On a Generalization of Euler's Constant

Surveys in Mathematics and its Applications

ISSN 1842-6298 (electronic), 1843-7265 (print)

Volume 16 (2021), 259 – 274

ON A GENERALIZATION OF EULER’S CONSTANT

Stephen Kaczkowski

Abstract. A one parameter generalization of Euler’s constant γ from [Numer. Algorithms 46(2)

(2007) 141–151] is investigated, and additional expressions for γ are derived. Included are forms

involving the Gregory coefficients and the Hurwitz Zeta function, and expressions of the later are

shown to lead to an alternative proof of Lerch’s limit formula for the gamma function.

1 Introduction

Euler’s constant γ is defined by the straightforward limit

γ = limn→∞

[n∑

k=1

1

k− ln(n)

]= 0.5772156649 · · · , (1.1)

and is seen in a variety of diverse areas of mathematics while being featured prominentlyin both analysis and number theory. Its exact nature is still unknown – to this dayno legitimate mathematical proof of its rationality or irrationally has been found.Given its ubiquity and mystery, it is not surprising that a variety of generalizationsof γ have been put forth.

In this paper the following generalization of γ will be explored in detail:

γ (a) = limn→∞

(n∑

k=1

1

k + a− 1−∫ n

1

dx

x+ a− 1

), (1.2)

where a > 0, and for which γ (1) = γ. This one parameter function γ (a) from (1.2)was formulated by Sıntamarian in [21], and can be related to some other well-knowngeneralizations of γ. One of these is the generalized Stieltjes constants defined by

γn (a) = limm→∞

[m∑k=0

(ln (k + a))n

k + a− (ln (m+ a))n+1

n+ 1

],

2020 Mathematics Subject Classification: 30B10; 33B15; 11M35; 30E20; 11M06Keywords: Euler’s Constant; Gamma Function; Hurwitz Zeta Function

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260 Stephen Kaczkowski

where a > 0 and n is a non-negative integer. The special case of a = 1 arethe standard Stieltjes constants, which appear in the regular part of the Laurentexpansion of the Hurwitz Zeta function given by

ζ (s, a) =∞∑n=0

(n+ a)−s (1.3)

for Re(s) > 1 and Re(a) > 0. The Laurent expansion, which is valid under the sameconditions, is

ζ (s, a) =1

s− 1+

∞∑n=0

(−1)n

n!γn (a) (s− 1)n.

In terms of the Stieltjes notation (1.2) becomes γ (a) = γ0 (a) + ln(a) for a > 0.Another prominent generalization of γ which can be related to γ (a) is the Euler-

Lehmer constants [17] given by

γ(a, q) = limn→∞

n∑0 < k ≤ n

k ≡ a mod q

[1

k− ln (n)

q

], (1.4)

where a and q are integers satisfying 0 < a ≤ q. In [19] Murty and Saradhademonstrated the usefulness of these constants with regard to the nature of γ byproving that at least one number in the following infinite list is algebraic:

γ (a, q) , 1 ≤ a < q, q ≥ 2.

Using this notation our expression for γ (a) from (1.2) can be expressed by

γ (a) = γ (a, 1) + ln(a) (1.5)

for positive integral a.Other generalizations include the q-analogue of the Euler constant discussed in

[14] and denoted here by [γ]q for q > 1. Its expression is

[γ]q =∞∑n=1

1

[n]q+(q − 1) ln (q − 1)

ln (q)−(q − 1)

2,

where [n]q = (qn − 1)/(q − 1). In [14] it is shown that limq→1+ [γ]q = γ, and that[γ]2 is irrational. Additionally, there are the generalized Euler’s constants associatedwith the function y = 1/xα for α > 0 which are given by

γα = limn→∞

[n∑

k=1

1

kα−∫ n

1

1

xαdx

], (1.6)

******************************************************************************Surveys in Mathematics and its Applications 16 (2021), 259 – 274




Euler’s Constant 261

and where γ1 = γ. Because this generalization and the Stieltjes constants contain asummation along with a corresponding definite integral, they are sometimes addressedin the literature in the context of the Euler-Maclaurin summation formula [2],[6].

In contrast to some of these other generalizations, γ (a), as it is given in (1.2),has not been examined in as great detail in the literature related to Euler’s constant.In the following analysis a variety of expressions for γ (a) will be derived and shownto be consistent with analogous results for Euler’s constant as the parameter aapproaches 1. These expressions include generalizations of a few of the most well-known expressions for γ, and in many cases, references explaining other knownmethods for proving a given result will be provided. The various expressions for γ (a)given here reveal its relationship to the Gregory coefficients and the Hurwitz zetafunction. Several identities involving γ (a) are brought together from the literaturethrough its close connection to the digamma function ψ, and other straightforwardresults are derived. Additionally, asymptotic formulas for (1.2) are derived for bothsmall and large values of the parameter a, and other derived results are used toevaluate a class of definite integrals. This work will also motivate and lead to otherexpressions for Euler’s constant, as well as to another avenue to prove a limit formulaattributed to Lerch given in [14] and [18].

2 A Classic Definite Integral

A definite integral expression for γ (a) can be derived using a modification of amethod that starts from the definition of γ in (1.1) and leads to the following identitywhich has been attributed to Euler [15]:

γ =

∫ 1

0

(1

z+

1

ln (1− z)

)dz. (2.1)

One way to arrive at (2.1) is through the Frullani integral [12, p. 406] given by

ln

(b

a

)=

∫ ∞

0

(e−at − e−bt

)t

dt, (2.2)

where a and b are arbitrary positive real numbers. This expression can be obtainedby separately evaluating each of the following integrals:∫ b

a

∫ ∞

0e−txdxdt =

∫ ∞

0

∫ b

ae−txdtdx.

Leaving a unchanged and letting b = n+ a− 1 in the Frullani integral expressionleads to

ln

(n+ a− 1

a

)=

∫ ∞

0

e−at − e−(n+a−1)t

tdt. (2.3)






Define Ha,n =∑n

k=1 (k + a− 1)−1. Then

Hn,a =n−1∑k=0

(∫ ∞

0e−(k+a)tdt

)=

∫ ∞

0

(e−at

n−1∑k=0

e−kt

)dt =

∫ ∞

0

e−at(1− e−nt

)1− e−t dt.

Subtracting (2.3) from this expression for Hn,a, and using the definition in (1.2)yields

γ (a) = limn→∞

∫ ∞

0

[e−at

(1− e−nt

)1− e−t −

(e−at − e−(n+a−1)t

)t

]dt.

Simplifying and moving the limit into the integral leads to

γ (a) =

∫ ∞

0e−at

(1

1− e−t− 1

t

)dt =

∫ 1

0(1− x)a−1

(1

x+

1

ln (1− x)

)dx, (2.4)

where the second integral in (2.4) arises from the substitution x = 1 − e−t, and asexpected, the case of a = 1 leads back to (2.1). It will be seen that the first integralin (2.4) can also be obtained from (5.6) along with Binet’s formula for the digammafunction in [4, Vol. 1, Sect. 1.7, Eq. 22].

Another integral expression that can be reduced to the rightmost equation of(2.4) is

γ(a) =

∫ 1

0

∫ 1

0

ta−1 (1− tx)

1− tdtdx.

This equivalence can be seen by starting with this double integral, switching theorder of integration, and completing only the integration with respect to x. Incidentally,the case of a = 1 yields the concise result

γ =

∫ 1

0Hxdx

where Hx =∫ 10 (1− tx)/(1− t)dt is Euler’s generalization of the harmonic numbers

for real x. This function is so defined because when n is a positive integer

Hn =n∑

k=1

1

k=

∫ 1

0

(n−1∑k=0

tk

)dt =

∫ 1

0

1− tn

1− tdt.

3 Another Expression for γ

Two methods for obtaining integral expressions for the harmonic numbers Hn havebeen used in the previous work. A key component in these procedures is the use ofelementary definite integral expressions for 1/k such as

1

k=

∫ ∞

0e−ktdt =

∫ 1

0tk−1dt.






In a similar manner, by integrating functions of the form 1/xk+1, another expressionfor Hn can be found. This method will be used in the proof of the following theorem.

Theorem 1. Euler’s constant can be obtained from the following limit:

γ = limn→∞

∫ n

1

2− x− x−n

x (x− 1)dx. (3.1)

Proof. Express the harmonic numbers Hn in the form

Hn =n∑

k=1

(∫ n

1

1

xk+1dx

)+ En,

where En =∑n

k=1 1/(knk

). From here, interchange the integral and the summation,

simplify the resulting geometric series, and use (1.1) to obtain

Hn − ln(n) =

∫ n

1

(1− x−n

x (x− 1)− 1

x

)dx+ En. (3.2)

Because 0 < En <∑∞

k=1 1/(knk

)= ln (n/(n− 1)) , the limn→∞En = 0. Equation

(3.1) follows after simplifying (3.2) and appealing to (1.1).

This unusual expression for Euler’s constant is not found among the standard listsof expressions for γ in references such as [7],[8],[11], and [25].

A γ (a) analog of (3.1) can be obtained by carrying out the same steps of theabove process except that the integration is performed on terms of the form 1/x(k+a).After summing and utilizing (1.2) the result is

γ (a) = limn→∞

∫ n

1

(1− x−n) (x+ a− 1)− xa (x− 1)

xa (x− 1) (x+ a− 1)dx .

4 Gregory Coefficients

In addition to forms involving definite integrals, there are several series expansionsfor Euler’s constant that can be generalized to expansions for γ(a). In this sectionone particular series for γ is derived from first principles, and this result will bereferred to again in the next section where an analogous form for γ(a) is obtainedfrom a result in the literature for the digamma function ψ.

A series for γ consisting of only positive terms can be obtained by expanding theintegrand of (2.1) into a Maclaurin series and dividing to obtain

f (z) =ln (1− z) + z

zln (1− z)=

∑∞n=0 z

n/(n+ 2)∑∞n=0 z

n/(n+ 1)=

∞∑n=0

anzn.






Upon clearing fractions(1 +

z

2+z2

3+ . . .

)( ∞∑n=0

anzn

)=

1

2+z

3+z2

4+ . . . ,

and after multiplying and equating powers, the coefficients an can be shown to satisfy

an =1

(n+ 2)−

n−1∑k=0

akn+ 1− k

,

where the first four values of the sequence {an} are

a0 =1

2, a1 =

1

12, a2 =

1

24, a3 =

19

720, a4 =

3

160.

These are the absolute values of the Gregory coefficients which are also referredto as Bernoulli numbers of the second kind [13, p. 285]. (This sequence was firstintroduced by James Gregory (1638–1675) in the context of numerical integration.)If the Gregory coefficients are denoted by Gn, then an = |Gn+1| and Gn+1 =(−1)nan. Integrating f (z) over the interval [0, 1] yields the Fontana-Mascheroniseries

γ =1

2+

1

24+

1

72+

19

2880+

3

800+ · · · =

∞∑n=1

|Gn|n

. (4.1)

Scott in [20] provides an alternative derivation of this formula by expanding f (z)after rewriting it in the following form:

f (z) =1

z−∫ 1

0

(1− z)t

zdt =

1

z

[1−

∫ 1

0

(1− tz + t (t− 1)

z2

2− · · ·

)dt

]. (4.2)

Equating like powers of z in this expression with those of the series∑∞

n=0 anzn

demonstrates that the coefficient an can be expressed by the definite integral

an =1

(n+ 1)!

∫ 1

0t (1− t) (2− t) · · · (n− t) dt.

5 The Connection Between γ (a) and ψ (a)

Some useful expressions for γ (a) can be established through its relationship to thegamma and digamma functions Γ and ψ. To proceed in this direction, we state forreference the following well-known definite integral form of Γ(z) valid for Re (z) > 0

Γ (z) =

∫ ∞

0tz−1e−tdt. (5.1)






The digamma function ψ can be defined by

ψ (z) =d [ln Γ (z)]

dz=

Γ′ (z)

Γ (z). (5.2)

With these definitions in place, we can proceed to a straightforward expression whichcasts γ (a) within the framework of the gamma function. Differentiating (5.1) withrespect to z and applying the definition in (5.2) yields

ψ (z) =1

Γ (z)

∫ ∞

0t(z−1)ln (t) e−tdt. (5.3)

Then from (5.6)

γ (a) = ln(a)− 1

Γ (a)

∫ ∞

0t(a−1)ln (t) e−tdt. (5.4)

Another relationship between γ (a) and the gamma function is given in thefollowing theorem:

Theorem 2. For Re(z) = −1,−2,−3, . . .

Γ (z + 1) = aze−γ(a)z · limn→∞

[n∏

k=1

k

z + kexp

(z

k + a− 1

)].

Proof. The function Γ (z + 1) can be expressed, as in [26, p. 231], by

Γ (z + 1) = limn→∞

nz ·n∏

k=1

(k

z + k

).

Since limn→∞ ln ((n+ a− 1)/n) = 0 , definition (1.2) is equivalent to

γ (a) = limn→∞

[Ha,n − ln (n/a)] . (5.5)

Since∏n

k=1 exp (−z/(k + a− 1)) = e−zHa,n ,

Γ (z + 1) = limn→∞

ezln(a) · e−z(Ha,n−ln(n)+ln(a))n∏

k=1

(k

z + k

)exp

(z

k + a− 1

),

and after evaluating the limit in the exponential function, the theorem follows.

This theorem is a generalization of the Weierstrass definition of the gamma function

Γ (z + 1) = e−γz∞∏k=1

(1 +

z

k

)−1ez/n ,






which can be found in [26, p. 230] and [1, p. 255, Eq. 6.1.3]. Taking the naturallogarithm, differentiating the result in Theorem 1, and using (5.2) yields

ψ (z + 1) = ln(a)− γ (a) +∞∑k=1

(1

k + a− 1− 1

z + k

),

and upon substituting z = a− 1 into the above we obtain

γ (a) = ln(a)− ψ (a) . (5.6)

Letting a = 1 in (5.6) brings us to the well-known result γ = −ψ (1). Another wayto arrive at (5.6) is to start with the following expression for ψ (a) found in [4, Vol.1, Sec. 1.7, Eq. 2]:

ψ (a) = limn→∞

[ln (n) −

n∑k=0

1

k + a

]. (5.7)

Then express the definition for γ (a) in (1.2) with the integral portion evaluated.That is,

γ (a) = limn→∞

[n−1∑k=0

1

k + a− ln (n+ a− 1)

]+ ln (a).

Adding and subtracting an ln (n), and doing the same with an additional series term(n+ a)−1 yields

γ (a) = ln (a)− limn→∞

[(ln (n) −

n∑k=0

1

k + a

)+

1

n+ a+ln

(1 +

a− 1

n

)]. (5.8)

The last two terms within the square brackets vanish in the limit, and upon comparingthe resulting expression to (5.7) the identity in (5.6) follows.

Equation (5.6) reveals significant differences between ψ (a) and γ (a). The functionγ (a) is decreasing on (0,∞), and its range, like its domain, is the positive reals.In contrast ψ (a) is increasing on the same domain; its range is (−∞,∞); and itcrosses the horizontal axis at approximately 1.4616321. Furthermore, the integrandand summand in the expression for γ (a) from (1.2) share a convenient symmetrywhich is not found in the formula for ψ (a) given in (5.7) – as the integral portionthere is given by ln (n) =

∫ n1 (1/t) dt.

A result similar to (5.6) has been obtained in terms of the Euler-Lehmer coefficientsin Theorem 7 of [17]. This theorem states that −qγ (a, q) = ψ (a/q) + ln(q), whichalong with (1.5) can be combined to bring us to (5.6) for the case of q = 1 andpositive integral values of a.

Many other expressions involving γ (a) can be obtained by utilizing (5.6) inconjunction with established formulas for ψ found in references such as [1],[4],[9],






[23], and [26]. A few examples are listed below.

γ (1/2) = γ + ln (2) γ (1/3) = γ + ln√3 + π/

(2√3)

(5.9)

γ (n) = γ −Hn−1 + ln (n) , n ∈ N γ(a+ 1) = γ(a) + ln

(1 +

1

a

)− 1

a(5.10)

γ′(1) = 1− π2

6

1

2x≤ γ(x) ≤ 1

x, x ∈ R+ (5.11)

Gauss’s digamma theorem [9, Eq. 8.3636] can be used to obtain additional γ (a)values at rational a along with those given above for a = 1/2 and a = 1/3. Equations6.3.2, 6.3.5, and 6.4.2 of [1] can be used to produce the remaining equalities listedabove, and the inequality 1/(2x) ≤ γ (x) ≤ 1/x, follows from Eq. 2.2 in [3].

In a similar manner, asymptotic formulas series can also be obtained. Forexample, by combining (5.6) with the rational zeta series for ψ (x+ 1) found in8.363 of [9] along with the digamma recurrence relation of 8.365 in [9] leads to

γ (x) = γ +1

x+ ln (x) +

∞∑n=2

ζ (n)(−x)n−1,

and therefore

γ (x) ∼ γ + 1/x+ ln (x)− x(π2/6

), x→ 0.

Differentiating the Stirling series as it is expressed in Eq. 6.1.40 of [1], and usingthe definition in (5.2) along with (5.6) yields

γ (x) ∼ 1

2x+

N∑n=1

B2n

2nx2n+O

(x−2N−1

), x→ ∞

where the B2n are Bernoulli numbers.Another series for γ (a), which can be obtained from (5.6), is the following

generalization of the Gregory coefficient series from (4.1):

γ (a) =

∞∑n=1

(|Gn|n

·n∏

k=1

k

a+ k − 1

).

This result clearly reduces to γ in the case of a = 1, and was obtained using (5.6)along with the formula for ψ found in Eq. 93 of [5].

6 γ(a) in Terms of the Hurwitz Zeta Function

The Hurwitz Zeta function ζ (s, a) from (1.3) has been used to obtain an expressionsfor γ(a) in [16], which will be discussed in this section. Additionally, a form for γ(a)






in terms of a series of Hurwitz Zeta function values is also derived. All of the resultspresented below are analogous to the well-known Riemann Zeta function expressionsfor γ. The first of such forms for γ(a) is given in the following theorem:

Theorem 3. Sıntamarian’s expression for γ (a) in (1.2) can be obtained from thefollowing formula:

γ (a) =1

a−

∞∑n=2

1

n

(ζ (n, a)− 1

an

). (6.1)

Proof. By expressing (1.2) in the form

γ (a) = limn→∞

[1

a+

n−1∑k=1

(1

a+ k+ ln

(1− 1

a+ k

) )]= lim

n→∞γn−1 (a) , (6.2)

and expanding the logarithm we obtain

1

a− γn−1 (a) =

n−1∑k=1

∞∑m=2

1

m(a+ k)m.

The double series consists of positive terms and is convergent; therefore, summationscan be interchanged, and the above result expressed in terms of ζ (m, a) is as follows:

1

a− γn−1 (a) =

∞∑m=2

1

m

(ζ (m, a)− 1

am

)−

∞∑m=2

1

m

∞∑k=n

1

(a+ k)m. (6.3)

The inside sum of the last term is less than∫∞a+n−1 t

−mdt. Therefore

∞∑m=2

1

m

∞∑k=n

1

(a+ k)m<

∞∑m=2

1

m(m− 1)(a+ n− 1)m−1 <1

2 (a+ n− 2)

for n ≥ 2, where the last inequality was obtained by first fixing m to a value of 2in the m(m − 1) factor, and afterwards summing the geometric series. Then (6.1)follows after applying the limit as n→∞ to the expression in (6.3).

Lampert in [16] provides the following formula for γ (a) involving the HurwitzZeta function:

γ (a) =

∞∑n=2

(−1)n

nζ (n, a) .

This alternating series expression is proved in detail in [19] directly from Sıntamarian’sdefinition of γ(a) in reference [21].






An Alternative Proof of Lerch’s Limit FormulaForms involving limits of the Hurwitz zeta functions can also be used to obtainexpressions for γ (a) which can motivate different paths to some results in classicalanalysis. One way this can be done is through the following expression which wasobtained by adding the parameter x where x ≥ 1 within the definition from (1.2):

G (x, n; a) =

n∑k=1

1

(k + a− 1)x−∫ n

1

du

(u+ a− 1)x.

Evaluating this integral and applying a limit on n yields

limn→∞

G (x, n; a) = ζ (x, a)− a1−x(x− 1)−1, (6.4)

and this convergence is uniform for any x > 1. Whereas from (1.2)

γ (a) = limn→∞

[lim

x→1+G (x, n; a)

]because for each n, the limx→1+ G (x, n; a) = γn (a) . (Recall that γn (a) is definedin (6.2).) Therefore, conditions for the the Moore-Osgood Theorem [24, p. 140] aremet, and consequentially it is permissible to switch the order of the limits. Thus

γ (a) = limx→1+

[ζ (x, a)− 1

a1−x (x− 1)

].

This result is an analog of “Lerch’s limit formula” from [18] (and cited in [14]) whichcan be written as

−ψ (a) = limx→1+

[ζ (x, a)− 1

x− 1

].

The connection between these two limit formulas can be clarified further by (5.6)along with the fact that limx→1+

(1− a1−x

)/(x− 1) = ln (a) . In either case, setting

a = 1 yields the more familiar result

γ = limx→1+

[ζ (x)− 1

x− 1

].

(See e.g., [26, Sec. 13.21, p. 265].)There are other classical expressions for γ that consist of a limiting difference

between two terms, each of which increase without bound under a common limit.Examples include the definition of γ in (1.1) and the expression

γ = limx→0+

[1

x− Γ (x)

],

which is given in [11, p. 109]. Through a direct application of the previous work,another expression for γ along these lines can be found, and which is spelled out inthe next theorem.






Theorem 4. Euler’s constant is given by

γ = limx→1+

[−ln

(1− 1

x

)−

∞∑n=2

ζ (n, x)

n

]. (6.5)

Proof. If a > 1 in Theorem 3, then the sum of the 1/an terms in (6.1) convergesto −ln (1− 1/a) . Therefore

γ (a) =1

a−

∞∑n=2

1

n

(ζ (n, a)− 1

an

)= −ln

(1− 1

a

)−

∞∑n=2

ζ (n, a)

n.

Replacing a with x, and taking the limit as x approaches 1 completes the proof.

This expression for Euler’s constant is not listed among the standard collections offormulas for γ found in the references below.

Further confirmation of the validity of (6.5) can be found from the followingintegral form of Hurwitz Zeta function [26, Ch. 13, p. 260] which is valid for n > 1and x > 0:

ζ (n, x) =1

Γ (n)

∫ ∞

0

tn−1

ext (1− e−t)dt.

Using nΓ (n) = n! along with the above integral, the series term in (6.5) can beexpressed as a definite integral. That is,

∞∑n=2

ζ (n, x)

n=

∫ ∞

0

1

text (1− e−t)

∞∑n=2

(tn

n!

)dt =

∫ ∞

0

et − t− 1

text (1− e−t)dt. (6.6)

Next write the natural logarithm term in (6.5) in terms of a Frullani integral asfollows:

− ln

(1− 1

x

)= ln

(x

x− 1

)=

∫ ∞

0

e−xt − e−t(x−1)

tdt. (6.7)

After subtracting the integral in (6.6) from the result in (6.7) it can be seen that theexpression within the limit of (6.5) is equivalent to γ (x) as expressed by the firstintegral in (2.4). Therefore (6.5) follows from the fact that limx→1+ γ(x) = γ.

The sequence {γn−1(a)}∞n=2 as defined in (6.2) is monotonically increasing andconverges pointwise to γ (a). And since each function γn−1(a) is differentiable for a ∈(0,∞), the nth derivative of γ (a) can be obtained via term by term differentiationof (6.2)

γ(n) (a) = (−1)nn!

[ζ (n+ 1, a)− 1

nan

]. (6.8)

(This result also follows from a fundamental expression for the polygamma function(cf. Eq. 6.4.2 on p. 260 of [1].) An incidental application of the above work is






the evaluation of a family of definite integrals. That is, equating (6.8) with the nthderivative of (2.4) with respect to a leads to the following identity:∫ 1

0xa−1(lnx)n

(1

1− x+

1

lnx

)dx= (−1)nn!

[ζ (n+ 1, a)− 1

nan

].

In the appendix below, additional expressions for γ(a) are provided along withsome brief notes explaining how they can be derived.

References

[1] M. Abramowitz, I. A. Stegun, Handbook of Mathematical Functions: WithFormulas, Graphs, and Mathematical Tables Applied mathematics series,National Bureau of Standards, Washington, DC, 1964.

[2] T. M. Apostol, An elementary view of Euler’s summation formula, Amer. Math.Monthly 106(5) (1999) 409–418. MR1699259. Zbl 1076.41509.

[3] A. Alzer, On some inequalities for the gamma and psi functions, Math. Comp.66(217) (1997) 373–389. MR1388887. Zbl 0854.33001.

[4] H. Bateman, A. Erdelyi, Higher Transcendental Functions [in 3 volumes], McGraw Hill Book Company, 1955.

[5] I. V. Blagouchine, Two series expansions for the logarithm of the gammafunction involving Stirling numbers and containing only rational coefficients forcertain arguments related to π−1, J. Math. Anal. Appl. 442(2) (2016) 404–434.MR3504006. Zbl 1339.33003.

[6] J. Bohman, C. Froberg, The Stieltjes function—definition and properties, Math.Comp. 51(183) (1988) 281–289. MR0942155. Zbl 0654.10038.

[7] T. P. Dence, J. B. Dence, A survey of Euler’s constant, Mathematics Magazine82(4) (2009) 255–265. MR2571789. Zbl 1227.11128.

[8] X. Gourdon, P. Sebah, Numbers, constants and computation, Webpage(accessed April 8, 2020).

[9] I. S. Gradshteyn, I. M. Ryzhik, Tables of Integrals, Series, and Products, 6thed. San Diego, CA: Academic Press, 2000.

[10] J. Guillera, J. Sondow, Double integrals and infinite products for some classicalconstants via analytic continuations of Lerch’s transcendent, Ramanujan J.16(3) (2008) 247–270. MR2429900. Zbl 1216.11075.



https://mathscinet.ams.org/mathscinet-getitem?mr=1699259

https://zbmath.org/?q=an:1076.41509









http://numbers.computation.free.fr/Constants/Gamma/gammaFormulas.html






[11] J. Havil, Gamma: Exploring Euler’s Constant, Princeton University Press,Princeton, NJ, (2003).

[12] H. Jeffreys, B. S. Jeffreys, Frullani’s Integrals, §12.16 in Methods ofMathematical Physics, 3rd ed. Cambridge University Press, Cambridge,England, 1988.

[13] C. Jordan, K. Jordan, Calculus of Finite Differences 33, AmericanMathematical Soc., Providence, Rhode Island, 1965.

[14] N. Kurokawa, M. Wakayama, On q-analogues of the Euler constant and Lerch’slimit formula, Proc. Amer. Math. Soc. 132(4) (2004) 935–943. MR2045407. Zbl1114.33022.

[15] J. Lagarias, Euler’s constant: Euler’s work and modern developments, Bulletinof the American Mathematical Society 50(4) (2013) 527–628. MR3090422. Zbl1282.11002.

[16] V. Lampret, An interplay between a generalized-Euler-constant function andthe Hurwitz zeta function, Bull. Belg. Math. Soc., 17(4) (2010) 741–747.MR2778449. Zbl 1218.11081.

[17] D. Lehmer, Euler constants for arithmetic progressions, Acta Arith. 27 (1975)125–142. MR0369233. Zbl 0302.12003.

[18] M. Lerch, Dalsi studie v oboru Malmstenovskych rad, Rozpravy Ceske Akad,3(28) (1894) 1–61.

[19] M. R. Murty, N. Saradha, Euler–Lehmer constants and a conjecture of Erdos,J. Number Theory 130(12) (2010) 2671–2682. MR2684489. Zbl 1204.11114.

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[21] A. Sıntamarian, A Generalization of Euler’s Constant, Numer. Algorithms46(2) (2007) 141–151, https://doi.org/10.1007/s11075-007-9132-0. MR2358246.Zbl 1130.11075Z.

[22] J. Sondow, Double Integrals for Euler’s Constant and ln(4/pi) and anAnalog of Hadjicostas’s Formula, Amer. Math. Monthly 112(1) (2005) 61–65.MR2110113. Zbl 1138.11356.

[23] H. M. Srivastava, J. Choi, Zeta and q-Zeta Functions and Associated Series andIntegrals, Elsevier, 2012.

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Appendix: Other Expressions for γ (a)

In this appendix several additional expressions for γ (a) are given and explained.The first such expression can be obtained by applying the substitution x = e−t tothe integral in (5.4) to yield

γ(a) = ln(a)− 1

Γ (a)

∫ 1

0

(ln

(1

x

))a−1

ln

(ln

(1

x

))dx.

Another expression is

γ (a) =1

a−∫ ∞

1

(x− ⌊x⌋) dx(x+ a− 1)2

,

which can be verified by evaluating the integral over intervals of the form [n, n+ 1],summing over natural numbers n and, and then appealing to the definition in (1.2).

Using (5.6) along with known expressions for the digamma function yields

γ(a) =

∫ ∞

0

(1

(t+ 1)a− e−at

)dt

t=

1

2a+ 2

∫ ∞

0

t

(t2 + a2) (e2πt − 1)dt.

Additionally, we have the following expression for γ (a)− ln(a) given by

1

Γ (a)

∫ 1

0xa−2

(1− e−x

)(1 + (a− 1) lnx)dx− 1

Γ (a)

∫ 1

0

(1 + (1− a) lnx) e−1/x

xadx,

which can be derived using a procedure analogous to one used in [11, p. 107] forderiving the corresponding γ expression. Then applying a Maclaurin series expansionof (1− e−x) to the first integral term of the above expression yields

γ (a)=ln(a) +1

Γ (a)

[ ∞∑n=0

(−1)n

n! (a+ n)2−∫ ∞

1

(1 + (a− 1) lnx)

x(2−a)exdx

].

Finally, the double integral

γ (a) = −∫ 1

0

∫ 1

0

(1− x) (xy)a−1

ln (xy) (1− xy)dydx. (6.9)

can be obtained using a modified form of a procedure outlined in [22], which will beexplained in detail below.



http://mathworld.wolfram.com/Euler-MascheroniConstant.html




A path connecting (1.2) to (6.9) can be established through the expression

I (a) =

∞∑n=0

(∫ 1

0

∫ 1

0

∫ ∞

n(1− x)(xy)t+a−1dtdydx

),

by showing that the quantity I (a) is equivalent to γ (a). This can be accomplishedby first switching the order of integration in I (a) and evaluating the x and y portionsas follows:(∫ 1

0yt+a−1dy

)(∫ 1

0

(xt+a−1 − xt+a

)dx

)=

(1

t+ a

)(1

t+ a− 1

t+ a+ 1

).

After distributing and expanding, the resulting integral in the variable t becomes∫ ∞

n

[1

(t+ a)2− 1

t+ a+

1

t+ a+ 1

]dt =

1

n+ a+ ln

(n+ a

n+ a+ 1

).

Then applying a limit to the summation in the definition of I (a) yields

I(a) = limN→∞

N∑n=0

[1

n+ a+ ln

(n+ a

n+ a+ 1

)]= lim

N→∞

[Ha,N − ln

(N + a− 1

a

)],

where the expression on the right is γ (a). To complete the task of verifying (6.9),it is necessary to show that I (a) is also equivalent to the right hand side of (6.9).Integrating within I (a) with respect to t only, moving the summation into theintegral, and using the formula for the sum of a geometric series with ratio xy yields

∞∑n=0

[∫ ∞

n(1− x)(xy)t+a−1dt

]=

(x− 1)

ln (xy)

∞∑n=0

(xy)n+a−1 =(x− 1) (xy)a−1

(1− xy) ln (xy),

which is the integrand of (6.9). Applying the double integral in x and y to bothsides of the above completes the identity.

Equation (6.9) can also be obtained through the digamma expression in Corollary4.1 of [10], which was derived from the Lerch transcendent.

Stephen Kaczkowski

South Carolina Governor’s School for Science and Mathematics,

401 Railroad Avenue

Hartsville, SC 29550 USA.

e-mail: [email protected]

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On a Generalization of Euler's Constant

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