198651198941(119909) according to Lemma 5 we obtain

1198651198941 (119909) le 2

100381710038171003817100381710038171198901015840

119894119873

10038171003817100381710038171003817

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

119909minus120575119873

minus1

119889119905

119908119894 (119905) (119905 minus 119909)

100381610038161003816100381610038161003816100381610038161003816

le 2100381710038171003817100381710038171198901015840

119894119873

10038171003817100381710038171003817[ln 2120575119873

+120587

radic2]

(59)

In the similar way for 1198651198942(119909) we have

1198651198942 (119909) le 2

100381710038171003817100381710038171198901015840

119894119873

10038171003817100381710038171003817

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

1

119909+120575119873

119889119905

119908119894 (119905) (119905 minus 119909)

100381610038161003816100381610038161003816100381610038161003816

le 2100381710038171003817100381710038171198901015840

119894119873

10038171003817100381710038171003817[ln 2120575119873

+120587

radic2]

(60)

6 Abstract and Applied Analysis

Utilizing (30) and (31) into (59) and (60) we arrive at

1198651198941 (119909) = 1198651198942 (119909)

le(16)119872

2119873minus3 (119873 minus 3)[1 +

1198711119873

119873+

7119873 minus 4

(119873 minus 1) (119873 minus 2)]

times [ln 2120575119873

+120587

radic2]

(61)

For the error of 1198651198943(119909)

1198651198943 (119909) le 119908119894 (119909) int

119909+120590119873

119909minus120590119873

100381610038161003816100381610038161198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)10038161003816100381610038161003816

119908119894 (119905)1003816100381610038161003816120577119905 minus 119909

1003816100381610038161003816

119889119905 (62)

It follows from (29) and Lemma 6 that100381610038161003816100381610038161198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)10038161003816100381610038161003816

le1

2119873119873

times 1003816100381610038161003816119875119894119873 (120577119905) minus 119875119894119873 (119909)

1003816100381610038161003816

10038161003816100381610038161003816100381610038161003816

119889

119889119909ℎ(119873)(120577119905)

10038161003816100381610038161003816100381610038161003816

+10038161003816100381610038161003816ℎ(119873)(120577119905)10038161003816100381610038161003816

100381610038161003816100381610038161198751015840

119894119873(120577119905) minus 1198751015840

119894119873(119909)10038161003816100381610038161003816

le1

2119873119873

times 119871119895119894119873

1003816100381610038161003816120577119905 minus 1199091003816100381610038161003816

10038161003816100381610038161003816100381610038161003816

119889

119889119909ℎ(119873)(120577119905)

10038161003816100381610038161003816100381610038161003816+ 119871119895119894119873

10038161003816100381610038161003816ℎ(119873)(120577119905)10038161003816100381610038161003816

1003816100381610038161003816120577119905 minus 1199091003816100381610038161003816

le119872

480 sdot 2119873minus5 (119873 minus 5)

1198732(119873 + 1) (2119873 + 1)

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)

times [1 +1

119873+3

1198732]1003816100381610038161003816120577119905 minus 119909

1003816100381610038161003816

le119872

240 sdot 2119873minus5 (119873 minus 5)

times [1 +231198733minus 69119873

2+ 100119873 minus 48

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

times [1 +1

119873+3

1198732]1003816100381610038161003816120577119905 minus 119909

1003816100381610038161003816

le119872

240 sdot 2119873minus5 (119873 minus 5)

times [1 +1198712119873

119873+

231198733minus 69119873

2+ 100119873 minus 48

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

times1003816100381610038161003816120577119905 minus 119909

1003816100381610038161003816

(63)

where

1198712119873= 1 +

3

119873+231198733minus 69119873

2+ 100119873 minus 48

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)(1 +

3

119873)

(64)

Substituting (63) into (62) yields

1198651198943 (119909) le

119872

240 sdot 2119873minus5 (119873 minus 5)

times [1 +1198712119873

119873+

231198733minus 69119873

2+ 100119873 minus 48

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

times

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

119909+120575119873

119909minus120575119873

119889119905

119908119894 (119905)

100381610038161003816100381610038161003816100381610038161003816

(65)

Using Lemma 5 gives

1198651198943 (119909) le

337119872

240 sdot 2119873minus5 (119873 minus 5)

times [1+1198712119873

119873+231198733minus 69119873

2+ 100119873 minus 48

2 (119873minus1) (119873minus2) (119873minus3) (119873minus4)] 120575119873

(66)

By choosing 120575119873= 2119873

2 for119873 ge 5 we obtain

1198651198943 (119909) le

011119872

2119873minus3 (119873 minus 3)

times [1 +1198712119873

119873+

231198733minus 69119873

2+ 100119873 minus 48

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

(67)

By substituting (61) and (67) into (57) we arrive at

1198641198941 (ℎ 119909) le

064119872ln119873

120587 sdot 2119873minus3 (119873 minus 3)

times [1 +127

ln119873+211

ln119873(1198711119873+ 008119871

2119873)

+254 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)(1 +

111

ln119873)

+017 (23119873

3minus 69119873

2+ 100119873 minus 48)

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

(68)

1198641198941 (ℎ 119909) le

246119872ln119873

2119873minus3 (119873 minus 3)

times [1 +010

ln119873+1198711119873

119873+010119871

1119873

119873 ln119873

+2316 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)+005 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)

+1059 (23119873

3minus 69119873

2+ 100119873 minus 48)

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

(69)

Abstract and Applied Analysis 7

Table 1 Results of HSIs for 1199081(119905)

119873 119909 Exact solution Approximate solution Absolute error

2

minus099999 minus22360444988237596342 minus22360444988237596345 313 times 10minus16

minus091111 minus21535241487484869307 minus21535241487484869310 308 times 10minus18

minus071111 minus93563357403104129848 minus93563357403104129863 142 times 10minus18

minus031111 minus30963683860810919235 minus30963683860810919240 586 times 10minus19

000000 minus10000000000000000000 minus10000000000000000003 300 times 10minus19

031111 017718528456311131724 017718528456311131709 127 times 10minus19

071111 075786494952828070264 075786494952828070264 639 times 10minus21

091111 057031832778553277005 057031832778553277007 140 times 10minus20

099999 000670813126012938238 000670813126012938238 224 times 10minus22

minus1 minus05 0 05 1

Singular point values

minus500

minus1000

minus1500

minus2000

HSI

val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

30

20

10

times10minus16

Abso

lute

erro

r val

ues

(b)Figure 1 (a) Exact and AQS solutions when 119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5 and (b) absolute error between exact and AQS solution when119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5

Due to (31) and (56)1198641198942 (ℎ 119909) = 0

1198641198943 (ℎ 119909) le

2

radic1 minus 1199092

008119872

2119873minus3 (119873 minus 3)

times [1 +1198711119873

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)]

(70)

Since from the hypothesis in Lemma 5

1 minus 1199092ge 1 minus (1 minus 120575

119873)2

1

radic1 minus 1199092le

1

radic1 minus (1 minus 120575119873)2

leradic2

radic3120575119873

=119873

radic3

(71)

we have

1198641198943 (ℎ 119909) le

009119872

2119873minus3 (119873 minus 3)

times 119873[1 +1198711119873

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)]

(72)

Substituting (69)ndash(72) into (54) yields

1003816100381610038161003816119864119894 (ℎ 119909)1003816100381610038161003816 le

068119872 ln119873120587 sdot 2119873minus3 (119873 minus 3)

times [1 +127

ln119873+211

ln119873(1198711119873+ 008119871

2119873)

+254 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)(1 +

111

ln119873)

+017 (23119873

3minus69119873

2+100119873minus48)

(119873minus1) (119873 minus 2) (119873minus3) (119873minus4)+042119873

ln119873

+042

ln1198731198711119873+

042119873 (7119873minus4)

2 ln119873(119873minus1) (119873 minus 2)]

(73)

Choosing119873 ge 5 the statement ofTheorem 7 follows

8 Abstract and Applied Analysis

Table 2 Results of HSIs for 1199082(119905)

119873 119909 Exact solution Approximate solution Absolute error

2

minus099999 002012457266627356241 00201245726662735624 157 times 10minus21

minus091111 18643200698228626472 18643200698228626470 143 times 10minus19

minus071111 32232147018485970132 32232147018485970129 240 times 10minus19

minus031111 45263577087926478228 45263577087926478225 308 times 10minus19

000000 50000000000000000000 49999999999999999997 300 times 10minus19

031111 51810684429214884711 51810684429214884708 242 times 10minus19

071111 52460548398163357508 52460548398163357508 379 times 10minus20

091111 62854320328769430625 62854320328769430630 299 times 10minus19

099999 44723036596367022628 44723036596367022632 447 times 10minus17

minus1 minus05 0 05

400

300

200

100

1

Singular point values

HSI

val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

40

30

20

10

times10minus17

Abso

lute

erro

r val

ues

(b)

Figure 2 (a) Exact and AQS solutions when 119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5 and (b) absolute error between exact and AQS solution when119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5

5 Numerical Results and Discussion

In this session we evaluate the hypersingular integrals (4) fordifferent choices of ℎ(119905) and compare with the exact solution

Example 1 Consider the HSIs of the form

1198671 (ℎ 119909) =

1

120587radic1 minus 119909

1 + 119909=int

1

minus1

radic1 + 119905

1 minus 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (74)

where ℎ(119905) = 21199052 minus 3119905 + 5 The exact solution is 1198671(ℎ 119909) =

2radic(1 minus 119909)(1 + 119909)(minus1 + 4119909)

The numerical results in Table 1 show that the AQS givesa very good result for the quadratic function ℎ(119905) = 21199052 minus3119905 + 5 for the case 119894 = 1 and 119873 = 2 This is furtherillustrated in Figure 1(a) where the graphs of the exact andAQS solutions are superimposed for the same choices ofsingular point values Details of the accuracy of the proposedAQS is observed from the absolute error in Figure 1(b)

Example 2 Consider the HSIs of the form

1198672 (ℎ 119909) =

1

120587radic1 + 119909

1 minus 119909=int

1

minus1

radic1 minus 119905

1 + 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (75)

where ℎ(119905) = 21199052 minus 3119905 + 5 The exact solution is 1198672(ℎ 119909) =

radic(1 + 119909)(1 minus 119909)(5 minus 4119909)

The numerical results in Table 2 show that the AQS (21)gives a very good result for the quadratic function ℎ(119905) = 21199052minus3119905 + 5 for the case 119894 = 2 and 119873 = 2 Figure 2(a) indicatesthat the exact and AQS solutions are nearly the same Furtherdetails of the accuracy of the absolute error of the proposedAQS are demonstrated in Figure 2(b)

Example 3 Consider the HSIs of the form

1198671 (ℎ 119909) =

1

120587radic1 minus 119909

1 + 119909=int

1

minus1

radic1 + 119905

1 minus 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (76)

Abstract and Applied Analysis 9

Table 3 Results of HSIs for 1199081(119905)

119873 119909 Exact solution Approximate solution Absolute error

10

minus099999 minus25819308036170334212 minus25817083250027430060 223 times 10minus2

minus091111 minus22578118246552894121 minus22578564435672151572 446 times 10minus5

minus077111 minus10634637975870125656 minus10634476360338990858 162 times 10minus5

minus033111 minus02924243743021256602 minus02924196578457308097 472 times 10minus6

000000 minus01443375672974064411 minus01443397190188599320 215 times 10minus6

033111 minus00753154744680519215 minus00753153016422631300 173 times 10minus7

077111 minus00270285335549129334 minus00270280566695517436 477 times 10minus7

091111 minus00146928400579184367 minus00146941562302624467 132 times 10minus6

099999 minus00001434451425475821 minus00001431650832316927 280 times 10minus7

20

minus099999 minus25819308036170334212 minus25819308021411085378 148 times 10minus7

minus091111 minus22578118246552894121 minus22578118246107028644 446 times 10minus11

minus077111 minus10634637975870125656 minus10634637975433417458 437 times 10minus11

minus033111 minus02924243743021256602 minus02924243742855352783 166 times 10minus11

000000 minus01443375672974064411 minus01443375672897319217 768 times 10minus12

033111 minus00753154744680519215 minus00753154744695265617 148 times 10minus12

077111 minus00270285335549129334 minus00270285335485592653 635 times 10minus12

091111 minus00146928400579184367 minus00146928400450019102 129 times 10minus11

099999 minus00001434451425475821 minus00001434451390347442 351 times 10minus12

40

minus099999 minus25819308036170334212 minus25819308036170338346 404 times 10minus14

minus091111 minus22578118246552894121 minus22578118246552894116 450 times 10minus19

minus077111 minus10634637975870125656 minus10634637975870125637 185 times 10minus18

minus033111 minus02924243743021256602 minus02924243743021256612 967 times 10minus19

000000 minus01443375672974064411 minus01443375672974064412 113 times 10minus19

033111 minus00753154744680519215 minus00753154744680519211 400 times 10minus19

077111 minus00270285335549129334 minus00270285335549129336 172 times 10minus19

091111 minus00146928400579184367 minus00146928400579184373 573 times 10minus19

099999 minus00001434451425475821 minus00001434451425475826 486 times 10minus19

minus1 minus05 0 05 1

Singular point valuesminus50

minus100

minus150

minus200

minus250

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

0020

0015

0010

0005

Abso

lute

erro

r val

ues

(b)

Figure 3 (a) Exact and AQS solutions when 119873 = 10 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 10 and ℎ(119905) = 1(2 + 119905)

10 Abstract and Applied Analysis

minus1 minus05 0 05 1

Singular point valuesminus50

minus100

minus150

minus200

minus250

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

14

12

10

80

60

40

20

times10minus8

(b)

Figure 4 (a) Exact and AQS solutions when 119873 = 20 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 20 and ℎ(119905) = 1(2 + 119905)

minus1 minus05 0 05 1

Singular point values

minus50

minus100

minus150

minus250

HSI

s val

ues

minus200

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

40

30

20

10

times10minus14

(b)

Figure 5 (a) Exact and AQS solutions when 119873 = 40 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 40 and ℎ(119905) = 1(2 + 119905)

where ℎ(119905) = 1(2 + 119905) The exact solution is 1198671(ℎ 119909) =

minusradic(1 minus 119909)(1 + 119909)(radic33(2 + 119909)2)

The numerical results in Table 3 show that the AQS (21)is highly accurate for the rational function ℎ(119905) = 1(2 + 119905)for the case 119894 = 1 and119873 = 40 Figure 3(a) reveals that there jsgood agreement between the exact solution andAQS solutionwhich can be seen from absolute error in Figure 3(b) when119873 = 10 The absolute error is improved in Figure 4(b) when

119873 = 20 with the corresponding exact and AQS solution inFigure 4(a) Best result is obtained in Figures 5(a) and 5(b)when119873 = 40

Example 4 Consider the HSIs of the form

1198672 (ℎ 119909) =

1

120587radic1 + 119909

1 minus 119909=int

1

minus1

radic1 minus 119905

1 + 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (77)

Abstract and Applied Analysis 11

Table 4 Results of HSIs for 1199082(119905)

119873 119909 Exact solution Approximate solution Absolute error

10

minus099999 000387291557000340015 000387141593292902562 150 times 10minus6

minus091111 031504763162812503076 031505448317559772604 685 times 10minus6

minus077111 041231131092313516294 041230879229200872013 252 times 10minus6

minus033111 044083450592426358350 044083359884530457966 907 times 10minus7

000000 043301270189221932338 043301675286046974386 405 times 10minus6

033111 044963976686376800332 044963389568653105901 587 times 10minus6

077111 062742591722366873321 062741343576104182143 125 times 10minus5

091111 094767578680690190767 094770492350476572039 291 times 10minus5

099999 86066655193121619385 86053938870581623578 0127 times 10minus2

20

minus099999 000387291557000340015 000387291555159212453 184 times 10minus11

minus091111 031504763162812503076 031504763156789346359 602 times 10minus11

minus077111 041231131092313516294 041231131089736085386 258 times 10minus11

minus033111 044083450592426358350 044083450592691374976 265 times 10minus12

000000 043301270189221932338 043301270187830840205 139 times 10minus11

033111 044963976686376800332 044963976684337128944 204 times 10minus11

077111 062742591722366873321 062742591718912403402 346 times 10minus11

091111 094767578680690190767 094767578677558019222 313 times 10minus11

099999 86066655193121619385 86066655108152556362 850 times 10minus8

40

minus099999 000387291557000340015 000387291557000339911 104 times 10minus18

minus091111 031504763162812503076 031504763162812503215 140 times 10minus18

minus077111 041231131092313516294 041231131092313516515 228 times 10minus18

minus033111 044083450592426358350 044083450592426358289 614 times 10minus19

000000 043301270189221932338 043301270189221932269 675 times 10minus19

033111 044963976686376800332 044963976686376800195 131 times 10minus18

077111 062742591722366873321 062742591722366873124 191 times 10minus18

091111 094767578680690190767 094767578680690190858 131 times 10minus18

099999 86066655193121619385 86066655193121618680 705 times 10minus16

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

0012

0010

0008

0006

0004

0002

Abso

lute

erro

r val

ues

(b)

Figure 6 (a) Exact and AQS solutions when 119873 = 10 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 10 and ℎ(119905) = 1(2 + 119905)

12 Abstract and Applied Analysis

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

times10minus8

Abso

lute

erro

r val

ues

(b)

Figure 7 (a) Exact and AQS solutions when 119873 = 20 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 20 and ℎ(119905) = 1(2 + 119905)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

Exact solution plotAQS solution plot

80

70

60

50

40

30

20

10

(a)

minus1 minus05 0 05 1

Singular point values

70

60

50

40

30

20

10

times10minus16

Abso

lute

erro

r val

ues

(b)

Figure 8 (a) Exact and AQS solutions when 119873 = 40 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 40 and ℎ(119905) = 1(2 + 119905)

where ℎ(119905) = 1(2 + 119905) The exact solution is 1198672(ℎ 119909) =

radic(1 + 119909)(1 minus 119909)(radic3(2 + 119909)2)

The numerical results in Table 4 show that the AQS (21)is highly accurate for the rational function ℎ(119905) = 1(2 + 119905)for the case 119894 = 2 and119873 = 40 Figure 6(a) shows that there is

good agreement between the exact solution andAQS solutionwhich can be seen from absolute error in Figure 6(b) when119873 = 10 The absolute error is reduced in Figure 7(b) when119873 = 20 with the corresponding exact and AQS solutionin Figure 7(a) The best result is obtained and illustrated inFigures 8(a) and 8(b) when119873 = 40

Abstract and Applied Analysis 13

6 Conclusion

In this paper we have constructed the automatic quadratureschemes for the semibounded weighted Hadamard typehypersingular integrals using the truncated series of the thirdand fourth kinds of Chebyshev polynomials The unknowncoefficients are found by applying the discrete orthogonalitycondition and the choice of collocation points on (minus1 1)The error estimations for the semibounded solutions of thehypersingular integrals are obtained Numerical exampleshave shown the accuracy of the developed methods andagreed with theoretical findings

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Hadamard Le Probleme de Cauchy et les Equations auxDerivees Particelles Lineaires Hyperboliques Hermann ParisFrance 1932

[2] E Lutz L Gray and A Ingraffea ldquoAn overview of integrationmethods for hypersingular integralsrdquo in Boundary Elements CA Brebbia and G S Gipson Eds vol 8 pp 913ndash925 CMPSouthampton UK 1991

[3] S G Samko Hypersingular Integrals and Their ApplicationsTaylor amp Francis London UK 2002

[4] I V Boikov ldquoNumerical methods of computation of singularand hypersingular integralsrdquo International Journal of Mathe-matics and Mathematical Sciences vol 28 no 3 pp 127ndash1792001

[5] I K Lifanov L N Poltavskii and G M Vainikko Hypersin-gular Integral Equations and Their Applications Chapman ampHallCRC A CRC Press Company London UK 2004

[6] H R Kutt ldquoQuadrature formulae for finite part integralsrdquoReport WISK 178 The National Research Institute for Mathe-matical Sciences Pretoria South Africa 1975

[7] C-Y Hui and D Shia ldquoEvaluations of hypersingular integralsusingGaussian quadraturerdquo International Journal for NumericalMethods in Engineering vol 44 no 2 pp 205ndash214 1999

[8] Y Z Chen ldquoNumerical solution of a curved crack problem byusing hypersingular integral equation approachrdquo EngineeringFracture Mechanics vol 46 no 2 pp 275ndash283 1993

[9] N M A Nik Long and Z K Eshkuvatov ldquoHypersingularintegral equation for multiple curved cracks problem in planeelasticityrdquo International Journal of Solids and Structures vol 46no 13 pp 2611ndash2617 2009

[10] S J Obaiys Z K Eshkuvatov and N M Nik Long ldquoOn errorestimation of automatic quadrature scheme for the evaluationof Hadamard integral of second order singularityrdquo University ofBucharest Scientific Bulletin Series A Applied Mathematics andPhysics vol 75 no 1 pp 85ndash98 2013

[11] J CMason andDCHandscombChebyshev Polynomials CRCPress 2003

[12] M I Israilov Semi-Analytic Solution of Singular Integral Equa-tions of First Kind by Quadrature Formula Numerical Integra-tion and Related Problems FAN Press 1990 (Russian)

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Stochastic AnalysisInternational Journal of

Page 6: An Automatic Quadrature Schemes and Error Estimates for Semibounded Weighted Hadamard Type Hypersingular Integrals

6 Abstract and Applied Analysis

Utilizing (30) and (31) into (59) and (60) we arrive at

1198651198941 (119909) = 1198651198942 (119909)

le(16)119872

2119873minus3 (119873 minus 3)[1 +

1198711119873

119873+

7119873 minus 4

(119873 minus 1) (119873 minus 2)]

times [ln 2120575119873

+120587

radic2]

(61)

For the error of 1198651198943(119909)

1198651198943 (119909) le 119908119894 (119909) int

119909+120590119873

119909minus120590119873

100381610038161003816100381610038161198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)10038161003816100381610038161003816

119908119894 (119905)1003816100381610038161003816120577119905 minus 119909

1003816100381610038161003816

119889119905 (62)

It follows from (29) and Lemma 6 that100381610038161003816100381610038161198901015840

119894119873(120577119905) minus 1198901015840

119894119873(119909)10038161003816100381610038161003816

le1

2119873119873

times 1003816100381610038161003816119875119894119873 (120577119905) minus 119875119894119873 (119909)

1003816100381610038161003816

10038161003816100381610038161003816100381610038161003816

119889

119889119909ℎ(119873)(120577119905)

10038161003816100381610038161003816100381610038161003816

+10038161003816100381610038161003816ℎ(119873)(120577119905)10038161003816100381610038161003816

100381610038161003816100381610038161198751015840

119894119873(120577119905) minus 1198751015840

119894119873(119909)10038161003816100381610038161003816

le1

2119873119873

times 119871119895119894119873

1003816100381610038161003816120577119905 minus 1199091003816100381610038161003816

10038161003816100381610038161003816100381610038161003816

119889

119889119909ℎ(119873)(120577119905)

10038161003816100381610038161003816100381610038161003816+ 119871119895119894119873

10038161003816100381610038161003816ℎ(119873)(120577119905)10038161003816100381610038161003816

1003816100381610038161003816120577119905 minus 1199091003816100381610038161003816

le119872

480 sdot 2119873minus5 (119873 minus 5)

1198732(119873 + 1) (2119873 + 1)

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)

times [1 +1

119873+3

1198732]1003816100381610038161003816120577119905 minus 119909

1003816100381610038161003816

le119872

240 sdot 2119873minus5 (119873 minus 5)

times [1 +231198733minus 69119873

2+ 100119873 minus 48

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

times [1 +1

119873+3

1198732]1003816100381610038161003816120577119905 minus 119909

1003816100381610038161003816

le119872

240 sdot 2119873minus5 (119873 minus 5)

times [1 +1198712119873

119873+

231198733minus 69119873

2+ 100119873 minus 48

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

times1003816100381610038161003816120577119905 minus 119909

1003816100381610038161003816

(63)

where

1198712119873= 1 +

3

119873+231198733minus 69119873

2+ 100119873 minus 48

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)(1 +

3

119873)

(64)

Substituting (63) into (62) yields

1198651198943 (119909) le

119872

240 sdot 2119873minus5 (119873 minus 5)

times [1 +1198712119873

119873+

231198733minus 69119873

2+ 100119873 minus 48

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

times

100381610038161003816100381610038161003816100381610038161003816

119908119894 (119909) int

119909+120575119873

119909minus120575119873

119889119905

119908119894 (119905)

100381610038161003816100381610038161003816100381610038161003816

(65)

Using Lemma 5 gives

1198651198943 (119909) le

337119872

240 sdot 2119873minus5 (119873 minus 5)

times [1+1198712119873

119873+231198733minus 69119873

2+ 100119873 minus 48

2 (119873minus1) (119873minus2) (119873minus3) (119873minus4)] 120575119873

(66)

By choosing 120575119873= 2119873

2 for119873 ge 5 we obtain

1198651198943 (119909) le

011119872

2119873minus3 (119873 minus 3)

times [1 +1198712119873

119873+

231198733minus 69119873

2+ 100119873 minus 48

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

(67)

By substituting (61) and (67) into (57) we arrive at

1198641198941 (ℎ 119909) le

064119872ln119873

120587 sdot 2119873minus3 (119873 minus 3)

times [1 +127

ln119873+211

ln119873(1198711119873+ 008119871

2119873)

+254 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)(1 +

111

ln119873)

+017 (23119873

3minus 69119873

2+ 100119873 minus 48)

(119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

(68)

1198641198941 (ℎ 119909) le

246119872ln119873

2119873minus3 (119873 minus 3)

times [1 +010

ln119873+1198711119873

119873+010119871

1119873

119873 ln119873

+2316 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)+005 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)

+1059 (23119873

3minus 69119873

2+ 100119873 minus 48)

2 (119873 minus 1) (119873 minus 2) (119873 minus 3) (119873 minus 4)]

(69)

Abstract and Applied Analysis 7

Table 1 Results of HSIs for 1199081(119905)

119873 119909 Exact solution Approximate solution Absolute error

2

minus099999 minus22360444988237596342 minus22360444988237596345 313 times 10minus16

minus091111 minus21535241487484869307 minus21535241487484869310 308 times 10minus18

minus071111 minus93563357403104129848 minus93563357403104129863 142 times 10minus18

minus031111 minus30963683860810919235 minus30963683860810919240 586 times 10minus19

000000 minus10000000000000000000 minus10000000000000000003 300 times 10minus19

031111 017718528456311131724 017718528456311131709 127 times 10minus19

071111 075786494952828070264 075786494952828070264 639 times 10minus21

091111 057031832778553277005 057031832778553277007 140 times 10minus20

099999 000670813126012938238 000670813126012938238 224 times 10minus22

minus1 minus05 0 05 1

Singular point values

minus500

minus1000

minus1500

minus2000

HSI

val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

30

20

10

times10minus16

Abso

lute

erro

r val

ues

(b)Figure 1 (a) Exact and AQS solutions when 119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5 and (b) absolute error between exact and AQS solution when119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5

Due to (31) and (56)1198641198942 (ℎ 119909) = 0

1198641198943 (ℎ 119909) le

2

radic1 minus 1199092

008119872

2119873minus3 (119873 minus 3)

times [1 +1198711119873

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)]

(70)

Since from the hypothesis in Lemma 5

1 minus 1199092ge 1 minus (1 minus 120575

119873)2

1

radic1 minus 1199092le

1

radic1 minus (1 minus 120575119873)2

leradic2

radic3120575119873

=119873

radic3

(71)

we have

1198641198943 (ℎ 119909) le

009119872

2119873minus3 (119873 minus 3)

times 119873[1 +1198711119873

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)]

(72)

Substituting (69)ndash(72) into (54) yields

1003816100381610038161003816119864119894 (ℎ 119909)1003816100381610038161003816 le

068119872 ln119873120587 sdot 2119873minus3 (119873 minus 3)

times [1 +127

ln119873+211

ln119873(1198711119873+ 008119871

2119873)

+254 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)(1 +

111

ln119873)

+017 (23119873

3minus69119873

2+100119873minus48)

(119873minus1) (119873 minus 2) (119873minus3) (119873minus4)+042119873

ln119873

+042

ln1198731198711119873+

042119873 (7119873minus4)

2 ln119873(119873minus1) (119873 minus 2)]

(73)

Choosing119873 ge 5 the statement ofTheorem 7 follows

8 Abstract and Applied Analysis

Table 2 Results of HSIs for 1199082(119905)

119873 119909 Exact solution Approximate solution Absolute error

2

minus099999 002012457266627356241 00201245726662735624 157 times 10minus21

minus091111 18643200698228626472 18643200698228626470 143 times 10minus19

minus071111 32232147018485970132 32232147018485970129 240 times 10minus19

minus031111 45263577087926478228 45263577087926478225 308 times 10minus19

000000 50000000000000000000 49999999999999999997 300 times 10minus19

031111 51810684429214884711 51810684429214884708 242 times 10minus19

071111 52460548398163357508 52460548398163357508 379 times 10minus20

091111 62854320328769430625 62854320328769430630 299 times 10minus19

099999 44723036596367022628 44723036596367022632 447 times 10minus17

minus1 minus05 0 05

400

300

200

100

1

Singular point values

HSI

val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

40

30

20

10

times10minus17

Abso

lute

erro

r val

ues

(b)

Figure 2 (a) Exact and AQS solutions when 119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5 and (b) absolute error between exact and AQS solution when119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5

5 Numerical Results and Discussion

In this session we evaluate the hypersingular integrals (4) fordifferent choices of ℎ(119905) and compare with the exact solution

Example 1 Consider the HSIs of the form

1198671 (ℎ 119909) =

1

120587radic1 minus 119909

1 + 119909=int

1

minus1

radic1 + 119905

1 minus 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (74)

where ℎ(119905) = 21199052 minus 3119905 + 5 The exact solution is 1198671(ℎ 119909) =

2radic(1 minus 119909)(1 + 119909)(minus1 + 4119909)

The numerical results in Table 1 show that the AQS givesa very good result for the quadratic function ℎ(119905) = 21199052 minus3119905 + 5 for the case 119894 = 1 and 119873 = 2 This is furtherillustrated in Figure 1(a) where the graphs of the exact andAQS solutions are superimposed for the same choices ofsingular point values Details of the accuracy of the proposedAQS is observed from the absolute error in Figure 1(b)

Example 2 Consider the HSIs of the form

1198672 (ℎ 119909) =

1

120587radic1 + 119909

1 minus 119909=int

1

minus1

radic1 minus 119905

1 + 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (75)

where ℎ(119905) = 21199052 minus 3119905 + 5 The exact solution is 1198672(ℎ 119909) =

radic(1 + 119909)(1 minus 119909)(5 minus 4119909)

The numerical results in Table 2 show that the AQS (21)gives a very good result for the quadratic function ℎ(119905) = 21199052minus3119905 + 5 for the case 119894 = 2 and 119873 = 2 Figure 2(a) indicatesthat the exact and AQS solutions are nearly the same Furtherdetails of the accuracy of the absolute error of the proposedAQS are demonstrated in Figure 2(b)

Example 3 Consider the HSIs of the form

1198671 (ℎ 119909) =

1

120587radic1 minus 119909

1 + 119909=int

1

minus1

radic1 + 119905

1 minus 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (76)

Abstract and Applied Analysis 9

Table 3 Results of HSIs for 1199081(119905)

119873 119909 Exact solution Approximate solution Absolute error

10

minus099999 minus25819308036170334212 minus25817083250027430060 223 times 10minus2

minus091111 minus22578118246552894121 minus22578564435672151572 446 times 10minus5

minus077111 minus10634637975870125656 minus10634476360338990858 162 times 10minus5

minus033111 minus02924243743021256602 minus02924196578457308097 472 times 10minus6

000000 minus01443375672974064411 minus01443397190188599320 215 times 10minus6

033111 minus00753154744680519215 minus00753153016422631300 173 times 10minus7

077111 minus00270285335549129334 minus00270280566695517436 477 times 10minus7

091111 minus00146928400579184367 minus00146941562302624467 132 times 10minus6

099999 minus00001434451425475821 minus00001431650832316927 280 times 10minus7

20

minus099999 minus25819308036170334212 minus25819308021411085378 148 times 10minus7

minus091111 minus22578118246552894121 minus22578118246107028644 446 times 10minus11

minus077111 minus10634637975870125656 minus10634637975433417458 437 times 10minus11

minus033111 minus02924243743021256602 minus02924243742855352783 166 times 10minus11

000000 minus01443375672974064411 minus01443375672897319217 768 times 10minus12

033111 minus00753154744680519215 minus00753154744695265617 148 times 10minus12

077111 minus00270285335549129334 minus00270285335485592653 635 times 10minus12

091111 minus00146928400579184367 minus00146928400450019102 129 times 10minus11

099999 minus00001434451425475821 minus00001434451390347442 351 times 10minus12

40

minus099999 minus25819308036170334212 minus25819308036170338346 404 times 10minus14

minus091111 minus22578118246552894121 minus22578118246552894116 450 times 10minus19

minus077111 minus10634637975870125656 minus10634637975870125637 185 times 10minus18

minus033111 minus02924243743021256602 minus02924243743021256612 967 times 10minus19

000000 minus01443375672974064411 minus01443375672974064412 113 times 10minus19

033111 minus00753154744680519215 minus00753154744680519211 400 times 10minus19

077111 minus00270285335549129334 minus00270285335549129336 172 times 10minus19

091111 minus00146928400579184367 minus00146928400579184373 573 times 10minus19

099999 minus00001434451425475821 minus00001434451425475826 486 times 10minus19

minus1 minus05 0 05 1

Singular point valuesminus50

minus100

minus150

minus200

minus250

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

0020

0015

0010

0005

Abso

lute

erro

r val

ues

(b)

Figure 3 (a) Exact and AQS solutions when 119873 = 10 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 10 and ℎ(119905) = 1(2 + 119905)

10 Abstract and Applied Analysis

minus1 minus05 0 05 1

Singular point valuesminus50

minus100

minus150

minus200

minus250

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

14

12

10

80

60

40

20

times10minus8

(b)

Figure 4 (a) Exact and AQS solutions when 119873 = 20 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 20 and ℎ(119905) = 1(2 + 119905)

minus1 minus05 0 05 1

Singular point values

minus50

minus100

minus150

minus250

HSI

s val

ues

minus200

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

40

30

20

10

times10minus14

(b)

Figure 5 (a) Exact and AQS solutions when 119873 = 40 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 40 and ℎ(119905) = 1(2 + 119905)

where ℎ(119905) = 1(2 + 119905) The exact solution is 1198671(ℎ 119909) =

minusradic(1 minus 119909)(1 + 119909)(radic33(2 + 119909)2)

The numerical results in Table 3 show that the AQS (21)is highly accurate for the rational function ℎ(119905) = 1(2 + 119905)for the case 119894 = 1 and119873 = 40 Figure 3(a) reveals that there jsgood agreement between the exact solution andAQS solutionwhich can be seen from absolute error in Figure 3(b) when119873 = 10 The absolute error is improved in Figure 4(b) when

119873 = 20 with the corresponding exact and AQS solution inFigure 4(a) Best result is obtained in Figures 5(a) and 5(b)when119873 = 40

Example 4 Consider the HSIs of the form

1198672 (ℎ 119909) =

1

120587radic1 + 119909

1 minus 119909=int

1

minus1

radic1 minus 119905

1 + 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (77)

Abstract and Applied Analysis 11

Table 4 Results of HSIs for 1199082(119905)

119873 119909 Exact solution Approximate solution Absolute error

10

minus099999 000387291557000340015 000387141593292902562 150 times 10minus6

minus091111 031504763162812503076 031505448317559772604 685 times 10minus6

minus077111 041231131092313516294 041230879229200872013 252 times 10minus6

minus033111 044083450592426358350 044083359884530457966 907 times 10minus7

000000 043301270189221932338 043301675286046974386 405 times 10minus6

033111 044963976686376800332 044963389568653105901 587 times 10minus6

077111 062742591722366873321 062741343576104182143 125 times 10minus5

091111 094767578680690190767 094770492350476572039 291 times 10minus5

099999 86066655193121619385 86053938870581623578 0127 times 10minus2

20

minus099999 000387291557000340015 000387291555159212453 184 times 10minus11

minus091111 031504763162812503076 031504763156789346359 602 times 10minus11

minus077111 041231131092313516294 041231131089736085386 258 times 10minus11

minus033111 044083450592426358350 044083450592691374976 265 times 10minus12

000000 043301270189221932338 043301270187830840205 139 times 10minus11

033111 044963976686376800332 044963976684337128944 204 times 10minus11

077111 062742591722366873321 062742591718912403402 346 times 10minus11

091111 094767578680690190767 094767578677558019222 313 times 10minus11

099999 86066655193121619385 86066655108152556362 850 times 10minus8

40

minus099999 000387291557000340015 000387291557000339911 104 times 10minus18

minus091111 031504763162812503076 031504763162812503215 140 times 10minus18

minus077111 041231131092313516294 041231131092313516515 228 times 10minus18

minus033111 044083450592426358350 044083450592426358289 614 times 10minus19

000000 043301270189221932338 043301270189221932269 675 times 10minus19

033111 044963976686376800332 044963976686376800195 131 times 10minus18

077111 062742591722366873321 062742591722366873124 191 times 10minus18

091111 094767578680690190767 094767578680690190858 131 times 10minus18

099999 86066655193121619385 86066655193121618680 705 times 10minus16

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

0012

0010

0008

0006

0004

0002

Abso

lute

erro

r val

ues

(b)

Figure 6 (a) Exact and AQS solutions when 119873 = 10 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 10 and ℎ(119905) = 1(2 + 119905)

12 Abstract and Applied Analysis

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

times10minus8

Abso

lute

erro

r val

ues

(b)

Figure 7 (a) Exact and AQS solutions when 119873 = 20 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 20 and ℎ(119905) = 1(2 + 119905)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

Exact solution plotAQS solution plot

80

70

60

50

40

30

20

10

(a)

minus1 minus05 0 05 1

Singular point values

70

60

50

40

30

20

10

times10minus16

Abso

lute

erro

r val

ues

(b)

Figure 8 (a) Exact and AQS solutions when 119873 = 40 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 40 and ℎ(119905) = 1(2 + 119905)

where ℎ(119905) = 1(2 + 119905) The exact solution is 1198672(ℎ 119909) =

radic(1 + 119909)(1 minus 119909)(radic3(2 + 119909)2)

The numerical results in Table 4 show that the AQS (21)is highly accurate for the rational function ℎ(119905) = 1(2 + 119905)for the case 119894 = 2 and119873 = 40 Figure 6(a) shows that there is

good agreement between the exact solution andAQS solutionwhich can be seen from absolute error in Figure 6(b) when119873 = 10 The absolute error is reduced in Figure 7(b) when119873 = 20 with the corresponding exact and AQS solutionin Figure 7(a) The best result is obtained and illustrated inFigures 8(a) and 8(b) when119873 = 40

Abstract and Applied Analysis 13

6 Conclusion

In this paper we have constructed the automatic quadratureschemes for the semibounded weighted Hadamard typehypersingular integrals using the truncated series of the thirdand fourth kinds of Chebyshev polynomials The unknowncoefficients are found by applying the discrete orthogonalitycondition and the choice of collocation points on (minus1 1)The error estimations for the semibounded solutions of thehypersingular integrals are obtained Numerical exampleshave shown the accuracy of the developed methods andagreed with theoretical findings

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Hadamard Le Probleme de Cauchy et les Equations auxDerivees Particelles Lineaires Hyperboliques Hermann ParisFrance 1932

[2] E Lutz L Gray and A Ingraffea ldquoAn overview of integrationmethods for hypersingular integralsrdquo in Boundary Elements CA Brebbia and G S Gipson Eds vol 8 pp 913ndash925 CMPSouthampton UK 1991

[3] S G Samko Hypersingular Integrals and Their ApplicationsTaylor amp Francis London UK 2002

[4] I V Boikov ldquoNumerical methods of computation of singularand hypersingular integralsrdquo International Journal of Mathe-matics and Mathematical Sciences vol 28 no 3 pp 127ndash1792001

[5] I K Lifanov L N Poltavskii and G M Vainikko Hypersin-gular Integral Equations and Their Applications Chapman ampHallCRC A CRC Press Company London UK 2004

[6] H R Kutt ldquoQuadrature formulae for finite part integralsrdquoReport WISK 178 The National Research Institute for Mathe-matical Sciences Pretoria South Africa 1975

[7] C-Y Hui and D Shia ldquoEvaluations of hypersingular integralsusingGaussian quadraturerdquo International Journal for NumericalMethods in Engineering vol 44 no 2 pp 205ndash214 1999

[8] Y Z Chen ldquoNumerical solution of a curved crack problem byusing hypersingular integral equation approachrdquo EngineeringFracture Mechanics vol 46 no 2 pp 275ndash283 1993

[9] N M A Nik Long and Z K Eshkuvatov ldquoHypersingularintegral equation for multiple curved cracks problem in planeelasticityrdquo International Journal of Solids and Structures vol 46no 13 pp 2611ndash2617 2009

[10] S J Obaiys Z K Eshkuvatov and N M Nik Long ldquoOn errorestimation of automatic quadrature scheme for the evaluationof Hadamard integral of second order singularityrdquo University ofBucharest Scientific Bulletin Series A Applied Mathematics andPhysics vol 75 no 1 pp 85ndash98 2013

[11] J CMason andDCHandscombChebyshev Polynomials CRCPress 2003

[12] M I Israilov Semi-Analytic Solution of Singular Integral Equa-tions of First Kind by Quadrature Formula Numerical Integra-tion and Related Problems FAN Press 1990 (Russian)

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Stochastic AnalysisInternational Journal of

Page 7: An Automatic Quadrature Schemes and Error Estimates for Semibounded Weighted Hadamard Type Hypersingular Integrals

Abstract and Applied Analysis 7

Table 1 Results of HSIs for 1199081(119905)

119873 119909 Exact solution Approximate solution Absolute error

2

minus099999 minus22360444988237596342 minus22360444988237596345 313 times 10minus16

minus091111 minus21535241487484869307 minus21535241487484869310 308 times 10minus18

minus071111 minus93563357403104129848 minus93563357403104129863 142 times 10minus18

minus031111 minus30963683860810919235 minus30963683860810919240 586 times 10minus19

000000 minus10000000000000000000 minus10000000000000000003 300 times 10minus19

031111 017718528456311131724 017718528456311131709 127 times 10minus19

071111 075786494952828070264 075786494952828070264 639 times 10minus21

091111 057031832778553277005 057031832778553277007 140 times 10minus20

099999 000670813126012938238 000670813126012938238 224 times 10minus22

minus1 minus05 0 05 1

Singular point values

minus500

minus1000

minus1500

minus2000

HSI

val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

30

20

10

times10minus16

Abso

lute

erro

r val

ues

(b)Figure 1 (a) Exact and AQS solutions when 119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5 and (b) absolute error between exact and AQS solution when119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5

Due to (31) and (56)1198641198942 (ℎ 119909) = 0

1198641198943 (ℎ 119909) le

2

radic1 minus 1199092

008119872

2119873minus3 (119873 minus 3)

times [1 +1198711119873

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)]

(70)

Since from the hypothesis in Lemma 5

1 minus 1199092ge 1 minus (1 minus 120575

119873)2

1

radic1 minus 1199092le

1

radic1 minus (1 minus 120575119873)2

leradic2

radic3120575119873

=119873

radic3

(71)

we have

1198641198943 (ℎ 119909) le

009119872

2119873minus3 (119873 minus 3)

times 119873[1 +1198711119873

119873+

7119873 minus 4

2 (119873 minus 1) (119873 minus 2)]

(72)

Substituting (69)ndash(72) into (54) yields

1003816100381610038161003816119864119894 (ℎ 119909)1003816100381610038161003816 le

068119872 ln119873120587 sdot 2119873minus3 (119873 minus 3)

times [1 +127

ln119873+211

ln119873(1198711119873+ 008119871

2119873)

+254 (7119873 minus 4)

(119873 minus 1) (119873 minus 2)(1 +

111

ln119873)

+017 (23119873

3minus69119873

2+100119873minus48)

(119873minus1) (119873 minus 2) (119873minus3) (119873minus4)+042119873

ln119873

+042

ln1198731198711119873+

042119873 (7119873minus4)

2 ln119873(119873minus1) (119873 minus 2)]

(73)

Choosing119873 ge 5 the statement ofTheorem 7 follows

8 Abstract and Applied Analysis

Table 2 Results of HSIs for 1199082(119905)

119873 119909 Exact solution Approximate solution Absolute error

2

minus099999 002012457266627356241 00201245726662735624 157 times 10minus21

minus091111 18643200698228626472 18643200698228626470 143 times 10minus19

minus071111 32232147018485970132 32232147018485970129 240 times 10minus19

minus031111 45263577087926478228 45263577087926478225 308 times 10minus19

000000 50000000000000000000 49999999999999999997 300 times 10minus19

031111 51810684429214884711 51810684429214884708 242 times 10minus19

071111 52460548398163357508 52460548398163357508 379 times 10minus20

091111 62854320328769430625 62854320328769430630 299 times 10minus19

099999 44723036596367022628 44723036596367022632 447 times 10minus17

minus1 minus05 0 05

400

300

200

100

1

Singular point values

HSI

val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

40

30

20

10

times10minus17

Abso

lute

erro

r val

ues

(b)

Figure 2 (a) Exact and AQS solutions when 119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5 and (b) absolute error between exact and AQS solution when119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5

5 Numerical Results and Discussion

In this session we evaluate the hypersingular integrals (4) fordifferent choices of ℎ(119905) and compare with the exact solution

Example 1 Consider the HSIs of the form

1198671 (ℎ 119909) =

1

120587radic1 minus 119909

1 + 119909=int

1

minus1

radic1 + 119905

1 minus 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (74)

where ℎ(119905) = 21199052 minus 3119905 + 5 The exact solution is 1198671(ℎ 119909) =

2radic(1 minus 119909)(1 + 119909)(minus1 + 4119909)

The numerical results in Table 1 show that the AQS givesa very good result for the quadratic function ℎ(119905) = 21199052 minus3119905 + 5 for the case 119894 = 1 and 119873 = 2 This is furtherillustrated in Figure 1(a) where the graphs of the exact andAQS solutions are superimposed for the same choices ofsingular point values Details of the accuracy of the proposedAQS is observed from the absolute error in Figure 1(b)

Example 2 Consider the HSIs of the form

1198672 (ℎ 119909) =

1

120587radic1 + 119909

1 minus 119909=int

1

minus1

radic1 minus 119905

1 + 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (75)

where ℎ(119905) = 21199052 minus 3119905 + 5 The exact solution is 1198672(ℎ 119909) =

radic(1 + 119909)(1 minus 119909)(5 minus 4119909)

The numerical results in Table 2 show that the AQS (21)gives a very good result for the quadratic function ℎ(119905) = 21199052minus3119905 + 5 for the case 119894 = 2 and 119873 = 2 Figure 2(a) indicatesthat the exact and AQS solutions are nearly the same Furtherdetails of the accuracy of the absolute error of the proposedAQS are demonstrated in Figure 2(b)

Example 3 Consider the HSIs of the form

1198671 (ℎ 119909) =

1

120587radic1 minus 119909

1 + 119909=int

1

minus1

radic1 + 119905

1 minus 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (76)

Abstract and Applied Analysis 9

Table 3 Results of HSIs for 1199081(119905)

119873 119909 Exact solution Approximate solution Absolute error

10

minus099999 minus25819308036170334212 minus25817083250027430060 223 times 10minus2

minus091111 minus22578118246552894121 minus22578564435672151572 446 times 10minus5

minus077111 minus10634637975870125656 minus10634476360338990858 162 times 10minus5

minus033111 minus02924243743021256602 minus02924196578457308097 472 times 10minus6

000000 minus01443375672974064411 minus01443397190188599320 215 times 10minus6

033111 minus00753154744680519215 minus00753153016422631300 173 times 10minus7

077111 minus00270285335549129334 minus00270280566695517436 477 times 10minus7

091111 minus00146928400579184367 minus00146941562302624467 132 times 10minus6

099999 minus00001434451425475821 minus00001431650832316927 280 times 10minus7

20

minus099999 minus25819308036170334212 minus25819308021411085378 148 times 10minus7

minus091111 minus22578118246552894121 minus22578118246107028644 446 times 10minus11

minus077111 minus10634637975870125656 minus10634637975433417458 437 times 10minus11

minus033111 minus02924243743021256602 minus02924243742855352783 166 times 10minus11

000000 minus01443375672974064411 minus01443375672897319217 768 times 10minus12

033111 minus00753154744680519215 minus00753154744695265617 148 times 10minus12

077111 minus00270285335549129334 minus00270285335485592653 635 times 10minus12

091111 minus00146928400579184367 minus00146928400450019102 129 times 10minus11

099999 minus00001434451425475821 minus00001434451390347442 351 times 10minus12

40

minus099999 minus25819308036170334212 minus25819308036170338346 404 times 10minus14

minus091111 minus22578118246552894121 minus22578118246552894116 450 times 10minus19

minus077111 minus10634637975870125656 minus10634637975870125637 185 times 10minus18

minus033111 minus02924243743021256602 minus02924243743021256612 967 times 10minus19

000000 minus01443375672974064411 minus01443375672974064412 113 times 10minus19

033111 minus00753154744680519215 minus00753154744680519211 400 times 10minus19

077111 minus00270285335549129334 minus00270285335549129336 172 times 10minus19

091111 minus00146928400579184367 minus00146928400579184373 573 times 10minus19

099999 minus00001434451425475821 minus00001434451425475826 486 times 10minus19

minus1 minus05 0 05 1

Singular point valuesminus50

minus100

minus150

minus200

minus250

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

0020

0015

0010

0005

Abso

lute

erro

r val

ues

(b)

Figure 3 (a) Exact and AQS solutions when 119873 = 10 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 10 and ℎ(119905) = 1(2 + 119905)

10 Abstract and Applied Analysis

minus1 minus05 0 05 1

Singular point valuesminus50

minus100

minus150

minus200

minus250

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

14

12

10

80

60

40

20

times10minus8

(b)

Figure 4 (a) Exact and AQS solutions when 119873 = 20 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 20 and ℎ(119905) = 1(2 + 119905)

minus1 minus05 0 05 1

Singular point values

minus50

minus100

minus150

minus250

HSI

s val

ues

minus200

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

40

30

20

10

times10minus14

(b)

Figure 5 (a) Exact and AQS solutions when 119873 = 40 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 40 and ℎ(119905) = 1(2 + 119905)

where ℎ(119905) = 1(2 + 119905) The exact solution is 1198671(ℎ 119909) =

minusradic(1 minus 119909)(1 + 119909)(radic33(2 + 119909)2)

The numerical results in Table 3 show that the AQS (21)is highly accurate for the rational function ℎ(119905) = 1(2 + 119905)for the case 119894 = 1 and119873 = 40 Figure 3(a) reveals that there jsgood agreement between the exact solution andAQS solutionwhich can be seen from absolute error in Figure 3(b) when119873 = 10 The absolute error is improved in Figure 4(b) when

119873 = 20 with the corresponding exact and AQS solution inFigure 4(a) Best result is obtained in Figures 5(a) and 5(b)when119873 = 40

Example 4 Consider the HSIs of the form

1198672 (ℎ 119909) =

1

120587radic1 + 119909

1 minus 119909=int

1

minus1

radic1 minus 119905

1 + 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (77)

Abstract and Applied Analysis 11

Table 4 Results of HSIs for 1199082(119905)

119873 119909 Exact solution Approximate solution Absolute error

10

minus099999 000387291557000340015 000387141593292902562 150 times 10minus6

minus091111 031504763162812503076 031505448317559772604 685 times 10minus6

minus077111 041231131092313516294 041230879229200872013 252 times 10minus6

minus033111 044083450592426358350 044083359884530457966 907 times 10minus7

000000 043301270189221932338 043301675286046974386 405 times 10minus6

033111 044963976686376800332 044963389568653105901 587 times 10minus6

077111 062742591722366873321 062741343576104182143 125 times 10minus5

091111 094767578680690190767 094770492350476572039 291 times 10minus5

099999 86066655193121619385 86053938870581623578 0127 times 10minus2

20

minus099999 000387291557000340015 000387291555159212453 184 times 10minus11

minus091111 031504763162812503076 031504763156789346359 602 times 10minus11

minus077111 041231131092313516294 041231131089736085386 258 times 10minus11

minus033111 044083450592426358350 044083450592691374976 265 times 10minus12

000000 043301270189221932338 043301270187830840205 139 times 10minus11

033111 044963976686376800332 044963976684337128944 204 times 10minus11

077111 062742591722366873321 062742591718912403402 346 times 10minus11

091111 094767578680690190767 094767578677558019222 313 times 10minus11

099999 86066655193121619385 86066655108152556362 850 times 10minus8

40

minus099999 000387291557000340015 000387291557000339911 104 times 10minus18

minus091111 031504763162812503076 031504763162812503215 140 times 10minus18

minus077111 041231131092313516294 041231131092313516515 228 times 10minus18

minus033111 044083450592426358350 044083450592426358289 614 times 10minus19

000000 043301270189221932338 043301270189221932269 675 times 10minus19

033111 044963976686376800332 044963976686376800195 131 times 10minus18

077111 062742591722366873321 062742591722366873124 191 times 10minus18

091111 094767578680690190767 094767578680690190858 131 times 10minus18

099999 86066655193121619385 86066655193121618680 705 times 10minus16

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

0012

0010

0008

0006

0004

0002

Abso

lute

erro

r val

ues

(b)

Figure 6 (a) Exact and AQS solutions when 119873 = 10 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 10 and ℎ(119905) = 1(2 + 119905)

12 Abstract and Applied Analysis

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

times10minus8

Abso

lute

erro

r val

ues

(b)

Figure 7 (a) Exact and AQS solutions when 119873 = 20 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 20 and ℎ(119905) = 1(2 + 119905)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

Exact solution plotAQS solution plot

80

70

60

50

40

30

20

10

(a)

minus1 minus05 0 05 1

Singular point values

70

60

50

40

30

20

10

times10minus16

Abso

lute

erro

r val

ues

(b)

Figure 8 (a) Exact and AQS solutions when 119873 = 40 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 40 and ℎ(119905) = 1(2 + 119905)

where ℎ(119905) = 1(2 + 119905) The exact solution is 1198672(ℎ 119909) =

radic(1 + 119909)(1 minus 119909)(radic3(2 + 119909)2)

The numerical results in Table 4 show that the AQS (21)is highly accurate for the rational function ℎ(119905) = 1(2 + 119905)for the case 119894 = 2 and119873 = 40 Figure 6(a) shows that there is

good agreement between the exact solution andAQS solutionwhich can be seen from absolute error in Figure 6(b) when119873 = 10 The absolute error is reduced in Figure 7(b) when119873 = 20 with the corresponding exact and AQS solutionin Figure 7(a) The best result is obtained and illustrated inFigures 8(a) and 8(b) when119873 = 40

Abstract and Applied Analysis 13

6 Conclusion

In this paper we have constructed the automatic quadratureschemes for the semibounded weighted Hadamard typehypersingular integrals using the truncated series of the thirdand fourth kinds of Chebyshev polynomials The unknowncoefficients are found by applying the discrete orthogonalitycondition and the choice of collocation points on (minus1 1)The error estimations for the semibounded solutions of thehypersingular integrals are obtained Numerical exampleshave shown the accuracy of the developed methods andagreed with theoretical findings

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Hadamard Le Probleme de Cauchy et les Equations auxDerivees Particelles Lineaires Hyperboliques Hermann ParisFrance 1932

[2] E Lutz L Gray and A Ingraffea ldquoAn overview of integrationmethods for hypersingular integralsrdquo in Boundary Elements CA Brebbia and G S Gipson Eds vol 8 pp 913ndash925 CMPSouthampton UK 1991

[3] S G Samko Hypersingular Integrals and Their ApplicationsTaylor amp Francis London UK 2002

[4] I V Boikov ldquoNumerical methods of computation of singularand hypersingular integralsrdquo International Journal of Mathe-matics and Mathematical Sciences vol 28 no 3 pp 127ndash1792001

[5] I K Lifanov L N Poltavskii and G M Vainikko Hypersin-gular Integral Equations and Their Applications Chapman ampHallCRC A CRC Press Company London UK 2004

[6] H R Kutt ldquoQuadrature formulae for finite part integralsrdquoReport WISK 178 The National Research Institute for Mathe-matical Sciences Pretoria South Africa 1975

[7] C-Y Hui and D Shia ldquoEvaluations of hypersingular integralsusingGaussian quadraturerdquo International Journal for NumericalMethods in Engineering vol 44 no 2 pp 205ndash214 1999

[8] Y Z Chen ldquoNumerical solution of a curved crack problem byusing hypersingular integral equation approachrdquo EngineeringFracture Mechanics vol 46 no 2 pp 275ndash283 1993

[9] N M A Nik Long and Z K Eshkuvatov ldquoHypersingularintegral equation for multiple curved cracks problem in planeelasticityrdquo International Journal of Solids and Structures vol 46no 13 pp 2611ndash2617 2009

[10] S J Obaiys Z K Eshkuvatov and N M Nik Long ldquoOn errorestimation of automatic quadrature scheme for the evaluationof Hadamard integral of second order singularityrdquo University ofBucharest Scientific Bulletin Series A Applied Mathematics andPhysics vol 75 no 1 pp 85ndash98 2013

[11] J CMason andDCHandscombChebyshev Polynomials CRCPress 2003

[12] M I Israilov Semi-Analytic Solution of Singular Integral Equa-tions of First Kind by Quadrature Formula Numerical Integra-tion and Related Problems FAN Press 1990 (Russian)

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Mathematical Problems in Engineering

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Stochastic AnalysisInternational Journal of

Page 8: An Automatic Quadrature Schemes and Error Estimates for Semibounded Weighted Hadamard Type Hypersingular Integrals

8 Abstract and Applied Analysis

Table 2 Results of HSIs for 1199082(119905)

119873 119909 Exact solution Approximate solution Absolute error

2

minus099999 002012457266627356241 00201245726662735624 157 times 10minus21

minus091111 18643200698228626472 18643200698228626470 143 times 10minus19

minus071111 32232147018485970132 32232147018485970129 240 times 10minus19

minus031111 45263577087926478228 45263577087926478225 308 times 10minus19

000000 50000000000000000000 49999999999999999997 300 times 10minus19

031111 51810684429214884711 51810684429214884708 242 times 10minus19

071111 52460548398163357508 52460548398163357508 379 times 10minus20

091111 62854320328769430625 62854320328769430630 299 times 10minus19

099999 44723036596367022628 44723036596367022632 447 times 10minus17

minus1 minus05 0 05

400

300

200

100

1

Singular point values

HSI

val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

40

30

20

10

times10minus17

Abso

lute

erro

r val

ues

(b)

Figure 2 (a) Exact and AQS solutions when 119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5 and (b) absolute error between exact and AQS solution when119873 = 2 and ℎ(119905) = 21199052 minus 3119905 + 5

5 Numerical Results and Discussion

In this session we evaluate the hypersingular integrals (4) fordifferent choices of ℎ(119905) and compare with the exact solution

Example 1 Consider the HSIs of the form

1198671 (ℎ 119909) =

1

120587radic1 minus 119909

1 + 119909=int

1

minus1

radic1 + 119905

1 minus 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (74)

where ℎ(119905) = 21199052 minus 3119905 + 5 The exact solution is 1198671(ℎ 119909) =

2radic(1 minus 119909)(1 + 119909)(minus1 + 4119909)

The numerical results in Table 1 show that the AQS givesa very good result for the quadratic function ℎ(119905) = 21199052 minus3119905 + 5 for the case 119894 = 1 and 119873 = 2 This is furtherillustrated in Figure 1(a) where the graphs of the exact andAQS solutions are superimposed for the same choices ofsingular point values Details of the accuracy of the proposedAQS is observed from the absolute error in Figure 1(b)

Example 2 Consider the HSIs of the form

1198672 (ℎ 119909) =

1

120587radic1 + 119909

1 minus 119909=int

1

minus1

radic1 minus 119905

1 + 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (75)

where ℎ(119905) = 21199052 minus 3119905 + 5 The exact solution is 1198672(ℎ 119909) =

radic(1 + 119909)(1 minus 119909)(5 minus 4119909)

The numerical results in Table 2 show that the AQS (21)gives a very good result for the quadratic function ℎ(119905) = 21199052minus3119905 + 5 for the case 119894 = 2 and 119873 = 2 Figure 2(a) indicatesthat the exact and AQS solutions are nearly the same Furtherdetails of the accuracy of the absolute error of the proposedAQS are demonstrated in Figure 2(b)

Example 3 Consider the HSIs of the form

1198671 (ℎ 119909) =

1

120587radic1 minus 119909

1 + 119909=int

1

minus1

radic1 + 119905

1 minus 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (76)

Abstract and Applied Analysis 9

Table 3 Results of HSIs for 1199081(119905)

119873 119909 Exact solution Approximate solution Absolute error

10

minus099999 minus25819308036170334212 minus25817083250027430060 223 times 10minus2

minus091111 minus22578118246552894121 minus22578564435672151572 446 times 10minus5

minus077111 minus10634637975870125656 minus10634476360338990858 162 times 10minus5

minus033111 minus02924243743021256602 minus02924196578457308097 472 times 10minus6

000000 minus01443375672974064411 minus01443397190188599320 215 times 10minus6

033111 minus00753154744680519215 minus00753153016422631300 173 times 10minus7

077111 minus00270285335549129334 minus00270280566695517436 477 times 10minus7

091111 minus00146928400579184367 minus00146941562302624467 132 times 10minus6

099999 minus00001434451425475821 minus00001431650832316927 280 times 10minus7

20

minus099999 minus25819308036170334212 minus25819308021411085378 148 times 10minus7

minus091111 minus22578118246552894121 minus22578118246107028644 446 times 10minus11

minus077111 minus10634637975870125656 minus10634637975433417458 437 times 10minus11

minus033111 minus02924243743021256602 minus02924243742855352783 166 times 10minus11

000000 minus01443375672974064411 minus01443375672897319217 768 times 10minus12

033111 minus00753154744680519215 minus00753154744695265617 148 times 10minus12

077111 minus00270285335549129334 minus00270285335485592653 635 times 10minus12

091111 minus00146928400579184367 minus00146928400450019102 129 times 10minus11

099999 minus00001434451425475821 minus00001434451390347442 351 times 10minus12

40

minus099999 minus25819308036170334212 minus25819308036170338346 404 times 10minus14

minus091111 minus22578118246552894121 minus22578118246552894116 450 times 10minus19

minus077111 minus10634637975870125656 minus10634637975870125637 185 times 10minus18

minus033111 minus02924243743021256602 minus02924243743021256612 967 times 10minus19

000000 minus01443375672974064411 minus01443375672974064412 113 times 10minus19

033111 minus00753154744680519215 minus00753154744680519211 400 times 10minus19

077111 minus00270285335549129334 minus00270285335549129336 172 times 10minus19

091111 minus00146928400579184367 minus00146928400579184373 573 times 10minus19

099999 minus00001434451425475821 minus00001434451425475826 486 times 10minus19

minus1 minus05 0 05 1

Singular point valuesminus50

minus100

minus150

minus200

minus250

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

0020

0015

0010

0005

Abso

lute

erro

r val

ues

(b)

Figure 3 (a) Exact and AQS solutions when 119873 = 10 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 10 and ℎ(119905) = 1(2 + 119905)

10 Abstract and Applied Analysis

minus1 minus05 0 05 1

Singular point valuesminus50

minus100

minus150

minus200

minus250

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

14

12

10

80

60

40

20

times10minus8

(b)

Figure 4 (a) Exact and AQS solutions when 119873 = 20 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 20 and ℎ(119905) = 1(2 + 119905)

minus1 minus05 0 05 1

Singular point values

minus50

minus100

minus150

minus250

HSI

s val

ues

minus200

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

40

30

20

10

times10minus14

(b)

Figure 5 (a) Exact and AQS solutions when 119873 = 40 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 40 and ℎ(119905) = 1(2 + 119905)

where ℎ(119905) = 1(2 + 119905) The exact solution is 1198671(ℎ 119909) =

minusradic(1 minus 119909)(1 + 119909)(radic33(2 + 119909)2)

The numerical results in Table 3 show that the AQS (21)is highly accurate for the rational function ℎ(119905) = 1(2 + 119905)for the case 119894 = 1 and119873 = 40 Figure 3(a) reveals that there jsgood agreement between the exact solution andAQS solutionwhich can be seen from absolute error in Figure 3(b) when119873 = 10 The absolute error is improved in Figure 4(b) when

119873 = 20 with the corresponding exact and AQS solution inFigure 4(a) Best result is obtained in Figures 5(a) and 5(b)when119873 = 40

Example 4 Consider the HSIs of the form

1198672 (ℎ 119909) =

1

120587radic1 + 119909

1 minus 119909=int

1

minus1

radic1 minus 119905

1 + 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (77)

Abstract and Applied Analysis 11

Table 4 Results of HSIs for 1199082(119905)

119873 119909 Exact solution Approximate solution Absolute error

10

minus099999 000387291557000340015 000387141593292902562 150 times 10minus6

minus091111 031504763162812503076 031505448317559772604 685 times 10minus6

minus077111 041231131092313516294 041230879229200872013 252 times 10minus6

minus033111 044083450592426358350 044083359884530457966 907 times 10minus7

000000 043301270189221932338 043301675286046974386 405 times 10minus6

033111 044963976686376800332 044963389568653105901 587 times 10minus6

077111 062742591722366873321 062741343576104182143 125 times 10minus5

091111 094767578680690190767 094770492350476572039 291 times 10minus5

099999 86066655193121619385 86053938870581623578 0127 times 10minus2

20

minus099999 000387291557000340015 000387291555159212453 184 times 10minus11

minus091111 031504763162812503076 031504763156789346359 602 times 10minus11

minus077111 041231131092313516294 041231131089736085386 258 times 10minus11

minus033111 044083450592426358350 044083450592691374976 265 times 10minus12

000000 043301270189221932338 043301270187830840205 139 times 10minus11

033111 044963976686376800332 044963976684337128944 204 times 10minus11

077111 062742591722366873321 062742591718912403402 346 times 10minus11

091111 094767578680690190767 094767578677558019222 313 times 10minus11

099999 86066655193121619385 86066655108152556362 850 times 10minus8

40

minus099999 000387291557000340015 000387291557000339911 104 times 10minus18

minus091111 031504763162812503076 031504763162812503215 140 times 10minus18

minus077111 041231131092313516294 041231131092313516515 228 times 10minus18

minus033111 044083450592426358350 044083450592426358289 614 times 10minus19

000000 043301270189221932338 043301270189221932269 675 times 10minus19

033111 044963976686376800332 044963976686376800195 131 times 10minus18

077111 062742591722366873321 062742591722366873124 191 times 10minus18

091111 094767578680690190767 094767578680690190858 131 times 10minus18

099999 86066655193121619385 86066655193121618680 705 times 10minus16

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

0012

0010

0008

0006

0004

0002

Abso

lute

erro

r val

ues

(b)

Figure 6 (a) Exact and AQS solutions when 119873 = 10 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 10 and ℎ(119905) = 1(2 + 119905)

12 Abstract and Applied Analysis

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

times10minus8

Abso

lute

erro

r val

ues

(b)

Figure 7 (a) Exact and AQS solutions when 119873 = 20 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 20 and ℎ(119905) = 1(2 + 119905)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

Exact solution plotAQS solution plot

80

70

60

50

40

30

20

10

(a)

minus1 minus05 0 05 1

Singular point values

70

60

50

40

30

20

10

times10minus16

Abso

lute

erro

r val

ues

(b)

Figure 8 (a) Exact and AQS solutions when 119873 = 40 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 40 and ℎ(119905) = 1(2 + 119905)

where ℎ(119905) = 1(2 + 119905) The exact solution is 1198672(ℎ 119909) =

radic(1 + 119909)(1 minus 119909)(radic3(2 + 119909)2)

The numerical results in Table 4 show that the AQS (21)is highly accurate for the rational function ℎ(119905) = 1(2 + 119905)for the case 119894 = 2 and119873 = 40 Figure 6(a) shows that there is

good agreement between the exact solution andAQS solutionwhich can be seen from absolute error in Figure 6(b) when119873 = 10 The absolute error is reduced in Figure 7(b) when119873 = 20 with the corresponding exact and AQS solutionin Figure 7(a) The best result is obtained and illustrated inFigures 8(a) and 8(b) when119873 = 40

Abstract and Applied Analysis 13

6 Conclusion

In this paper we have constructed the automatic quadratureschemes for the semibounded weighted Hadamard typehypersingular integrals using the truncated series of the thirdand fourth kinds of Chebyshev polynomials The unknowncoefficients are found by applying the discrete orthogonalitycondition and the choice of collocation points on (minus1 1)The error estimations for the semibounded solutions of thehypersingular integrals are obtained Numerical exampleshave shown the accuracy of the developed methods andagreed with theoretical findings

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Hadamard Le Probleme de Cauchy et les Equations auxDerivees Particelles Lineaires Hyperboliques Hermann ParisFrance 1932

[2] E Lutz L Gray and A Ingraffea ldquoAn overview of integrationmethods for hypersingular integralsrdquo in Boundary Elements CA Brebbia and G S Gipson Eds vol 8 pp 913ndash925 CMPSouthampton UK 1991

[3] S G Samko Hypersingular Integrals and Their ApplicationsTaylor amp Francis London UK 2002

[4] I V Boikov ldquoNumerical methods of computation of singularand hypersingular integralsrdquo International Journal of Mathe-matics and Mathematical Sciences vol 28 no 3 pp 127ndash1792001

[5] I K Lifanov L N Poltavskii and G M Vainikko Hypersin-gular Integral Equations and Their Applications Chapman ampHallCRC A CRC Press Company London UK 2004

[6] H R Kutt ldquoQuadrature formulae for finite part integralsrdquoReport WISK 178 The National Research Institute for Mathe-matical Sciences Pretoria South Africa 1975

[7] C-Y Hui and D Shia ldquoEvaluations of hypersingular integralsusingGaussian quadraturerdquo International Journal for NumericalMethods in Engineering vol 44 no 2 pp 205ndash214 1999

[8] Y Z Chen ldquoNumerical solution of a curved crack problem byusing hypersingular integral equation approachrdquo EngineeringFracture Mechanics vol 46 no 2 pp 275ndash283 1993

[9] N M A Nik Long and Z K Eshkuvatov ldquoHypersingularintegral equation for multiple curved cracks problem in planeelasticityrdquo International Journal of Solids and Structures vol 46no 13 pp 2611ndash2617 2009

[10] S J Obaiys Z K Eshkuvatov and N M Nik Long ldquoOn errorestimation of automatic quadrature scheme for the evaluationof Hadamard integral of second order singularityrdquo University ofBucharest Scientific Bulletin Series A Applied Mathematics andPhysics vol 75 no 1 pp 85ndash98 2013

[11] J CMason andDCHandscombChebyshev Polynomials CRCPress 2003

[12] M I Israilov Semi-Analytic Solution of Singular Integral Equa-tions of First Kind by Quadrature Formula Numerical Integra-tion and Related Problems FAN Press 1990 (Russian)

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

Page 9: An Automatic Quadrature Schemes and Error Estimates for Semibounded Weighted Hadamard Type Hypersingular Integrals

Abstract and Applied Analysis 9

Table 3 Results of HSIs for 1199081(119905)

119873 119909 Exact solution Approximate solution Absolute error

10

minus099999 minus25819308036170334212 minus25817083250027430060 223 times 10minus2

minus091111 minus22578118246552894121 minus22578564435672151572 446 times 10minus5

minus077111 minus10634637975870125656 minus10634476360338990858 162 times 10minus5

minus033111 minus02924243743021256602 minus02924196578457308097 472 times 10minus6

000000 minus01443375672974064411 minus01443397190188599320 215 times 10minus6

033111 minus00753154744680519215 minus00753153016422631300 173 times 10minus7

077111 minus00270285335549129334 minus00270280566695517436 477 times 10minus7

091111 minus00146928400579184367 minus00146941562302624467 132 times 10minus6

099999 minus00001434451425475821 minus00001431650832316927 280 times 10minus7

20

minus099999 minus25819308036170334212 minus25819308021411085378 148 times 10minus7

minus091111 minus22578118246552894121 minus22578118246107028644 446 times 10minus11

minus077111 minus10634637975870125656 minus10634637975433417458 437 times 10minus11

minus033111 minus02924243743021256602 minus02924243742855352783 166 times 10minus11

000000 minus01443375672974064411 minus01443375672897319217 768 times 10minus12

033111 minus00753154744680519215 minus00753154744695265617 148 times 10minus12

077111 minus00270285335549129334 minus00270285335485592653 635 times 10minus12

091111 minus00146928400579184367 minus00146928400450019102 129 times 10minus11

099999 minus00001434451425475821 minus00001434451390347442 351 times 10minus12

40

minus099999 minus25819308036170334212 minus25819308036170338346 404 times 10minus14

minus091111 minus22578118246552894121 minus22578118246552894116 450 times 10minus19

minus077111 minus10634637975870125656 minus10634637975870125637 185 times 10minus18

minus033111 minus02924243743021256602 minus02924243743021256612 967 times 10minus19

000000 minus01443375672974064411 minus01443375672974064412 113 times 10minus19

033111 minus00753154744680519215 minus00753154744680519211 400 times 10minus19

077111 minus00270285335549129334 minus00270285335549129336 172 times 10minus19

091111 minus00146928400579184367 minus00146928400579184373 573 times 10minus19

099999 minus00001434451425475821 minus00001434451425475826 486 times 10minus19

minus1 minus05 0 05 1

Singular point valuesminus50

minus100

minus150

minus200

minus250

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

0020

0015

0010

0005

Abso

lute

erro

r val

ues

(b)

Figure 3 (a) Exact and AQS solutions when 119873 = 10 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 10 and ℎ(119905) = 1(2 + 119905)

10 Abstract and Applied Analysis

minus1 minus05 0 05 1

Singular point valuesminus50

minus100

minus150

minus200

minus250

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

14

12

10

80

60

40

20

times10minus8

(b)

Figure 4 (a) Exact and AQS solutions when 119873 = 20 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 20 and ℎ(119905) = 1(2 + 119905)

minus1 minus05 0 05 1

Singular point values

minus50

minus100

minus150

minus250

HSI

s val

ues

minus200

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

40

30

20

10

times10minus14

(b)

Figure 5 (a) Exact and AQS solutions when 119873 = 40 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 40 and ℎ(119905) = 1(2 + 119905)

where ℎ(119905) = 1(2 + 119905) The exact solution is 1198671(ℎ 119909) =

minusradic(1 minus 119909)(1 + 119909)(radic33(2 + 119909)2)

The numerical results in Table 3 show that the AQS (21)is highly accurate for the rational function ℎ(119905) = 1(2 + 119905)for the case 119894 = 1 and119873 = 40 Figure 3(a) reveals that there jsgood agreement between the exact solution andAQS solutionwhich can be seen from absolute error in Figure 3(b) when119873 = 10 The absolute error is improved in Figure 4(b) when

119873 = 20 with the corresponding exact and AQS solution inFigure 4(a) Best result is obtained in Figures 5(a) and 5(b)when119873 = 40

Example 4 Consider the HSIs of the form

1198672 (ℎ 119909) =

1

120587radic1 + 119909

1 minus 119909=int

1

minus1

radic1 minus 119905

1 + 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (77)

Abstract and Applied Analysis 11

Table 4 Results of HSIs for 1199082(119905)

119873 119909 Exact solution Approximate solution Absolute error

10

minus099999 000387291557000340015 000387141593292902562 150 times 10minus6

minus091111 031504763162812503076 031505448317559772604 685 times 10minus6

minus077111 041231131092313516294 041230879229200872013 252 times 10minus6

minus033111 044083450592426358350 044083359884530457966 907 times 10minus7

000000 043301270189221932338 043301675286046974386 405 times 10minus6

033111 044963976686376800332 044963389568653105901 587 times 10minus6

077111 062742591722366873321 062741343576104182143 125 times 10minus5

091111 094767578680690190767 094770492350476572039 291 times 10minus5

099999 86066655193121619385 86053938870581623578 0127 times 10minus2

20

minus099999 000387291557000340015 000387291555159212453 184 times 10minus11

minus091111 031504763162812503076 031504763156789346359 602 times 10minus11

minus077111 041231131092313516294 041231131089736085386 258 times 10minus11

minus033111 044083450592426358350 044083450592691374976 265 times 10minus12

000000 043301270189221932338 043301270187830840205 139 times 10minus11

033111 044963976686376800332 044963976684337128944 204 times 10minus11

077111 062742591722366873321 062742591718912403402 346 times 10minus11

091111 094767578680690190767 094767578677558019222 313 times 10minus11

099999 86066655193121619385 86066655108152556362 850 times 10minus8

40

minus099999 000387291557000340015 000387291557000339911 104 times 10minus18

minus091111 031504763162812503076 031504763162812503215 140 times 10minus18

minus077111 041231131092313516294 041231131092313516515 228 times 10minus18

minus033111 044083450592426358350 044083450592426358289 614 times 10minus19

000000 043301270189221932338 043301270189221932269 675 times 10minus19

033111 044963976686376800332 044963976686376800195 131 times 10minus18

077111 062742591722366873321 062742591722366873124 191 times 10minus18

091111 094767578680690190767 094767578680690190858 131 times 10minus18

099999 86066655193121619385 86066655193121618680 705 times 10minus16

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

0012

0010

0008

0006

0004

0002

Abso

lute

erro

r val

ues

(b)

Figure 6 (a) Exact and AQS solutions when 119873 = 10 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 10 and ℎ(119905) = 1(2 + 119905)

12 Abstract and Applied Analysis

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

times10minus8

Abso

lute

erro

r val

ues

(b)

Figure 7 (a) Exact and AQS solutions when 119873 = 20 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 20 and ℎ(119905) = 1(2 + 119905)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

Exact solution plotAQS solution plot

80

70

60

50

40

30

20

10

(a)

minus1 minus05 0 05 1

Singular point values

70

60

50

40

30

20

10

times10minus16

Abso

lute

erro

r val

ues

(b)

Figure 8 (a) Exact and AQS solutions when 119873 = 40 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 40 and ℎ(119905) = 1(2 + 119905)

where ℎ(119905) = 1(2 + 119905) The exact solution is 1198672(ℎ 119909) =

radic(1 + 119909)(1 minus 119909)(radic3(2 + 119909)2)

The numerical results in Table 4 show that the AQS (21)is highly accurate for the rational function ℎ(119905) = 1(2 + 119905)for the case 119894 = 2 and119873 = 40 Figure 6(a) shows that there is

good agreement between the exact solution andAQS solutionwhich can be seen from absolute error in Figure 6(b) when119873 = 10 The absolute error is reduced in Figure 7(b) when119873 = 20 with the corresponding exact and AQS solutionin Figure 7(a) The best result is obtained and illustrated inFigures 8(a) and 8(b) when119873 = 40

Abstract and Applied Analysis 13

6 Conclusion

In this paper we have constructed the automatic quadratureschemes for the semibounded weighted Hadamard typehypersingular integrals using the truncated series of the thirdand fourth kinds of Chebyshev polynomials The unknowncoefficients are found by applying the discrete orthogonalitycondition and the choice of collocation points on (minus1 1)The error estimations for the semibounded solutions of thehypersingular integrals are obtained Numerical exampleshave shown the accuracy of the developed methods andagreed with theoretical findings

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Hadamard Le Probleme de Cauchy et les Equations auxDerivees Particelles Lineaires Hyperboliques Hermann ParisFrance 1932

[2] E Lutz L Gray and A Ingraffea ldquoAn overview of integrationmethods for hypersingular integralsrdquo in Boundary Elements CA Brebbia and G S Gipson Eds vol 8 pp 913ndash925 CMPSouthampton UK 1991

[3] S G Samko Hypersingular Integrals and Their ApplicationsTaylor amp Francis London UK 2002

[4] I V Boikov ldquoNumerical methods of computation of singularand hypersingular integralsrdquo International Journal of Mathe-matics and Mathematical Sciences vol 28 no 3 pp 127ndash1792001

[5] I K Lifanov L N Poltavskii and G M Vainikko Hypersin-gular Integral Equations and Their Applications Chapman ampHallCRC A CRC Press Company London UK 2004

[6] H R Kutt ldquoQuadrature formulae for finite part integralsrdquoReport WISK 178 The National Research Institute for Mathe-matical Sciences Pretoria South Africa 1975

[7] C-Y Hui and D Shia ldquoEvaluations of hypersingular integralsusingGaussian quadraturerdquo International Journal for NumericalMethods in Engineering vol 44 no 2 pp 205ndash214 1999

[8] Y Z Chen ldquoNumerical solution of a curved crack problem byusing hypersingular integral equation approachrdquo EngineeringFracture Mechanics vol 46 no 2 pp 275ndash283 1993

[9] N M A Nik Long and Z K Eshkuvatov ldquoHypersingularintegral equation for multiple curved cracks problem in planeelasticityrdquo International Journal of Solids and Structures vol 46no 13 pp 2611ndash2617 2009

[10] S J Obaiys Z K Eshkuvatov and N M Nik Long ldquoOn errorestimation of automatic quadrature scheme for the evaluationof Hadamard integral of second order singularityrdquo University ofBucharest Scientific Bulletin Series A Applied Mathematics andPhysics vol 75 no 1 pp 85ndash98 2013

[11] J CMason andDCHandscombChebyshev Polynomials CRCPress 2003

[12] M I Israilov Semi-Analytic Solution of Singular Integral Equa-tions of First Kind by Quadrature Formula Numerical Integra-tion and Related Problems FAN Press 1990 (Russian)

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

Page 10: An Automatic Quadrature Schemes and Error Estimates for Semibounded Weighted Hadamard Type Hypersingular Integrals

10 Abstract and Applied Analysis

minus1 minus05 0 05 1

Singular point valuesminus50

minus100

minus150

minus200

minus250

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

14

12

10

80

60

40

20

times10minus8

(b)

Figure 4 (a) Exact and AQS solutions when 119873 = 20 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 20 and ℎ(119905) = 1(2 + 119905)

minus1 minus05 0 05 1

Singular point values

minus50

minus100

minus150

minus250

HSI

s val

ues

minus200

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

40

30

20

10

times10minus14

(b)

Figure 5 (a) Exact and AQS solutions when 119873 = 40 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 40 and ℎ(119905) = 1(2 + 119905)

where ℎ(119905) = 1(2 + 119905) The exact solution is 1198671(ℎ 119909) =

minusradic(1 minus 119909)(1 + 119909)(radic33(2 + 119909)2)

The numerical results in Table 3 show that the AQS (21)is highly accurate for the rational function ℎ(119905) = 1(2 + 119905)for the case 119894 = 1 and119873 = 40 Figure 3(a) reveals that there jsgood agreement between the exact solution andAQS solutionwhich can be seen from absolute error in Figure 3(b) when119873 = 10 The absolute error is improved in Figure 4(b) when

119873 = 20 with the corresponding exact and AQS solution inFigure 4(a) Best result is obtained in Figures 5(a) and 5(b)when119873 = 40

Example 4 Consider the HSIs of the form

1198672 (ℎ 119909) =

1

120587radic1 + 119909

1 minus 119909=int

1

minus1

radic1 minus 119905

1 + 119905

ℎ (119905)

(119905 minus 119909)2119889119905 (77)

Abstract and Applied Analysis 11

Table 4 Results of HSIs for 1199082(119905)

119873 119909 Exact solution Approximate solution Absolute error

10

minus099999 000387291557000340015 000387141593292902562 150 times 10minus6

minus091111 031504763162812503076 031505448317559772604 685 times 10minus6

minus077111 041231131092313516294 041230879229200872013 252 times 10minus6

minus033111 044083450592426358350 044083359884530457966 907 times 10minus7

000000 043301270189221932338 043301675286046974386 405 times 10minus6

033111 044963976686376800332 044963389568653105901 587 times 10minus6

077111 062742591722366873321 062741343576104182143 125 times 10minus5

091111 094767578680690190767 094770492350476572039 291 times 10minus5

099999 86066655193121619385 86053938870581623578 0127 times 10minus2

20

minus099999 000387291557000340015 000387291555159212453 184 times 10minus11

minus091111 031504763162812503076 031504763156789346359 602 times 10minus11

minus077111 041231131092313516294 041231131089736085386 258 times 10minus11

minus033111 044083450592426358350 044083450592691374976 265 times 10minus12

000000 043301270189221932338 043301270187830840205 139 times 10minus11

033111 044963976686376800332 044963976684337128944 204 times 10minus11

077111 062742591722366873321 062742591718912403402 346 times 10minus11

091111 094767578680690190767 094767578677558019222 313 times 10minus11

099999 86066655193121619385 86066655108152556362 850 times 10minus8

40

minus099999 000387291557000340015 000387291557000339911 104 times 10minus18

minus091111 031504763162812503076 031504763162812503215 140 times 10minus18

minus077111 041231131092313516294 041231131092313516515 228 times 10minus18

minus033111 044083450592426358350 044083450592426358289 614 times 10minus19

000000 043301270189221932338 043301270189221932269 675 times 10minus19

033111 044963976686376800332 044963976686376800195 131 times 10minus18

077111 062742591722366873321 062742591722366873124 191 times 10minus18

091111 094767578680690190767 094767578680690190858 131 times 10minus18

099999 86066655193121619385 86066655193121618680 705 times 10minus16

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

0012

0010

0008

0006

0004

0002

Abso

lute

erro

r val

ues

(b)

Figure 6 (a) Exact and AQS solutions when 119873 = 10 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 10 and ℎ(119905) = 1(2 + 119905)

12 Abstract and Applied Analysis

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

times10minus8

Abso

lute

erro

r val

ues

(b)

Figure 7 (a) Exact and AQS solutions when 119873 = 20 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 20 and ℎ(119905) = 1(2 + 119905)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

Exact solution plotAQS solution plot

80

70

60

50

40

30

20

10

(a)

minus1 minus05 0 05 1

Singular point values

70

60

50

40

30

20

10

times10minus16

Abso

lute

erro

r val

ues

(b)

Figure 8 (a) Exact and AQS solutions when 119873 = 40 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 40 and ℎ(119905) = 1(2 + 119905)

where ℎ(119905) = 1(2 + 119905) The exact solution is 1198672(ℎ 119909) =

radic(1 + 119909)(1 minus 119909)(radic3(2 + 119909)2)

The numerical results in Table 4 show that the AQS (21)is highly accurate for the rational function ℎ(119905) = 1(2 + 119905)for the case 119894 = 2 and119873 = 40 Figure 6(a) shows that there is

good agreement between the exact solution andAQS solutionwhich can be seen from absolute error in Figure 6(b) when119873 = 10 The absolute error is reduced in Figure 7(b) when119873 = 20 with the corresponding exact and AQS solutionin Figure 7(a) The best result is obtained and illustrated inFigures 8(a) and 8(b) when119873 = 40

Abstract and Applied Analysis 13

6 Conclusion

In this paper we have constructed the automatic quadratureschemes for the semibounded weighted Hadamard typehypersingular integrals using the truncated series of the thirdand fourth kinds of Chebyshev polynomials The unknowncoefficients are found by applying the discrete orthogonalitycondition and the choice of collocation points on (minus1 1)The error estimations for the semibounded solutions of thehypersingular integrals are obtained Numerical exampleshave shown the accuracy of the developed methods andagreed with theoretical findings

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Hadamard Le Probleme de Cauchy et les Equations auxDerivees Particelles Lineaires Hyperboliques Hermann ParisFrance 1932

[2] E Lutz L Gray and A Ingraffea ldquoAn overview of integrationmethods for hypersingular integralsrdquo in Boundary Elements CA Brebbia and G S Gipson Eds vol 8 pp 913ndash925 CMPSouthampton UK 1991

[3] S G Samko Hypersingular Integrals and Their ApplicationsTaylor amp Francis London UK 2002

[4] I V Boikov ldquoNumerical methods of computation of singularand hypersingular integralsrdquo International Journal of Mathe-matics and Mathematical Sciences vol 28 no 3 pp 127ndash1792001

[5] I K Lifanov L N Poltavskii and G M Vainikko Hypersin-gular Integral Equations and Their Applications Chapman ampHallCRC A CRC Press Company London UK 2004

[6] H R Kutt ldquoQuadrature formulae for finite part integralsrdquoReport WISK 178 The National Research Institute for Mathe-matical Sciences Pretoria South Africa 1975

[7] C-Y Hui and D Shia ldquoEvaluations of hypersingular integralsusingGaussian quadraturerdquo International Journal for NumericalMethods in Engineering vol 44 no 2 pp 205ndash214 1999

[8] Y Z Chen ldquoNumerical solution of a curved crack problem byusing hypersingular integral equation approachrdquo EngineeringFracture Mechanics vol 46 no 2 pp 275ndash283 1993

[9] N M A Nik Long and Z K Eshkuvatov ldquoHypersingularintegral equation for multiple curved cracks problem in planeelasticityrdquo International Journal of Solids and Structures vol 46no 13 pp 2611ndash2617 2009

[10] S J Obaiys Z K Eshkuvatov and N M Nik Long ldquoOn errorestimation of automatic quadrature scheme for the evaluationof Hadamard integral of second order singularityrdquo University ofBucharest Scientific Bulletin Series A Applied Mathematics andPhysics vol 75 no 1 pp 85ndash98 2013

[11] J CMason andDCHandscombChebyshev Polynomials CRCPress 2003

[12] M I Israilov Semi-Analytic Solution of Singular Integral Equa-tions of First Kind by Quadrature Formula Numerical Integra-tion and Related Problems FAN Press 1990 (Russian)

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

Page 11: An Automatic Quadrature Schemes and Error Estimates for Semibounded Weighted Hadamard Type Hypersingular Integrals

Abstract and Applied Analysis 11

Table 4 Results of HSIs for 1199082(119905)

119873 119909 Exact solution Approximate solution Absolute error

10

minus099999 000387291557000340015 000387141593292902562 150 times 10minus6

minus091111 031504763162812503076 031505448317559772604 685 times 10minus6

minus077111 041231131092313516294 041230879229200872013 252 times 10minus6

minus033111 044083450592426358350 044083359884530457966 907 times 10minus7

000000 043301270189221932338 043301675286046974386 405 times 10minus6

033111 044963976686376800332 044963389568653105901 587 times 10minus6

077111 062742591722366873321 062741343576104182143 125 times 10minus5

091111 094767578680690190767 094770492350476572039 291 times 10minus5

099999 86066655193121619385 86053938870581623578 0127 times 10minus2

20

minus099999 000387291557000340015 000387291555159212453 184 times 10minus11

minus091111 031504763162812503076 031504763156789346359 602 times 10minus11

minus077111 041231131092313516294 041231131089736085386 258 times 10minus11

minus033111 044083450592426358350 044083450592691374976 265 times 10minus12

000000 043301270189221932338 043301270187830840205 139 times 10minus11

033111 044963976686376800332 044963976684337128944 204 times 10minus11

077111 062742591722366873321 062742591718912403402 346 times 10minus11

091111 094767578680690190767 094767578677558019222 313 times 10minus11

099999 86066655193121619385 86066655108152556362 850 times 10minus8

40

minus099999 000387291557000340015 000387291557000339911 104 times 10minus18

minus091111 031504763162812503076 031504763162812503215 140 times 10minus18

minus077111 041231131092313516294 041231131092313516515 228 times 10minus18

minus033111 044083450592426358350 044083450592426358289 614 times 10minus19

000000 043301270189221932338 043301270189221932269 675 times 10minus19

033111 044963976686376800332 044963976686376800195 131 times 10minus18

077111 062742591722366873321 062742591722366873124 191 times 10minus18

091111 094767578680690190767 094767578680690190858 131 times 10minus18

099999 86066655193121619385 86066655193121618680 705 times 10minus16

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

0012

0010

0008

0006

0004

0002

Abso

lute

erro

r val

ues

(b)

Figure 6 (a) Exact and AQS solutions when 119873 = 10 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 10 and ℎ(119905) = 1(2 + 119905)

12 Abstract and Applied Analysis

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

times10minus8

Abso

lute

erro

r val

ues

(b)

Figure 7 (a) Exact and AQS solutions when 119873 = 20 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 20 and ℎ(119905) = 1(2 + 119905)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

Exact solution plotAQS solution plot

80

70

60

50

40

30

20

10

(a)

minus1 minus05 0 05 1

Singular point values

70

60

50

40

30

20

10

times10minus16

Abso

lute

erro

r val

ues

(b)

Figure 8 (a) Exact and AQS solutions when 119873 = 40 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 40 and ℎ(119905) = 1(2 + 119905)

where ℎ(119905) = 1(2 + 119905) The exact solution is 1198672(ℎ 119909) =

radic(1 + 119909)(1 minus 119909)(radic3(2 + 119909)2)

The numerical results in Table 4 show that the AQS (21)is highly accurate for the rational function ℎ(119905) = 1(2 + 119905)for the case 119894 = 2 and119873 = 40 Figure 6(a) shows that there is

good agreement between the exact solution andAQS solutionwhich can be seen from absolute error in Figure 6(b) when119873 = 10 The absolute error is reduced in Figure 7(b) when119873 = 20 with the corresponding exact and AQS solutionin Figure 7(a) The best result is obtained and illustrated inFigures 8(a) and 8(b) when119873 = 40

Abstract and Applied Analysis 13

6 Conclusion

In this paper we have constructed the automatic quadratureschemes for the semibounded weighted Hadamard typehypersingular integrals using the truncated series of the thirdand fourth kinds of Chebyshev polynomials The unknowncoefficients are found by applying the discrete orthogonalitycondition and the choice of collocation points on (minus1 1)The error estimations for the semibounded solutions of thehypersingular integrals are obtained Numerical exampleshave shown the accuracy of the developed methods andagreed with theoretical findings

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Hadamard Le Probleme de Cauchy et les Equations auxDerivees Particelles Lineaires Hyperboliques Hermann ParisFrance 1932

[2] E Lutz L Gray and A Ingraffea ldquoAn overview of integrationmethods for hypersingular integralsrdquo in Boundary Elements CA Brebbia and G S Gipson Eds vol 8 pp 913ndash925 CMPSouthampton UK 1991

[3] S G Samko Hypersingular Integrals and Their ApplicationsTaylor amp Francis London UK 2002

[4] I V Boikov ldquoNumerical methods of computation of singularand hypersingular integralsrdquo International Journal of Mathe-matics and Mathematical Sciences vol 28 no 3 pp 127ndash1792001

[5] I K Lifanov L N Poltavskii and G M Vainikko Hypersin-gular Integral Equations and Their Applications Chapman ampHallCRC A CRC Press Company London UK 2004

[6] H R Kutt ldquoQuadrature formulae for finite part integralsrdquoReport WISK 178 The National Research Institute for Mathe-matical Sciences Pretoria South Africa 1975

[7] C-Y Hui and D Shia ldquoEvaluations of hypersingular integralsusingGaussian quadraturerdquo International Journal for NumericalMethods in Engineering vol 44 no 2 pp 205ndash214 1999

[8] Y Z Chen ldquoNumerical solution of a curved crack problem byusing hypersingular integral equation approachrdquo EngineeringFracture Mechanics vol 46 no 2 pp 275ndash283 1993

[9] N M A Nik Long and Z K Eshkuvatov ldquoHypersingularintegral equation for multiple curved cracks problem in planeelasticityrdquo International Journal of Solids and Structures vol 46no 13 pp 2611ndash2617 2009

[10] S J Obaiys Z K Eshkuvatov and N M Nik Long ldquoOn errorestimation of automatic quadrature scheme for the evaluationof Hadamard integral of second order singularityrdquo University ofBucharest Scientific Bulletin Series A Applied Mathematics andPhysics vol 75 no 1 pp 85ndash98 2013

[11] J CMason andDCHandscombChebyshev Polynomials CRCPress 2003

[12] M I Israilov Semi-Analytic Solution of Singular Integral Equa-tions of First Kind by Quadrature Formula Numerical Integra-tion and Related Problems FAN Press 1990 (Russian)

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

Page 12: An Automatic Quadrature Schemes and Error Estimates for Semibounded Weighted Hadamard Type Hypersingular Integrals

12 Abstract and Applied Analysis

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

HSI

s val

ues

Exact solution plotAQS solution plot

(a)

minus1 minus05 0 05 1

Singular point values

80

70

60

50

40

30

20

10

times10minus8

Abso

lute

erro

r val

ues

(b)

Figure 7 (a) Exact and AQS solutions when 119873 = 20 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 20 and ℎ(119905) = 1(2 + 119905)

minus1 minus05 0 05 1

Singular point values

HSI

s val

ues

Exact solution plotAQS solution plot

80

70

60

50

40

30

20

10

(a)

minus1 minus05 0 05 1

Singular point values

70

60

50

40

30

20

10

times10minus16

Abso

lute

erro

r val

ues

(b)

Figure 8 (a) Exact and AQS solutions when 119873 = 40 and ℎ(119905) = 1(2 + 119905) and (b) absolute error between exact and AQS solution when119873 = 40 and ℎ(119905) = 1(2 + 119905)

where ℎ(119905) = 1(2 + 119905) The exact solution is 1198672(ℎ 119909) =

radic(1 + 119909)(1 minus 119909)(radic3(2 + 119909)2)

The numerical results in Table 4 show that the AQS (21)is highly accurate for the rational function ℎ(119905) = 1(2 + 119905)for the case 119894 = 2 and119873 = 40 Figure 6(a) shows that there is

good agreement between the exact solution andAQS solutionwhich can be seen from absolute error in Figure 6(b) when119873 = 10 The absolute error is reduced in Figure 7(b) when119873 = 20 with the corresponding exact and AQS solutionin Figure 7(a) The best result is obtained and illustrated inFigures 8(a) and 8(b) when119873 = 40

Abstract and Applied Analysis 13

6 Conclusion

In this paper we have constructed the automatic quadratureschemes for the semibounded weighted Hadamard typehypersingular integrals using the truncated series of the thirdand fourth kinds of Chebyshev polynomials The unknowncoefficients are found by applying the discrete orthogonalitycondition and the choice of collocation points on (minus1 1)The error estimations for the semibounded solutions of thehypersingular integrals are obtained Numerical exampleshave shown the accuracy of the developed methods andagreed with theoretical findings

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Hadamard Le Probleme de Cauchy et les Equations auxDerivees Particelles Lineaires Hyperboliques Hermann ParisFrance 1932

[2] E Lutz L Gray and A Ingraffea ldquoAn overview of integrationmethods for hypersingular integralsrdquo in Boundary Elements CA Brebbia and G S Gipson Eds vol 8 pp 913ndash925 CMPSouthampton UK 1991

[3] S G Samko Hypersingular Integrals and Their ApplicationsTaylor amp Francis London UK 2002

[4] I V Boikov ldquoNumerical methods of computation of singularand hypersingular integralsrdquo International Journal of Mathe-matics and Mathematical Sciences vol 28 no 3 pp 127ndash1792001

[5] I K Lifanov L N Poltavskii and G M Vainikko Hypersin-gular Integral Equations and Their Applications Chapman ampHallCRC A CRC Press Company London UK 2004

[6] H R Kutt ldquoQuadrature formulae for finite part integralsrdquoReport WISK 178 The National Research Institute for Mathe-matical Sciences Pretoria South Africa 1975

[7] C-Y Hui and D Shia ldquoEvaluations of hypersingular integralsusingGaussian quadraturerdquo International Journal for NumericalMethods in Engineering vol 44 no 2 pp 205ndash214 1999

[8] Y Z Chen ldquoNumerical solution of a curved crack problem byusing hypersingular integral equation approachrdquo EngineeringFracture Mechanics vol 46 no 2 pp 275ndash283 1993

[9] N M A Nik Long and Z K Eshkuvatov ldquoHypersingularintegral equation for multiple curved cracks problem in planeelasticityrdquo International Journal of Solids and Structures vol 46no 13 pp 2611ndash2617 2009

[10] S J Obaiys Z K Eshkuvatov and N M Nik Long ldquoOn errorestimation of automatic quadrature scheme for the evaluationof Hadamard integral of second order singularityrdquo University ofBucharest Scientific Bulletin Series A Applied Mathematics andPhysics vol 75 no 1 pp 85ndash98 2013

[11] J CMason andDCHandscombChebyshev Polynomials CRCPress 2003

[12] M I Israilov Semi-Analytic Solution of Singular Integral Equa-tions of First Kind by Quadrature Formula Numerical Integra-tion and Related Problems FAN Press 1990 (Russian)

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

Page 13: An Automatic Quadrature Schemes and Error Estimates for Semibounded Weighted Hadamard Type Hypersingular Integrals

Abstract and Applied Analysis 13

6 Conclusion

In this paper we have constructed the automatic quadratureschemes for the semibounded weighted Hadamard typehypersingular integrals using the truncated series of the thirdand fourth kinds of Chebyshev polynomials The unknowncoefficients are found by applying the discrete orthogonalitycondition and the choice of collocation points on (minus1 1)The error estimations for the semibounded solutions of thehypersingular integrals are obtained Numerical exampleshave shown the accuracy of the developed methods andagreed with theoretical findings

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] J Hadamard Le Probleme de Cauchy et les Equations auxDerivees Particelles Lineaires Hyperboliques Hermann ParisFrance 1932

[2] E Lutz L Gray and A Ingraffea ldquoAn overview of integrationmethods for hypersingular integralsrdquo in Boundary Elements CA Brebbia and G S Gipson Eds vol 8 pp 913ndash925 CMPSouthampton UK 1991

[3] S G Samko Hypersingular Integrals and Their ApplicationsTaylor amp Francis London UK 2002

[4] I V Boikov ldquoNumerical methods of computation of singularand hypersingular integralsrdquo International Journal of Mathe-matics and Mathematical Sciences vol 28 no 3 pp 127ndash1792001

[5] I K Lifanov L N Poltavskii and G M Vainikko Hypersin-gular Integral Equations and Their Applications Chapman ampHallCRC A CRC Press Company London UK 2004

[6] H R Kutt ldquoQuadrature formulae for finite part integralsrdquoReport WISK 178 The National Research Institute for Mathe-matical Sciences Pretoria South Africa 1975

[7] C-Y Hui and D Shia ldquoEvaluations of hypersingular integralsusingGaussian quadraturerdquo International Journal for NumericalMethods in Engineering vol 44 no 2 pp 205ndash214 1999

[8] Y Z Chen ldquoNumerical solution of a curved crack problem byusing hypersingular integral equation approachrdquo EngineeringFracture Mechanics vol 46 no 2 pp 275ndash283 1993

[9] N M A Nik Long and Z K Eshkuvatov ldquoHypersingularintegral equation for multiple curved cracks problem in planeelasticityrdquo International Journal of Solids and Structures vol 46no 13 pp 2611ndash2617 2009

[10] S J Obaiys Z K Eshkuvatov and N M Nik Long ldquoOn errorestimation of automatic quadrature scheme for the evaluationof Hadamard integral of second order singularityrdquo University ofBucharest Scientific Bulletin Series A Applied Mathematics andPhysics vol 75 no 1 pp 85ndash98 2013

[11] J CMason andDCHandscombChebyshev Polynomials CRCPress 2003

[12] M I Israilov Semi-Analytic Solution of Singular Integral Equa-tions of First Kind by Quadrature Formula Numerical Integra-tion and Related Problems FAN Press 1990 (Russian)

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

Page 14: An Automatic Quadrature Schemes and Error Estimates for Semibounded Weighted Hadamard Type Hypersingular Integrals

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


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Abscissas and weights for Guassian quadrature for N=2 to ...

Abscissas and weights for Guassian quadrature for N=2 to ...

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Discrete Walsh-Hadamard Transform in Signal Processing

Hermite-Hadamard inequalities for relative semi-convex functions and applications

Hermite-Hadamard inequalities for relative semi-convex functions and applications

Hilbert spaces from path integrals

Hilbert spaces from path integrals

Conditional Moment Generating Functions for Integrals and Stochastic Integrals

Conditional Moment Generating Functions for Integrals and Stochastic Integrals

12 Multiple Integrals

12 Multiple Integrals

On good matrices, skew Hadamard matrices and optimal designs

On good matrices, skew Hadamard matrices and optimal designs

Iterative controller tuning using Bode's integrals

Iterative controller tuning using Bode's integrals

A general treatment of crack tip contour integrals

A general treatment of crack tip contour integrals

Differential Quadrature Method for Two-Dimensional Burgers' Equations

Differential Quadrature Method for Two-Dimensional Burgers' Equations

Poisson integrals of regular functions

Poisson integrals of regular functions

Higher differentiability of minimizers of convex variational integrals

Higher differentiability of minimizers of convex variational integrals

TRIGONOMETRIC FUNCTION INTEGRALS

TRIGONOMETRIC FUNCTION INTEGRALS

Leading singularities and off-shell conformal integrals

Leading singularities and off-shell conformal integrals

A regularity theorem for minimizers of quasiconvex integrals

A regularity theorem for minimizers of quasiconvex integrals

Feynman-Jackson integrals

Feynman-Jackson integrals

Hyperbolicity, transversality and analytic first integrals

Hyperbolicity, transversality and analytic first integrals

Hadamard-type integrals as G-transforms

Hadamard-type integrals as G-transforms