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# Certain integral representations of Stieltjes constants $γ n$

## Abstract

A remarkably large number of integral formulas for the Euler-Mascheroni constant γ have been presented. The Stieltjes constants (or generalized Euler-Mascheroni constants) $γ n$ and $γ 0 =γ$, which arise from the coefficients of the Laurent series expansion of the Riemann zeta function $ζ(s)$ at $s=1$, have been investigated in various ways, especially for their integral representations. Here we aim at presenting certain integral representations for $γ n$ by choosing to use three known integral representations for the Riemann zeta function $ζ(s)$. Our method used here is similar to those in some earlier works, but our results seem a little simpler. Some relevant connections of some special cases of our results presented here with those in earlier works are also pointed out.

MSC:11M06, 11M35, 11Y60, 33B15.

## 1 Introduction and preliminaries

The Riemann zeta function $ζ(s)$ is defined by (see, e.g., [, Section 2.3])

$ζ(s):={ ∑ n = 1 ∞ 1 n s = 1 1 − 2 − s ∑ n = 1 ∞ 1 ( 2 n − 1 ) s ( ℜ ( s ) > 1 ) , 1 1 − 2 1 − s ∑ n = 1 ∞ ( − 1 ) n − 1 n s ( ℜ ( s ) > 0 ; s ≠ 1 ) ,$
(1.1)

which is an obvious special case of the Hurwitz (or generalized) zeta function $ζ(s,a)$ defined by

$ζ(s,a):= ∑ k = 0 ∞ ( k + a ) − s ( ℜ ( s ) > 1 ; a ∈ C ∖ Z 0 − ) ,$
(1.2)

where and $Z 0 −$ denote the sets of complex numbers and nonpositive integers, respectively. Both the Riemann zeta function $ζ(s)$ and the Hurwitz zeta function $ζ(s,a)$ can be continued meromorphically to the whole complex s-plane, except for a simple pole only at $s=1$, with their respective residue 1, in many different ways. The Stieltjes constants $γ n$ for $n∈ N 0 :=N∪{0}$, $N:={1,2,3,…}$, arise from the following Laurent expansion of the Riemann zeta function $ζ(s)$ about $s=1$ (see, e.g., [, pp.166-169], [, p.255] and [, p.165]):

$ζ(s)= 1 s − 1 + ∑ n = 0 ∞ ( − 1 ) n n ! γ n ( s − 1 ) n ,$
(1.3)

where

$γ n = lim m → ∞ { ∑ k = 1 m ( log k ) n k − ∫ 1 m ( log x ) n x d x } = lim m → ∞ { ∑ k = 1 m ( log k ) n k − ( log m ) n + 1 n + 1 } ( n ∈ N 0 )$
(1.4)

and, in particular, $γ 0$ (denoted by γ) is the Euler-Mascheroni constant (see, for details, [, Section 1.5] and [, Section 1.2]):

$γ:= lim m → ∞ ( ∑ k = 1 m 1 k − log m ) ≅0.5772156649⋯.$
(1.5)

The Stieltjes constants $γ n$ are named after Thomas Jan Stieltjes and often referred to as generalized Euler-Mascheroni constants. Liang and Todd  computed numerical approximations of the first 20 Stieltjes constants in 1972. In 1985, using contour integration, Ainsworth and Howell  showed that

$γ n =2ℜ { ∫ 0 ∞ ( x − i ) ( log ( 1 − i x ) ) n ( 1 + x 2 ) ( e 2 π x − 1 ) d x } (n∈N)$
(1.6)

and

$γ = γ 0 = 1 2 + 2 ℜ { ∫ 0 ∞ ( x − i ) ( 1 + x 2 ) ( e 2 π x − 1 ) d x } = 1 2 + 2 ∫ 0 ∞ x ( 1 + x 2 ) ( e 2 π x − 1 ) d x .$
(1.7)

By using binomial theorem, we have

$( log ( 1 − i x ) ) 2 m = { 1 2 log ( 1 + x 2 ) − i arctan x } 2 m = A m ( x ) + i B m ( x ) ( m ∈ N ) ,$
(1.8)

where, for convenience and simplicity,

$A m (x):= ∑ k = 0 m ( − 1 ) k 2 2 m − 2 k ( 2 m 2 k ) ( arctan x ) 2 k ( ln ( 1 + x 2 ) ) 2 m − 2 k$

and

$B m (x):= ∑ k = 0 m − 1 ( − 1 ) k + 1 2 2 m − 2 k − 1 ( 2 m 2 k + 1 ) ( arctan x ) 2 k + 1 ( ln ( 1 + x 2 ) ) 2 m − 2 k − 1 .$

From (1.6) and (1.8), we obtain a more explicit integral representation for the Stieltjes constants $γ 2 m$:

$γ 2 m =2 ∫ 0 ∞ x A m ( x ) + B m ( x ) ( 1 + x 2 ) ( e 2 π x − 1 ) dx(m∈N),$
(1.9)

where $A m (x)$ and $B m (x)$ are given in (1.8). Similarly, we have

$( log ( 1 − i x ) ) 2 m + 1 = C m (x)+i D m (x)(m∈ N 0 ),$
(1.10)

where, for convenience and simplicity,

$C m (x):= ∑ k = 0 m ( − 1 ) k 2 2 m + 1 − 2 k ( 2 m + 1 2 k ) ( arctan x ) 2 k ( ln ( 1 + x 2 ) ) 2 m + 1 − 2 k$

and

$D m (x):= ∑ k = 0 m ( − 1 ) k + 1 2 2 m − 2 k ( 2 m + 1 2 k + 1 ) ( arctan x ) 2 k + 1 ( ln ( 1 + x 2 ) ) 2 m − 2 k .$

From (1.6) and (1.10), we get a more explicit integral representation for the Stieltjes constants $γ 2 m + 1$:

$γ 2 m + 1 =2 ∫ 0 ∞ x C m ( x ) + D m ( x ) ( 1 + x 2 ) ( e 2 π x − 1 ) dx(m∈ N 0 ),$
(1.11)

where $C m (x)$ and $D m (x)$ are given in (1.10). Connon (see, e.g., cf., [, Eq. (4.3)]; see also [, Eq. (1.5)]) presented an integral representation of the Stieltjes constants $γ n$ of a similar nature in (1.6):

$γ n =i ∫ 0 ∞ ( 1 − i x ) ( log ( 1 + i x ) ) n − ( 1 + i x ) ( log ( 1 − i x ) ) n ( 1 + x 2 ) ( e 2 π x − 1 ) dx(n∈N).$
(1.12)

We recall the polygamma functions $ψ ( n ) (s)$ ($n∈N$) defined by

$ψ ( n ) (s):= d n + 1 d z n + 1 logΓ(s)= d n d s n ψ(s) ( n ∈ N 0 ; s ∈ C ∖ Z 0 − ) ,$
(1.13)

where $ψ(s)$ denotes the psi (or digamma) function defined by

$ψ(s):= d d s logΓ(s)and ψ ( 0 ) (s)=ψ(s) ( s ∈ C ∖ Z 0 − ) .$
(1.14)

Connon [, Eq. (4.27)] also obtained an integral representation of the Stieltjes constants $γ n$:

$γ n = ( − 1 ) n ∑ k = 0 n ( n k ) Y k ( − ψ ( 1 ) , − ψ ( 1 ) ( 1 ) , … , − ψ ( k − 1 ) ( 1 ) ) ⋅ ∫ 0 ∞ ( log t ) n − k ⋅ ( 1 e t − 1 − 1 t ) e − t d t ( n ∈ N ) ,$
(1.15)

where $Y n ( x 1 ,…, x n )$ are the complete Bell polynomials defined by $Y 0 =1$ and

$Y n ( x 1 ,…, x n )= ∑ π ( n ) n ! k 1 ! k 2 ! ⋯ k n ! ( x 1 1 ! ) k 1 ( x 2 2 ! ) k 2 ⋯ ( x n n ! ) k n (n∈N),$
(1.16)

the sum being taken over all partitions $π(n)$ of n, i.e., over all sets of $k j ∈ N 0$ such that

$k 1 +2 k 2 +⋯+n k n =n.$

Connon [, Eq. (4.11)] presented an integral representation of $γ 1$ which may be a special case of (1.15):

$γ 1 =γ− γ 2 − ∫ 0 ∞ logt⋅ ( 1 e t − 1 − 1 t ) e − t dt.$
(1.17)

A remarkably large number of integral formulas for the Euler-Mascheroni constant γ have been presented (see, e.g., [9, 10], and [, Section 1.2]). The Stieltjes constants $γ n$ ($n∈ N 0$) have been investigated in various ways, especially for their integral representations (see, e.g., ; see also [, Section 2.21] and the references cited therein). Here we aim at presenting certain integral representations for $γ n$ by choosing to use three known integral representations for the Riemann zeta function $ζ(s)$. Our method used here is similar to those in some earlier works, but our results seem a little simpler. Some relevant connections of some special cases of our results presented here with those in earlier works are also pointed out.

To do this, we first observe a simple property asserted in the following lemma.

Lemma 1 If some representations of the Riemann zeta function $ζ(s)$ are analytic in a deleted neighborhood of $s=1$, except for a simple pole at $s=1$ with its residue 1, then the following function $Z(s)$ defined by

$Z(s):=ζ(s)− 1 s − 1$
(1.18)

is analytic at $s=1$ if we define

$Z(1):=γ= lim s → 1 ζ(s)− 1 s − 1 .$
(1.19)

Furthermore, we have

$Z ( n ) (1)= ( − 1 ) n γ n (n∈ N 0 ).$
(1.20)

Proof We prove only (1.20). If the above-defined $Z(s)$ is analytic at $s=1$, then the Taylor series expansion of $Z(s)$ is given as follows:

$Z(s)= ∑ n = 0 ∞ Z ( n ) ( 1 ) n ! ( s − 1 ) n$

in a neighborhood of $s=1$. In view of (1.3), by uniqueness of Taylor (or Laurent) series expansion of a function, (1.20) is proved. The other argument is obvious from a well-known property of the Riemann zeta function $ζ(s)$. □

A well-known (and potentially useful) relationship between the polygamma functions $ψ ( n ) (s)$ and the generalized zeta function $ζ(s,a)$ is also given by

$ψ ( n ) (s)= ( − 1 ) n + 1 n! ∑ k = 0 ∞ 1 ( k + s ) n + 1 = ( − 1 ) n + 1 n!ζ(n+1,s) ( n ∈ N ; s ∈ C ∖ Z 0 − ) .$
(1.21)

In particular, we have

$ψ ( n ) (1)= ( − 1 ) n + 1 n!ζ(n+1)(n∈N).$
(1.22)

## 2 Integral representations for $γ n$

We begin by presenting an integral representation for the Stieltjes constants $γ n$ given in the following theorem.

Theorem 1 The following integral representation for $γ n$ holds true:

$γ n = ( − 1 ) n ∫ 0 ∞ ( ψ ′ ( 1 + t ) − 1 1 + t ) ⋅ { ∑ k = 0 n ( n k ) ( − 1 ) [ k / 2 ] + 1 1 − ( − 1 ) k + 1 2 π k k + 1 ( log t ) n − k } d t ( n ∈ N 0 ) .$
(2.1)

We note that $Z(s)$ in (2.7) below is analytic in a neighborhood of $s=1$. So we can use the relation (1.20) for the integral representation of $Z(s)$. In this regard, we first try to get the following formulas asserted by Lemma 2 below.

Lemma 2 Each of the following formulas holds true:

$lim s → 1 ( d n d s n 1 t 1 − s ) = ( log t ) n (n∈ N 0 )$
(2.2)

and

$lim s → 1 { d n d s n ( sin ( π s ) π ( s − 1 ) ) } = ( − 1 ) [ n / 2 ] + 1 1 − ( − 1 ) n + 1 2 π n n + 1 (n∈ N 0 ),$
(2.3)

where $[x]$ denotes the greatest integer less than or equal to a real number x.

Proof The formula (2.2) is obvious. For (2.3), we recall the Maclaurin series expansion of sint:

$sint= ∑ k = 0 ∞ ( − 1 ) k ( 2 k + 1 ) ! t 2 k + 1 ( | t | < ∞ ) .$
(2.4)

By using (2.4), we have

$lim s → 1 { d 2 n d s 2 n ( sin ( π s ) π ( s − 1 ) ) } = − lim t → 0 { d 2 n d t 2 n ( sin ( π t ) π t ) } = ( − 1 ) n + 1 ( 2 n + 1 ) ! lim t → 0 d 2 n d t 2 n ( π t ) 2 n = ( − 1 ) n + 1 π 2 n 2 n + 1 ( n ∈ N 0 ) .$
(2.5)

Similarly, we obtain

$lim s → 1 { d 2 n − 1 d s 2 n − 1 ( sin ( π s ) π ( s − 1 ) ) } =0(n∈N).$
(2.6)

Now it is not difficult to combine the two formulas (2.5) and (2.6) to see the unified formula (2.3). □

Proof of Theorem 1 We choose to recall the following integral representation of $ζ(s)$ (see, e.g., [, p.172, Eq. (47)]):

$Z(s)=ζ(s)− 1 s − 1 = sin ( π s ) π ( s − 1 ) ∫ 0 ∞ ( ψ ′ ( 1 + t ) − 1 1 + t ) d t t 1 − s ( 0 < ℜ ( s ) < 2 ) .$
(2.7)

To get the n th derivative of a product of the two involved functions in (2.7),

$sin ( π s ) π ( s − 1 ) ⋅ 1 t 1 − s ,$

we apply Leibniz’s generalization of the product rule for differentiation and use the results in Lemma 2, in view of (1.20), to yield (2.1). □

Recall an integral representation for $ζ(s)$ (see, e.g., [, p.169, Eq. (31)]):

$Z(s)=ζ(s)− 1 s − 1 = 1 2 + 1 Γ ( s ) ∫ 0 ∞ ( 1 e t − 1 − 1 t + 1 2 ) e − t t s − 1 dt ( ℜ ( s ) > − 1 ) .$
(2.8)

In order to use (2.8) to get an integral representation for $γ n$, we first find the following formula given in Lemma 3.

Lemma 3 If we define $α j$ by

$α j := lim s → 1 ( 1 Γ ( s ) ) ( j ) (j∈ N 0 ),$
(2.9)

then we have a recurrence formula for $α j$

$α k + 1 = ∑ j = 0 k − 1 ( − 1 ) k − j k ! j ! ζ(k+1−j) α j +γ α k (k∈ N 0 ),$
(2.10)

where

$α 0 =1and α 1 =−ψ(1)=γ,$
(2.11)

and an empty sum (as usual) is understood to be nil throughout this paper.

In addition to the formulas in (2.11), the next several $α j$ are given as follows:

$α 2 = γ 2 − ζ ( 2 ) , α 3 = γ 3 − 3 γ ζ ( 2 ) + 2 ζ ( 3 ) , α 4 = γ 4 − 6 γ 2 ζ ( 2 ) + 8 γ ζ ( 3 ) + 3 ( ζ ( 2 ) ) 2 − 6 ζ ( 4 ) , α 5 = γ 5 − 10 γ 3 ζ ( 2 ) + 20 γ 2 ζ ( 3 ) − 20 ζ ( 2 ) ζ ( 3 ) α 5 = + 15 γ ( ζ ( 2 ) ) 2 − 30 γ ζ ( 4 ) + 24 ζ ( 5 ) .$
(2.12)

Proof of Lemma 3 Taking the logarithmic derivative of $1/Γ(s)$, we have

$( 1 Γ ( s ) ) ′ =− 1 Γ ( s ) ⋅ψ(s).$

Using Leibniz’s generalization of the product rule for differentiation when we differentiate the last formula k times and taking the limit $s→1$ on the resulting identity, and applying (1.22) to the last resulting formula, we obtain

$α k + 1 = − ∑ j = 0 k − 1 ( k j ) ψ ( k − j ) ( 1 ) + γ α k = ∑ j = 0 k − 1 ( − 1 ) k − j k ! j ! ζ ( k + 1 − j ) α j + γ α k .$

This completes the proof of Lemma 3. □

Using Leibniz’s generalization of the product rule for differentiation to differentiate both sides of $Z(s)$ in (2.8) with respect to s, n times, and taking the limit $s→1$, $Z(s)$ being analytic at $s=1$ on the resulting identity, and finally using the $α j$ in (2.9) and the relation (1.20), we obtain an integral formula for $γ n$ asserted by Theorem 2 below.

Theorem 2 The following integral representation for $γ n$ holds true:

$γ = γ 0 = 1 + ∫ 0 ∞ ( 1 e t − 1 − 1 t ) e − t d t = ∫ 0 ∞ ( 1 1 − e − t − 1 t ) e − t d t$
(2.13)

and

$γ n = ∫ 0 ∞ ( 1 e t − 1 − 1 t + 1 2 ) e − t ( ∑ k = 0 n ( n k ) α k ( log t ) n − k ) dt(n∈N),$
(2.14)

where $α k$ are given in Lemma 3.

The first three of $γ n$ in (2.14) are given in Corollary 1 below.

Corollary 1 Each of the following integral formulas holds true:

$γ 1 =γ− γ 2 − ∫ 0 ∞ ( 1 e t − 1 − 1 t ) e − t logtdt;$
(2.15)
$γ 2 = γ 3 − γ 2 − γ ζ ( 2 ) + ζ ( 2 ) + ∫ 0 ∞ ( 1 e t − 1 − 1 t ) e − t { 2 γ log t + ( log t ) 2 } d t ;$
(2.16)
$γ 3 = γ 3 − γ 4 + 3 γ 2 ζ ( 2 ) − 3 γ ζ ( 2 ) − 2 γ ζ ( 3 ) + 2 ζ ( 3 ) − ∫ 0 ∞ ( 1 e t − 1 − 1 t ) e − t { 3 ( γ 2 − ζ ( 2 ) ) log t + 3 γ ( log t ) 2 + ( log t ) 3 } d t .$
(2.17)

Proof It is enough to apply (2.13) and a known recurrence formula (see, e.g., [, pp.369-371]) for

$Γ ( n ) (1)= ∫ 0 ∞ e − t ( log t ) n dt(n∈ N 0 )$
(2.18)

to the first three of $γ n$ in (2.14). For easy reference, we record here the first three of $Γ ( n ) (1)$:

$Γ ′ (1)=−γ; Γ ( 2 ) (1)= γ 2 +ζ(2); Γ ( 3 ) (1)=− γ 3 −3γζ(2)−2ζ(3).$
(2.19)

□

We recall Hermite’s integral formula for $ζ(s)$ (see, e.g., [, p.169, Eq. (34)]):

$Z(s)=ζ(s)− 1 s − 1 = 1 2 +2 ∫ 0 ∞ sin ( s arctan t ) ( 1 + t 2 ) 1 2 s d t e 2 π t − 1 .$
(2.20)

Applying Leiniz’s generalization of the product rule for differentiation to (2.20), similarly as in Theorems 1 and 2, we get an integral representation for $γ n$ given in Theorem 3 below.

Theorem 3 The following integral representation for $γ n$ holds true:

$γ= γ 0 = 1 2 +2 ∫ 0 ∞ t ( 1 + t 2 ) ( e 2 π t − 1 ) dt$
(2.21)

and

$γ n = ( − 1 ) n ∫ 0 ∞ { ∑ k = 0 n ( n k ) ( − 1 2 ) n − k ( arctan t ) k sin ( arctan t + k π 2 ) ⋅ ( log ( 1 + t 2 ) ) n − k 1 + t 2 } d t e 2 π t − 1 ( n ∈ N ) ,$
(2.22)

where

$sin(arctant)= t 1 + t 2 andcos(arctant)= 1 1 + t 2 .$
(2.23)

The first three of $γ n$ in (2.22) are given in Corollary 2 below.

Corollary 2 Each of the following integral formulas holds true:

$γ 1 =− ∫ 0 ∞ 2 arctan t − t log ( 1 + t 2 ) 1 + t 2 d t e 2 π t − 1 ;$
(2.24)
$γ 2 = ∫ 0 ∞ − 4 t arctan 2 t − 4 arctan t log ( 1 + t 2 ) + t ( log ( 1 + t 2 ) ) 2 2 ( 1 + t 2 ) d t e 2 π t − 1 ;$
(2.25)
$γ 3 = ∫ 0 ∞ { 8 arctan 3 t − 12 t arctan 2 t log ( 1 + t 2 ) − 6 arctan t ( log ( 1 + t 2 ) ) 2 + t ( log ( 1 + t 2 ) ) 3 } 1 4 ( 1 + t 2 ) d t e 2 π t − 1 .$
(2.26)

Remark Setting $n=0$ in (2.1), in view of relation (1.21), we obtain an integral representation for γ:

$γ = γ 0 = ∫ 0 ∞ ( 1 1 + t − ψ ′ ( 1 + t ) ) d t = ∫ 0 ∞ ( 1 1 + t − ζ ( 2 , 1 + t ) ) d t ,$
(2.27)

which is a known formula (see, e.g., [, Eq. (3.67)]). Equation (1.7) is equal to Equation (2.21), which is recorded, for example, in [, p.17, Eq. (31)]. Equation (2.13) is a known result (see, e.g., [, p.355, Entry 3.427-2]). The result (2.24) is equal to the special case of (1.11) when $m=0$. Connon’s result (1.17) is equal to the integral representation (2.15) for $γ 1$. It is also interesting to compare Connon’s result with our one (2.14).

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## Acknowledgements

The author would like to express his deep gratitude for the reviewers’ helpful comments via their rather detailed reading to make this paper more clear. This research was supported by the Basic Science Research Program through the National Research Foundation of Korea funded by the Ministry of Education, Science and Technology (2010-0011005).

## Author information

Correspondence to Junesang Choi.

### Competing interests

The author declares that he has no competing interests.

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