Certain integral representations of Stieltjes constants ${\gamma }_n$

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) ${\gamma }_n$ and ${\gamma }_0=\gamma$, which arise from the coefficients of the Laurent series expansion of the Riemann zeta function $\zeta (s)$ at $s=1$, have been investigated in various ways, especially for their integral representations. Here we aim at presenting certain integral representations for ${\gamma }_n$ by choosing to use three known integral representations for the Riemann zeta function $\zeta (s)$. 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 $\zeta (s)$ is defined by (see, e.g., [[1], Section 2.3])

$$\zeta (s):=\{\begin{array}{cc}{\sum }_{n=1}^{\mathrm{\infty }}\frac{1}{n^s}=\frac{1}{1-2^{-s}}{\sum }_{n=1}^{\mathrm{\infty }}\frac{1}{{(2n-1)}^s}\hfill & (\mathrm{\Re }(s)>1),\hfill \\ \frac{1}{1-2^{1-s}}{\sum }_{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n^s}\hfill & (\mathrm{\Re }(s)>0;s\ne 1),\hfill \end{array}$$

(1.1)

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

$$\zeta (s,a):=\sum _{k=0}^{\mathrm{\infty }}{(k+a)}^{-s}(\mathrm{\Re }(s)>1;a\in \mathbb{C}\setminus {\mathbb{Z}}_0^{-}),$$

(1.2)

where ℂ and ${\mathbb{Z}}_0^{-}$ denote the sets of complex numbers and nonpositive integers, respectively. Both the Riemann zeta function $\zeta (s)$ and the Hurwitz zeta function $\zeta (s,a)$ 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 ${\gamma }_n$ for $n\in {\mathbb{N}}_0:=\mathbb{N}\cup \{0\}$, $\mathbb{N}:=\{1,2,3,\dots \}$, arise from the following Laurent expansion of the Riemann zeta function $\zeta (s)$ about $s=1$ (see, e.g., [[2], pp.166-169], [[3], p.255] and [[1], p.165]):

$$\zeta (s)=\frac{1}{s-1}+\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n!}{\gamma }_n{(s-1)}^n,$$

(1.3)

where

$$\begin{array}{rl}{\gamma }_n& =\underset{m\to \mathrm{\infty }}{lim}\{\sum _{k=1}^m\frac{{(logk)}^n}{k}-{\int }_1^m\frac{{(logx)}^n}{x}dx\}\\ =\underset{m\to \mathrm{\infty }}{lim}\{\sum _{k=1}^m\frac{{(logk)}^n}{k}-\frac{{(logm)}^{n+1}}{n+1}\}(n\in {\mathbb{N}}_0)\end{array}$$

(1.4)

and, in particular, ${\gamma }_0$ (denoted by γ) is the Euler-Mascheroni constant (see, for details, [[2], Section 1.5] and [[1], Section 1.2]):

$$\gamma :=\underset{m\to \mathrm{\infty }}{lim}(\sum _{k=1}^m\frac{1}{k}-logm)\cong 0.5772156649\cdots .$$

(1.5)

The Stieltjes constants ${\gamma }_n$ are named after Thomas Jan Stieltjes and often referred to as generalized Euler-Mascheroni constants. Liang and Todd [4] computed numerical approximations of the first 20 Stieltjes constants in 1972. In 1985, using contour integration, Ainsworth and Howell [5] showed that

$${\gamma }_n=2\mathrm{\Re }\left\{{\int }_0^{\mathrm{\infty }}\frac{(x-i){(log(1-ix))}^n}{(1+x^2)(e^{2\pi x}-1)}dx\right\}(n\in \mathbb{N})$$

(1.6)

and

$$\begin{array}{rl}\gamma & ={\gamma }_0=\frac{1}{2}+2\mathrm{\Re }\left\{{\int }_0^{\mathrm{\infty }}\frac{(x-i)}{(1+x^2)(e^{2\pi x}-1)}dx\right\}\\ =\frac{1}{2}+2{\int }_0^{\mathrm{\infty }}\frac{x}{(1+x^2)(e^{2\pi x}-1)}dx.\end{array}$$

(1.7)

By using binomial theorem, we have

$$\begin{array}{rl}{(log(1-ix))}^{2m}& ={\{\frac{1}{2}log(1+x^2)-iarctanx\}}^{2m}\\ ={\mathcal{A}}_m(x)+i{\mathcal{B}}_m(x)(m\in \mathbb{N}),\end{array}$$

(1.8)

where, for convenience and simplicity,

$${\mathcal{A}}_m(x):=\sum _{k=0}^m\frac{{(-1)}^k}{2^{2m-2k}}\left(\genfrac{}{}{0ex}{}{2m}{2k}\right){(arctanx)}^{2k}{(ln(1+x^2))}^{2m-2k}$$

and

$${\mathcal{B}}_m(x):=\sum _{k=0}^{m-1}\frac{{(-1)}^{k+1}}{2^{2m-2k-1}}\left(\genfrac{}{}{0ex}{}{2m}{2k+1}\right){(arctanx)}^{2k+1}{(ln(1+x^2))}^{2m-2k-1}.$$

From (1.6) and (1.8), we obtain a more explicit integral representation for the Stieltjes constants ${\gamma }_{2m}$:

$${\gamma }_{2m}=2{\int }_0^{\mathrm{\infty }}\frac{x{\mathcal{A}}_m(x)+{\mathcal{B}}_m(x)}{(1+x^2)(e^{2\pi x}-1)}dx(m\in \mathbb{N}),$$

(1.9)

where ${\mathcal{A}}_m(x)$ and ${\mathcal{B}}_m(x)$ are given in (1.8). Similarly, we have

$${(log(1-ix))}^{2m+1}={\mathcal{C}}_m(x)+i{\mathcal{D}}_m(x)(m\in {\mathbb{N}}_0),$$

(1.10)

where, for convenience and simplicity,

$${\mathcal{C}}_m(x):=\sum _{k=0}^m\frac{{(-1)}^k}{2^{2m+1-2k}}\left(\genfrac{}{}{0ex}{}{2m+1}{2k}\right){(arctanx)}^{2k}{(ln(1+x^2))}^{2m+1-2k}$$

and

$${\mathcal{D}}_m(x):=\sum _{k=0}^m\frac{{(-1)}^{k+1}}{2^{2m-2k}}\left(\genfrac{}{}{0ex}{}{2m+1}{2k+1}\right){(arctanx)}^{2k+1}{(ln(1+x^2))}^{2m-2k}.$$

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

$${\gamma }_{2m+1}=2{\int }_0^{\mathrm{\infty }}\frac{x{\mathcal{C}}_m(x)+{\mathcal{D}}_m(x)}{(1+x^2)(e^{2\pi x}-1)}dx(m\in {\mathbb{N}}_0),$$

(1.11)

where ${\mathcal{C}}_m(x)$ and ${\mathcal{D}}_m(x)$ are given in (1.10). Connon (see, e.g., cf., [[6], Eq. (4.3)]; see also [[7], Eq. (1.5)]) presented an integral representation of the Stieltjes constants ${\gamma }_n$ of a similar nature in (1.6):

$${\gamma }_n=i{\int }_0^{\mathrm{\infty }}\frac{(1-ix){(log(1+ix))}^n-(1+ix){(log(1-ix))}^n}{(1+x^2)(e^{2\pi x}-1)}dx(n\in \mathbb{N}).$$

(1.12)

We recall the polygamma functions ${\psi }^{(n)}(s)$ ($n\in \mathbb{N}$) defined by

$${\psi }^{(n)}(s):=\frac{d^{n+1}}{d{z}^{n+1}}log\mathrm{\Gamma }(s)=\frac{d^n}{d{s}^n}\psi (s)(n\in {\mathbb{N}}_0;s\in \mathbb{C}\setminus {\mathbb{Z}}_0^{-}),$$

(1.13)

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

$$\psi (s):=\frac{d}{ds}log\mathrm{\Gamma }(s)\text{and}{\psi }^{(0)}(s)=\psi (s)(s\in \mathbb{C}\setminus {\mathbb{Z}}_0^{-}).$$

(1.14)

Connon [[8], Eq. (4.27)] also obtained an integral representation of the Stieltjes constants ${\gamma }_n$:

$$\begin{array}{rl}{\gamma }_n=& {(-1)}^n\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)Y_k(-\psi (1),-{\psi }^{(1)}(1),\dots ,-{\psi }^{(k-1)}(1))\\ \cdot {\int }_0^{\mathrm{\infty }}{(logt)}^{n-k}\cdot (\frac{1}{e^t-1}-\frac{1}{t})e^{-t}dt(n\in \mathbb{N}),\end{array}$$

(1.15)

where $Y_n(x_1,\dots ,x_n)$ are the complete Bell polynomials defined by $Y_0=1$ and

$$Y_n(x_1,\dots ,x_n)=\sum _{\pi (n)}\frac{n!}{k_1!k_2!\cdots k_n!}{\left(\frac{x_1}{1!}\right)}^{k_1}{\left(\frac{x_2}{2!}\right)}^{k_2}\cdots {\left(\frac{x_n}{n!}\right)}^{k_n}(n\in \mathbb{N}),$$

(1.16)

the sum being taken over all partitions $\pi (n)$ of n, i.e., over all sets of $k_j\in {\mathbb{N}}_0$ such that

$$k_1+2k_2+\cdots +n{k}_n=n.$$

Connon [[8], Eq. (4.11)] presented an integral representation of ${\gamma }_1$ which may be a special case of (1.15):

$${\gamma }_1=\gamma -{\gamma }^2-{\int }_0^{\mathrm{\infty }}logt\cdot (\frac{1}{e^t-1}-\frac{1}{t})e^{-t}dt.$$

(1.17)

A remarkably large number of integral formulas for the Euler-Mascheroni constant γ have been presented (see, e.g., [9, 10], and [[1], Section 1.2]). The Stieltjes constants ${\gamma }_n$ ($n\in {\mathbb{N}}_0$) have been investigated in various ways, especially for their integral representations (see, e.g., [4–8]; see also [[2], Section 2.21] and the references cited therein). Here we aim at presenting certain integral representations for ${\gamma }_n$ by choosing to use three known integral representations for the Riemann zeta function $\zeta (s)$. 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 $\zeta (s)$ 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

$$Z(s):=\zeta (s)-\frac{1}{s-1}$$

(1.18)

is analytic at $s=1$ if we define

$$Z(1):=\gamma =\underset{s\to 1}{lim}\zeta (s)-\frac{1}{s-1}.$$

(1.19)

Furthermore, we have

$$Z^{(n)}(1)={(-1)}^n{\gamma }_n(n\in {\mathbb{N}}_0).$$

(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:

$$Z(s)=\sum _{n=0}^{\mathrm{\infty }}\frac{Z^{(n)}(1)}{n!}{(s-1)}^n$$

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 $\zeta (s)$. □

A well-known (and potentially useful) relationship between the polygamma functions ${\psi }^{(n)}(s)$ and the generalized zeta function $\zeta (s,a)$ is also given by

$${\psi }^{(n)}(s)={(-1)}^{n+1}n!\sum _{k=0}^{\mathrm{\infty }}\frac{1}{{(k+s)}^{n+1}}={(-1)}^{n+1}n!\zeta (n+1,s)(n\in \mathbb{N};s\in \mathbb{C}\setminus {\mathbb{Z}}_0^{-}).$$

(1.21)

In particular, we have

$${\psi }^{(n)}(1)={(-1)}^{n+1}n!\zeta (n+1)(n\in \mathbb{N}).$$

(1.22)

2 Integral representations for ${\gamma }_n$

We begin by presenting an integral representation for the Stieltjes constants ${\gamma }_n$ given in the following theorem.

Theorem 1 The following integral representation for ${\gamma }_n$ holds true:

$$\begin{array}{rl}{\gamma }_n=& {(-1)}^n{\int }_0^{\mathrm{\infty }}({\psi }^{\prime }(1+t)-\frac{1}{1+t})\\ \cdot \left\{\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right){(-1)}^{[k/2]+1}\frac{1-{(-1)}^{k+1}}{2}\frac{{\pi }^k}{k+1}{(logt)}^{n-k}\right\}dt(n\in {\mathbb{N}}_0).\end{array}$$

(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:

$$\underset{s\to 1}{lim}\left(\frac{d^n}{d{s}^n}\frac{1}{t^{1-s}}\right)={(logt)}^n(n\in {\mathbb{N}}_0)$$

(2.2)

and

$$\underset{s\to 1}{lim}\left\{\frac{d^n}{d{s}^n}\left(\frac{sin(\pi s)}{\pi (s-1)}\right)\right\}={(-1)}^{[n/2]+1}\frac{1-{(-1)}^{n+1}}{2}\frac{{\pi }^n}{n+1}(n\in {\mathbb{N}}_0),$$

(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:

$$sint=\sum _{k=0}^{\mathrm{\infty }}\frac{{(-1)}^k}{(2k+1)!}{t}^{2k+1}(|t|<\mathrm{\infty }).$$

(2.4)

By using (2.4), we have

$$\begin{array}{r}\underset{s\to 1}{lim}\left\{\frac{d^{2n}}{d{s}^{2n}}\left(\frac{sin(\pi s)}{\pi (s-1)}\right)\right\}\\ =-\underset{t\to 0}{lim}\left\{\frac{d^{2n}}{d{t}^{2n}}\left(\frac{sin(\pi t)}{\pi t}\right)\right\}\\ =\frac{{(-1)}^{n+1}}{(2n+1)!}\underset{t\to 0}{lim}\frac{d^{2n}}{d{t}^{2n}}{(\pi t)}^{2n}={(-1)}^{n+1}\frac{{\pi }^{2n}}{2n+1}(n\in {\mathbb{N}}_0).\end{array}$$

(2.5)

Similarly, we obtain

$$\underset{s\to 1}{lim}\left\{\frac{d^{2n-1}}{d{s}^{2n-1}}\left(\frac{sin(\pi s)}{\pi (s-1)}\right)\right\}=0(n\in \mathbb{N}).$$

(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 $\zeta (s)$ (see, e.g., [[1], p.172, Eq. (47)]):

$$Z(s)=\zeta (s)-\frac{1}{s-1}=\frac{sin(\pi s)}{\pi (s-1)}{\int }_0^{\mathrm{\infty }}({\psi }^{\prime }(1+t)-\frac{1}{1+t})\frac{dt}{t^{1-s}}(0<\mathrm{\Re }(s)<2).$$

(2.7)

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

$$\frac{sin(\pi s)}{\pi (s-1)}\cdot \frac{1}{t^{1-s}},$$

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 $\zeta (s)$ (see, e.g., [[1], p.169, Eq. (31)]):

$$Z(s)=\zeta (s)-\frac{1}{s-1}=\frac{1}{2}+\frac{1}{\mathrm{\Gamma }(s)}{\int }_0^{\mathrm{\infty }}(\frac{1}{e^t-1}-\frac{1}{t}+\frac{1}{2})e^{-t}{t}^{s-1}dt(\mathrm{\Re }(s)>-1).$$

(2.8)

In order to use (2.8) to get an integral representation for ${\gamma }_n$, we first find the following formula given in Lemma 3.

Lemma 3 If we define ${\alpha }_j$ by

$${\alpha }_j:=\underset{s\to 1}{lim}{\left(\frac{1}{\mathrm{\Gamma }(s)}\right)}^{(j)}(j\in {\mathbb{N}}_0),$$

(2.9)

then we have a recurrence formula for ${\alpha }_j$

$${\alpha }_{k+1}=\sum _{j=0}^{k-1}{(-1)}^{k-j}\frac{k!}{j!}\zeta (k+1-j){\alpha }_j+\gamma {\alpha }_k(k\in {\mathbb{N}}_0),$$

(2.10)

where

$${\alpha }_0=1\mathit{\text{and}}{\alpha }_1=-\psi (1)=\gamma ,$$

(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 ${\alpha }_j$ are given as follows:

$$\begin{array}{r}{\alpha }_2={\gamma }^2-\zeta (2),{\alpha }_3={\gamma }^3-3\gamma \zeta (2)+2\zeta (3),\\ {\alpha }_4={\gamma }^4-6{\gamma }^2\zeta (2)+8\gamma \zeta (3)+3{(\zeta (2))}^2-6\zeta (4),\\ {\alpha }_5={\gamma }^5-10{\gamma }^3\zeta (2)+20{\gamma }^2\zeta (3)-20\zeta (2)\zeta (3)\\ \phantom{{\alpha }_5=}+15\gamma {(\zeta (2))}^2-30\gamma \zeta (4)+24\zeta (5).\end{array}$$

(2.12)

Proof of Lemma 3 Taking the logarithmic derivative of $1/\mathrm{\Gamma }(s)$, we have

$${\left(\frac{1}{\mathrm{\Gamma }(s)}\right)}^{\prime }=-\frac{1}{\mathrm{\Gamma }(s)}\cdot \psi (s).$$

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

$$\begin{array}{rl}{\alpha }_{k+1}& =-\sum _{j=0}^{k-1}\left(\genfrac{}{}{0ex}{}{k}{j}\right){\psi }^{(k-j)}(1)+\gamma {\alpha }_k\\ =\sum _{j=0}^{k-1}{(-1)}^{k-j}\frac{k!}{j!}\zeta (k+1-j){\alpha }_j+\gamma {\alpha }_k.\end{array}$$

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 $s\to 1$, $Z(s)$ being analytic at $s=1$ on the resulting identity, and finally using the ${\alpha }_j$ in (2.9) and the relation (1.20), we obtain an integral formula for ${\gamma }_n$ asserted by Theorem 2 below.

Theorem 2 The following integral representation for ${\gamma }_n$ holds true:

$$\begin{array}{rl}\gamma & ={\gamma }_0=1+{\int }_0^{\mathrm{\infty }}(\frac{1}{e^t-1}-\frac{1}{t})e^{-t}dt\\ ={\int }_0^{\mathrm{\infty }}(\frac{1}{1-e^{-t}}-\frac{1}{t})e^{-t}dt\end{array}$$

(2.13)

and

$${\gamma }_n={\int }_0^{\mathrm{\infty }}(\frac{1}{e^t-1}-\frac{1}{t}+\frac{1}{2})e^{-t}\left(\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right){\alpha }_k{(logt)}^{n-k}\right)dt(n\in \mathbb{N}),$$

(2.14)

where ${\alpha }_k$ are given in Lemma 3.

The first three of ${\gamma }_n$ in (2.14) are given in Corollary 1 below.

Corollary 1 Each of the following integral formulas holds true:

$${\gamma }_1=\gamma -{\gamma }^2-{\int }_0^{\mathrm{\infty }}(\frac{1}{e^t-1}-\frac{1}{t})e^{-t}logtdt;$$

(2.15)

$$\begin{array}{rl}{\gamma }_2=& {\gamma }^3-{\gamma }^2-\gamma \zeta (2)+\zeta (2)\\ +{\int }_0^{\mathrm{\infty }}(\frac{1}{e^t-1}-\frac{1}{t})e^{-t}\{2\gamma logt+{(logt)}^2\}dt;\end{array}$$

(2.16)

$$\begin{array}{rl}{\gamma }_3=& {\gamma }^3-{\gamma }^4+3{\gamma }^2\zeta (2)-3\gamma \zeta (2)-2\gamma \zeta (3)+2\zeta (3)\\ -{\int }_0^{\mathrm{\infty }}(\frac{1}{e^t-1}-\frac{1}{t})e^{-t}\{3({\gamma }^2-\zeta (2))logt+3\gamma {(logt)}^2+{(logt)}^3\}dt.\end{array}$$

(2.17)

Proof It is enough to apply (2.13) and a known recurrence formula (see, e.g., [[1], pp.369-371]) for

$${\mathrm{\Gamma }}^{(n)}(1)={\int }_0^{\mathrm{\infty }}{e}^{-t}{(logt)}^{n}dt(n\in {\mathbb{N}}_0)$$

(2.18)

to the first three of ${\gamma }_n$ in (2.14). For easy reference, we record here the first three of ${\mathrm{\Gamma }}^{(n)}(1)$:

$${\mathrm{\Gamma }}^{\prime }(1)=-\gamma ;{\mathrm{\Gamma }}^{(2)}(1)={\gamma }^2+\zeta (2);{\mathrm{\Gamma }}^{(3)}(1)=-{\gamma }^3-3\gamma \zeta (2)-2\zeta (3).$$

(2.19)

 □

We recall Hermite’s integral formula for $\zeta (s)$ (see, e.g., [[1], p.169, Eq. (34)]):

$$Z(s)=\zeta (s)-\frac{1}{s-1}=\frac{1}{2}+2{\int }_0^{\mathrm{\infty }}\frac{sin(sarctant)}{{(1+t^2)}^{\frac{1}{2}s}}\frac{dt}{e^{2\pi t}-1}.$$

(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 ${\gamma }_n$ given in Theorem 3 below.

Theorem 3 The following integral representation for ${\gamma }_n$ holds true:

$$\gamma ={\gamma }_0=\frac{1}{2}+2{\int }_0^{\mathrm{\infty }}\frac{t}{(1+t^2)(e^{2\pi t}-1)}dt$$

(2.21)

and

$$\begin{array}{rl}{\gamma }_n=& {(-1)}^n{\int }_0^{\mathrm{\infty }}\{\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right){(-\frac{1}{2})}^{n-k}{(arctant)}^{k}sin(arctant+\frac{k\pi }{2})\\ \cdot \frac{{(log(1+t^2))}^{n-k}}{\sqrt{1+t^2}}\}\frac{dt}{e^{2\pi t}-1}(n\in \mathbb{N}),\end{array}$$

(2.22)

where

$$sin(arctant)=\frac{t}{\sqrt{1+t^2}}\mathit{\text{and}}cos(arctant)=\frac{1}{\sqrt{1+t^2}}.$$

(2.23)

The first three of ${\gamma }_n$ in (2.22) are given in Corollary 2 below.

Corollary 2 Each of the following integral formulas holds true:

$${\gamma }_1=-{\int }_0^{\mathrm{\infty }}\frac{2arctant-tlog(1+t^2)}{1+t^2}\frac{dt}{e^{2\pi t}-1};$$

(2.24)

$${\gamma }_2={\int }_0^{\mathrm{\infty }}\frac{-4t{arctan}^{2}t-4arctantlog(1+t^2)+t{(log(1+t^2))}^2}{2(1+t^2)}\frac{dt}{e^{2\pi t}-1};$$

(2.25)

$$\begin{array}{rl}{\gamma }_3=& {\int }_0^{\mathrm{\infty }}\{8{arctan}^{3}t-12t{arctan}^{2}tlog(1+t^2)\\ -6arctant{(log(1+t^2))}^2+t{(log(1+t^2))}^3\}\frac{1}{4(1+t^2)}\frac{dt}{e^{2\pi t}-1}.\end{array}$$

(2.26)

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

$$\begin{array}{rl}\gamma & ={\gamma }_0={\int }_0^{\mathrm{\infty }}(\frac{1}{1+t}-{\psi }^{\prime }(1+t))dt\\ ={\int }_0^{\mathrm{\infty }}(\frac{1}{1+t}-\zeta (2,1+t))dt,\end{array}$$

(2.27)

which is a known formula (see, e.g., [[9], Eq. (3.67)]). Equation (1.7) is equal to Equation (2.21), which is recorded, for example, in [[1], p.17, Eq. (31)]. Equation (2.13) is a known result (see, e.g., [[10], 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 ${\gamma }_1$. It is also interesting to compare Connon’s result with our one (2.14).