1 Introduction and statement of results

The Stieltjes (or generalized Euler) constants \(\gamma _k(a)\) appear as expansion coefficients in the Laurent series for the Hurwitz zeta function \(\zeta (s,a)\) about its simple pole at \(s=1\) [5–7, 10, 18, 22, 26, 28, 31],

$$\begin{aligned} \zeta (s,a)={1 \over {s-1}}+\sum _{n=0}^\infty {{(-1)^n} \over {n!}}\gamma _n(a)(s-1)^n. \end{aligned}$$
(1.1)

These constants are important in analytic number theory and elsewhere, where they appear in various estimations and as a result of asymptotic analyses, being given by the limit relation

$$\begin{aligned} \gamma _k(a)=\lim _{N \rightarrow \infty } \left[ \sum _{j=1}^N {{\ln ^k (j+a)} \over j}-{{\ln ^{k+1}(N+a)} \over {k+1}}\right] . \end{aligned}$$

In particular, \(\gamma _0(a)=-\psi (a)\), where \(\psi (z)=\Gamma '(z)/\Gamma (z)\) is the digamma function, with \(\Gamma (z)\) as the Gamma function. With \(\gamma \) as the Euler constant and \(\gamma _1=\gamma _1(1)\) and \(\gamma _2=\gamma _2(1)\), we recall the connection with sums of reciprocal powers of the nontrivial zeros \(\rho \) of the Riemann zeta function,

$$\begin{aligned} \sum _\rho {1 \over \rho ^2}=1-{\pi ^2 \over 8}+2\gamma _1+\gamma ^2, \quad \sum _\rho {1 \over \rho ^3}=1-{7 \over 8}\zeta (3)+\gamma ^3+3\gamma \gamma _1+{3 \over 2}\gamma _2; \end{aligned}$$

such relations follow from the Hadamard factorization [20], Ch. 1.3].

An effective asymptotic expression for \(\gamma _k\) [23] and \(\gamma _k(a)\) [24] for \(k \gg 1\) has recently been given. From this expression, previously known results on sign changes within the sequence of Stieltjes constants follow.

In this paper, we first evaluate the first and second Stieltjes constants at rational arguments. These decompositions are effectively Fourier series, thus implying many extensions and applications, and they supplement the relations presented in [9]. We then present multiplication formulas for the zeroth, first, and second Stieltjes constants, and certain log–log integrals with integer parameters. The latter integral evaluations also provide explicit expressions for the differences \(\gamma _1(j/m) -\gamma _1\) and \(\gamma _2(j/m)-\gamma _2\). Besides elaborating on a multiplication formula for the Stieltjes constants, the Discussion section provides examples of integrals evaluating in terms of differences of the first and second of these constants. In addition, presented there is a novel method of determining log–log integrals with a certain polynomial denominator integrand.

We recall the connection of differences of Stieltjes constants with logarithmic sums,

$$\begin{aligned} \gamma _\ell (a)-\gamma _\ell (b)=\sum _{n=0}^\infty \left[ {{\ln ^\ell (n+a)} \over {n+a}}-{{\ln ^\ell (n+b)} \over {n+b}}\right] . \end{aligned}$$
(1.2)

Very recently an evaluation of \(\gamma _1(j/m)-\gamma _1\) was also performed [4]. However, the method of proof is circuitous—integrals are used in addition to multiple applications of functional equations.

The Hurwitz zeta function, initially defined by \(\zeta (s,a)=\sum _{n=0}^\infty (n+a)^{-s}\) for \(\text{ Re } ~s>1\), has an analytic continuation to the whole complex plane [3, 15, 21, 29]. In the case of \(a=1\), \(\zeta (s,a)\) reduces to the Riemann zeta function \(\zeta (s)\) [13, 20, 27]. In this instance, by convention, the Stieltjes constants \(\gamma _k(1)\) are simply denoted as \(\gamma _k\) [5, 18, 22, 25, 26, 32]. We recall that \(\gamma _k(a+1)=\gamma _k(a)-(\ln ^k a)/a\), and more generally that for \(n \ge 1\) an integer

$$\begin{aligned} \gamma _k(a+n)=\gamma _k(a)-\sum _{j=0}^{n-1} {{\ln ^k(a+j)} \over {a+j}}, \end{aligned}$$

as follows from the functional equation \(\zeta (s,a+n)=\zeta (s,a)-\sum _{j=0}^{n-1} (a+j)^{-s}\). In fact, an interval of length \(1/2\) is sufficient to characterize the \(\gamma _k(a)\)’s [17].

Unless specified otherwise below, letters \(j\), \(k\), \(\ell \), \(m\), \(n\), and \(r\) denote positive integers. The Euler constant is given by \(\gamma =-\psi (1)=\gamma _0(1)\). The polygamma functions are denoted as \(\psi ^{(n)}(z)\) and we note that \(\psi ^{(n)}(z)=(-1)^{n+1}n!\zeta (n+1,z)\) [1, 16].

Proposition 1

For \(m>1\) and \(j<m\), (a)

$$\begin{aligned} \gamma _1\left( {j \over m}\right)&=\gamma _1+\gamma ^2+\gamma \ln 2\pi m+\ln (2\pi )\ln m+ {{\ln ^2 m} \over 2}+(\gamma +\ln 2\pi m)\psi \left( {j \over m}\right) \nonumber \\&\quad +\pi \sum _{r=1}^{m-1}\sin {{2\pi jr} \over m}\ln \Gamma \left( {r \over m}\right) + \sum _{r=1}^{m-1} \cos {{2\pi jr} \over m} \zeta ''\left( 0,{r \over m}\right) \end{aligned}$$
(1.3)

and (b)

$$\begin{aligned} {1 \over 2}\gamma _2\left( {j \over m}\right)&={\gamma _2 \over 2}-\gamma _1\ln m+{\gamma \over 8}\pi ^2+{\pi ^2 \over 8}\ln m+{\gamma \over 2}\ln ^2 m+{{\ln ^3 m} \over 6}\nonumber \\&\quad +A+\sum _{r=1}^{m-1}\cos {{2\pi jr} \over m}\left\{ \left[ {\pi ^2 \over 6}+(\gamma +\ln 2\pi m)^2 \right] \zeta '\left( 0,{r \over m}\right) \right. \nonumber \\&\quad \left. -(\gamma +\ln 2\pi m) \zeta ''\left( 0,{r \over m}\right) +{1 \over 3}\zeta '''\left( 0,{r \over m}\right) \right\} , \end{aligned}$$
(1.4)

where

and \('\) indicates differentiation with respect to the first argument.

Both parts (a) and (b) may be written in many alternative forms. For instance, for (b), the well-known relation \(\zeta '(0,a)=\ln \Gamma (a)-\ln (2\pi )/2\) may be used, and as well, the right member may be modified by introducing \(\gamma _1(j/m)\).

Various summation results are known for the Stieltjes constants, including [6]

$$\begin{aligned} \sum _{r=1}^q \gamma _k\left( {r \over q}\right) =q(-1)^k {{\ln ^{k+1}q} \over {k+1}} +q\sum _{j=0}^k (-1)^j{k \atopwithdelims ()j}(\ln ^j q) \gamma _{k-j}. \end{aligned}$$

As we briefly indicate, the \(k=1\) and \(2\) cases follow from Proposition 1.

Corollary 1

$$\begin{aligned} \sum _{r=1}^q \gamma _1\left( {r \over q}\right) =-{q \over 2}\ln ^2 q+q(-\gamma \ln q+\gamma _1) \end{aligned}$$

and

$$\begin{aligned} \sum _{r=1}^q \gamma _2\left( {r \over q}\right) ={q \over 3}\ln ^3 q+q\left( \gamma \ln ^2q -2\gamma _1\ln q+ \gamma _2\right) . \end{aligned}$$

Proof

The summation for \(\gamma _1\) follows from (1.3) as the sums over the cosine and sine terms are zero, and we have the readily verified relation \(\sum _{r=1}^q\psi \left( {r \over q}\right) =-q(\gamma +\ln q)\). Similarly for the summation for \(\gamma _2\), the sine and cosine terms do not contribute, the just-mentioned summation of \(\psi (r/q)\) holds, and \(\sum _{j=1}^{q-1} \cot {{\pi j} \over q}=0\). \(\square \)

Corollary 2

For \(\ell =1,2,\ldots ,m-1\), (a)

$$\begin{aligned}&{1 \over 2}\sum _{j=1}^{m-1}\gamma _2\left( {j \over m}\right) \sin {{2\pi j \ell } \over m} \nonumber \\&={\pi ^3 \over {48}}(2\ell -m) -{{\pi m} \over 4}\left\{ \left( {\pi ^2 \over 6}+(\gamma +\ln 2\pi m)^2 \right) \left[ \zeta \left( 0,1-{\ell \over m}\right) -\zeta \left( 0,{\ell \over m}\right) \right] \right. \nonumber \\&\quad \left. + 2(\gamma +\ln 2\pi m)\left[ \zeta '\left( 0,{\ell \over m}\right) -\zeta '\left( 0,1-{\ell \over m}\right) \right] \right. \nonumber \\&\quad \left. +\zeta ''\left( 0,1-{\ell \over m}\right) -\zeta ''\left( 0,{\ell \over m}\right) \right\} \end{aligned}$$

and (b)

$$\begin{aligned}&{1 \over 2}\sum _{j=1}^{m-1}\gamma _2\left( {j \over m}\right) \cos {{2\pi j \ell } \over m}\!=\!-\left[ {\gamma _2 \over 2}\!-\!\gamma _1\ln m\!+\!{\gamma \over 8}\pi ^2\!+\!{\pi ^2 \over 8}\ln m\!+\!{\gamma \over 2}\ln ^2 m\!+\!{1 \over 6}\ln ^3 m\right] \nonumber \\&\quad +{\pi ^2 \over 8}\left[ m\ln \left( 2\sin {{\pi \ell } \over m}\right) +\gamma \right] -\sum _{r=1}^{m-1}\left\{ \left( {\pi ^2 \over 6}+(\gamma +\ln 2\pi m)^2\right) \zeta '\left( 0,{r \over m}\right) \right. \nonumber \\&\quad \left. -(\gamma +\ln 2\pi m)\zeta ''\left( 0,{r \over m}\right) +{1 \over 3}\zeta ''' \left( 0,{r \over m}\right) \right\} \nonumber \\&\quad +{m \over 2}\left\{ \left( {\pi ^2 \over 6}+(\gamma +\ln 2\pi m)^2\right) \left[ \zeta '\left( 0,{\ell \over m}\right) +\zeta '\left( 0,1-{\ell \over m}\right) \right] \right. \nonumber \\&\quad \left. -(\gamma +\ln 2\pi m)\left[ \zeta ''\left( 0,{\ell \over m}\right) +\zeta ''\left( 0,1-{\ell \over m}\right) \right] \right. \nonumber \\&\quad \left. +{1 \over 3}\left[ \zeta '''\left( 0,{\ell \over m}\right) +\zeta '''\left( 0,1-{\ell \over m}\right) \right] \right\} . \end{aligned}$$

Part (b) may be rewritten by using the three sums

$$\begin{aligned} \sum _{\ell =1}^{m-1}\zeta '\left( 0,{\ell \over m}\right) =-{{\ln m} \over 2}, \quad \sum _{\ell =1}^{m-1}\zeta ''\left( 0,{\ell \over m}\right) =-{1 \over 2}\ln ^2 m-(\ln m)\ln (2\pi ), \end{aligned}$$

and

$$\begin{aligned} \sum _{\ell =1}^{m-1}\zeta '''\left( 0,{\ell \over m}\right)&=-{{\ln ^3 m} \over 2}-{3 \over 2} \ln ^2 m \ln (2\pi ) \nonumber \\&\quad +3(\ln m)\left( {\gamma ^2 \over 2}-{\pi ^2 \over {24}}-{1 \over 2}\ln ^2(2\pi ) +\gamma _1\right) . \end{aligned}$$

Corollary 3

(Asymptotic expressions) For \(m\rightarrow \infty \),

$$\begin{aligned} \gamma _1\left( {1 \over m}\right) \sim -m\ln m+\gamma _1 \end{aligned}$$

and

$$\begin{aligned} \gamma _2\left( {1 \over m}\right) \sim m\ln ^2m +\gamma _2. \end{aligned}$$

The general situation for \(\gamma _k(a)\) with \(a\rightarrow 0\) is more conveniently proved otherwise, and is presented in Proposition 5.

Proposition 2

For Re \(z>0\) and \(0<k<2\),

$$\begin{aligned} \psi (kz)=\sum _{n=0}^\infty (k-1)^n {z^n \over {n!}}\psi ^{(n)}(z). \end{aligned}$$

Proposition 3

For Re \(z>0\) and \(0<k<2\), (a)

$$\begin{aligned} \gamma _1(kz)&=\gamma _1(z)+\gamma [\psi (kz)-\psi (z)] \nonumber \\&\quad -\sum _{n=1}^\infty (1-k)^n z^n \left[ {{(-1)^{n+1}} \over {n!}} \psi ^{(n)}(z)\psi (n+1)+\left. {{\partial \zeta (s,z)} \over {\partial s}}\right| _{s=n+1}\right] \end{aligned}$$

and (b)

$$\begin{aligned} \gamma _2(kz)= & {} \gamma _2(z)+[\gamma ^2-\zeta (2)][\psi (z)-\psi (kz)] \nonumber \\&+2\sum _{n=1}^\infty (1-k)^n z^n\left[ \left( \gamma +{1 \over 2}\psi (n+1)\right) \psi (n+1) \zeta (n+1,z) \right. \nonumber \\&\quad \left. +{1 \over 2}\psi '(n+1)\zeta (n+1,z) \right. \nonumber \\&\quad \left. +[\gamma +\psi (n+1)]\left. {{\partial \zeta (s,z)} \over {\partial s}}\right| _{s=n+1}+{1 \over 2}\left. {{\partial ^2 \zeta (s,z)} \over {\partial s^2}}\right| _{s=n+1}\right] . \end{aligned}$$

The following result, wherein the differences \(\gamma _1(j/m)-\gamma _1\) and \(\gamma _2(j/m)-\gamma _2\) appear, is a generalization of Gauss’ formula for the digamma function at rational argument. For integers \(p\) and \(q\) with \(0<p<q\), we put

$$\begin{aligned} I^k_{pq}\equiv q\int _0^1 \left( {{x^{q-1}-x^{p-1}} \over {1-x^q}}\right) \ln ^k(-\ln x)\mathrm{{d}}x. \end{aligned}$$

Proposition 4

Let \(\omega _k\equiv \exp (2\pi ik/q)\). Then (a)

$$\begin{aligned} I_{pq}^1&=-\sum _{k=1}^{q-1}(\omega _k^p-1)\left[ \gamma \ln \left( {{\omega _k-1} \over \omega _k} \right) +\sum _{n=1}^\infty \zeta '(1-n){{\ln ^n(\omega _k^{-1})} \over {n!}} \right. \\&\qquad \left. -\gamma \ln \left( -\ln \omega _k^{-1}\right) -{1 \over 2}\ln ^2 \left( -\ln \omega _k^{-1}\right) \right] -q\left( \gamma _1+{\gamma ^2 \over 2}+{\pi ^2 \over {12}}\right) \nonumber \\&=-(\gamma +\ln q)\left[ \gamma +\psi \left( {p \over q}\right) \right] +\gamma _1\left( {p \over q} \right) -\gamma _1 \end{aligned}$$

and (b)

$$\begin{aligned} I_{pq}^2&=\sum _{k=1}^{q-1}(\omega _k^p-1)\left[ \left( \gamma ^2+{\pi ^2 \over 6}\right) \ln \left( {{\omega _k-1} \over \omega _k}\right) \right. \\&\quad \left. +2\gamma \left. {{\partial \text{ Li }_s} \over {\partial s}}(\omega _k^{-1})\right| _{s=1}-\left. {{\partial ^2 \text{ Li }_s} \over {\partial s^2}}(\omega _k^{-1})\right| _{s=1}\right] \nonumber \\&=\left( \gamma ^2+{\pi ^2 \over 6}+2\gamma \ln q+\ln ^2 q\right) \left[ \gamma +\psi \left( {p \over q}\right) \right] \\&\quad -2(\gamma +\ln q)\left[ \gamma _1\left( {p \over q}\right) -\gamma _1\right] +\gamma _2- \gamma _2\left( {p \over q}\right) . \end{aligned}$$

In (b), the analytically continuable polylogarithm (or Jonquière) function \(Li_s(z)=\sum _{k=1}^\infty z^k/k^s\). Explicit expressions for the partial derivatives appearing there are provided.

Proposition 5

(Asymptotic expression) For \(a \rightarrow 0\),

$$\begin{aligned} \gamma _k(a) \sim {1 \over a}\ln ^k a + \gamma _k. \end{aligned}$$

Dirichlet \(L\) functions with characters \(\chi _k\) modulo \(k\) may be written as

$$\begin{aligned} L(s)=\sum _{n=1}^\infty {{\chi _k(n)} \over n^s}={1 \over k^s}\sum _{m=1}^k \chi _k(m) \zeta \left( s,{m \over k}\right) , \quad \text {Re} ~s >1 \end{aligned}$$

If \(\chi _k\) is a nonprincipal character, then convergence additionally holds for Re \(s>0\). Such \(L\) functions may be analytically continued throughout the whole complex plane and satisfy a functional equation relating \(L(s)\) to \(L(1-s)\). Then the results of this paper show that the values \(L'(p/q)\) and \(L'(1-p/q)\) may be expressed in closed form.

2 Proof of Propositions

Proposition 1. We will be expanding a functional equation due to Hurwitz ([2], p. 261], [19], p. 93]),

$$\begin{aligned} \zeta \left( s,{j \over m}\right) ={{2^s \pi ^{s-1}} \over m^{1-s}}\Gamma (1-s)\sum _{r=1}^m \sin \left( {{\pi s} \over 2}+{{2\pi j r} \over m}\right) \zeta \left( 1-s,{r \over m}\right) , \end{aligned}$$
(2.1)

holding for \(1 \le j \le m\), about \(s=1\). We will use two elementary trigonometric identities

$$\begin{aligned} -2\sum _{r=1}^m \cos {{2\pi j r} \over m}\zeta \left( 0,{r \over m}\right) =-2\sum _{r=1}^m \left( {1 \over 2}-{r \over m}\right) \cos {{2\pi j r} \over m}=1, \quad 1 \le j<m,\nonumber \\ \end{aligned}$$
(2.2a)

and

$$\begin{aligned} \sum _{r=1}^m \sin {{2\pi j r} \over m}\zeta \left( 0,{r \over m}\right) =\sum _{r=1}^m \left( {1 \over 2}-{r \over m}\right) \sin {{2\pi j r} \over m}={1 \over 2} \cot {{\pi j} \over m}, \quad 1 \le j<m,\nonumber \\ \end{aligned}$$
(2.2b)

as well as the values

$$\begin{aligned} \zeta ''(0)=\gamma _1+{\gamma ^2 \over 2}-{\pi ^2 \over {24}}-{{\ln ^2 2\pi } \over 2}, \end{aligned}$$
(2.3a)

and

$$\begin{aligned} \zeta '''(0)\!=\!\gamma ^3\!+\!{3 \over 2}\gamma ^2 \ln 2\pi -{\pi ^2 \over 8}\ln 2\pi -{1 \over 2} \ln ^3 2\pi \!+\!3(\gamma \!+\!\ln 2\pi ) \gamma _1\!+\!{3 \over 2}\gamma _2-\zeta (3).\nonumber \\ \end{aligned}$$
(2.3b)

These derivative values may be obtained via (1.1) and the functional equation of the zeta function, \(\zeta (s)=2(2\pi )^{s-1} \sin (\pi s/2)\Gamma (1-s)\zeta (1-s)\). For (2.2) we have used the well-known relation \(\zeta (0,a)=1/2-a=-B_1(a)\), where \(B_1(a)\) is the first-degree Bernoulli polynomial. We recall that about \(s=1\), \(\Gamma (1-s)\) has a simple pole and that

$$\begin{aligned} \Gamma (1-s)&=-{1 \over {s-1}}-\gamma -\left( {\pi ^2 \over {12}}+{\gamma ^2 \over 2}\right) (s-1) \\&\quad \, -{1 \over 6}\left( \gamma ^3+{\gamma \over 2}\pi ^2-\psi ''(1)\right) (s-1)^2+O[(s-1)^3], \end{aligned}$$

wherein the tetragamma function value \(\psi ''(1)=-2\zeta (3)\), and recall the expansion \((2\pi m)^{s-1}=\sum _{\ell =0}^\infty {{\ln ^\ell 2\pi m} \over {\ell !}}(s-1)^\ell \). For the sine factor on the right side of (2.1), we have

$$\begin{aligned} \sin \left( {{\pi s} \over 2}+{{2\pi j r} \over m}\right) =&\cos {{2\pi jr} \over m} \sum _{\ell =0}^\infty {{(-1)^\ell \pi ^{2\ell }} \over {(2\ell )!2^{2\ell }}}(s-1)^{2\ell } \\&+\sin {{2\pi jr} \over m}\sum _{\ell =0}^\infty {{(-1)^{\ell +1} \pi ^{2\ell +1}} \over {(2\ell +1)!2^{2\ell +1}}}(s-1)^{2\ell +1}. \end{aligned}$$

The left side of (2.1) expands as

$$\begin{aligned} \zeta \left( s,{j \over m}\right) \!=\!{1 \over {s-1}}\!-\!\psi \left( {j \over m}\right) \!-\!\gamma _1 \left( {j \over m}\right) (s-1)\!+\!{1 \over 2}\gamma _2\left( {j \over m}\right) (s\!-\!1)^2 \!+\! O[(s-1)^3]. \end{aligned}$$

The polar contributions on the two sides of (2.1) cancel due to the cosine sum (2.2a). At order \((s-1)^0\), one finds

$$\begin{aligned} \psi \left( {j \over m}\right)= & {} \sum _{r=1}^m\left\{ -\pi \zeta \left( 0,{r \over m}\right) \sin \left( {{2\pi jr} \over m}\right) \right. \nonumber \\&\qquad \quad \left. +2\cos {{2\pi jr} \over m}\left[ \zeta \left( 0,{r \over m}\right) (\gamma \!+\!\ln 2\pi m)\!-\!\zeta '\left( 0,{r \over m}\right) \right] \right\} , \end{aligned}$$
(2.4)

where both of the sums of (2.2) apply. With \(\zeta '(0,r/m)=\ln \Gamma (r/m)-\ln (2\pi )/2\),

$$\begin{aligned} \psi \left( {j \over m}\right) =-\gamma -\ln 2\pi m-{\pi \over 2}\cot {{\pi j} \over m} -2\sum _{r=1}^{m-1} \cos {{2\pi jr} \over m}\ln \Gamma \left( {r \over m}\right) , \end{aligned}$$

being a form of one of Gauss’ formulas for \(\psi \) at rational argument.

At order \((s-1)^1\), one finds

$$\begin{aligned} -\gamma _1\left( {j \over m}\right) =&\sum _{r=1}^m\left\{ {\pi ^2 \over 4}\cos {{2\pi jr} \over m}\zeta \left( 0,{r \over m}\right) \right. \nonumber \\&\quad \left. -{\pi \over 2}\sin {{2\pi jr} \over m} \left[ \zeta \left( 0,{r \over m}\right) (\gamma +\ln 2\pi m) +2\zeta '\left( 0,{r \over m}\right) \right] \right. \nonumber \\&\quad \left. +\cos {{2\pi jr} \over m}\left[ -\zeta \left( 0,{r \over m}\right) \left( {\pi ^2 \over 6} +(\gamma +\ln 2\pi m)^2\right) \right. \right. \nonumber \\&\quad \left. \left. +2(\gamma +\ln 2\pi m)\zeta '\left( 0,{r \over m}\right) - \zeta ''\left( 0,{r \over m}\right) \right] \right\} , \end{aligned}$$
(2.5)

and again the trigonometric summations of (2.2) apply. We add and subtract \((\gamma +\ln 2\pi m)\cos (2\pi jr/m)\zeta (0,r/m)\) to introduce the \(\psi (j/m)\) expression (2.4). Then separating the \(r=m\) terms of the sums and using \(\zeta ''(0)\) from (2.3a) gives part (a).

For part (b), at order \((s-1)^2\) in (2.1) we have

$$\begin{aligned} {1 \over 2}\gamma _2\left( {j \over m}\right) =A+B+C, \end{aligned}$$

where

$$\begin{aligned} A=&\sum _{r=1}^m\left\{ -{\pi ^3 \over {24}}\zeta \left( 0,{r \over m}\right) \sin {{2\pi jr} \over m} \right. \nonumber \\&\left. \quad +{\pi ^2 \over 4}\cos {{2\pi jr} \over m} \left[ \zeta \left( 0,{r \over m}\right) (\gamma + \ln 2\pi m)-\zeta '\left( 0,{r \over m}\right) \right] \right. \nonumber \\&\quad -{\pi \over 2}\sin {{2\pi jr} \over m}\left[ -\zeta \left( 0,{r \over m}\right) \left( {\pi ^2 \over 6}+(\gamma +\ln 2\pi m)^2\right) \right. \\&\quad \left. \left. +2(\gamma +\ln 2\pi m)\zeta '\left( 0,{r \over m}\right) -\zeta ''\left( 0,{r \over m}\right) \right] \right\} , \end{aligned}$$
$$\begin{aligned} B&=-\sum _{r=1}^m \cos {{2\pi jr} \over m}\zeta \left( 0,{r \over m}\right) \left[ {1 \over 3} \ln ^3 2\pi m-\left( \gamma ^2+{\pi ^2 \over 6}\right) \ln 2\pi m\right. \nonumber \\&\quad \left. -\gamma \ln ^2 2\pi m \right. \left. +{1 \over 3} \left( \gamma ^3-{\gamma \over 2}\pi ^2+2\zeta (3)\right) \right] , \end{aligned}$$

and

$$\begin{aligned} C&=\sum _{r=1}^m \cos {{2\pi jr} \over m}\left\{ \left[ {\pi ^2 \over 6}+(\gamma +\ln 2\pi m)^2 \right] \zeta '\left( 0,{r \over m}\right) \right. \nonumber \\&\quad \, \left. -(\gamma +\ln 2\pi m)\zeta ''\left( 0,{r \over m}\right) \right. \left. +{1 \over 3}\zeta '''\left( 0,{r \over m}\right) \right\} . \end{aligned}$$

The sum in \(B\) is immediately evaluated with the aid of (2.2a). The expression for \(A\) is rewritten by using (2.4), making use of the values \(\psi (j/m)\), and (2.2b). In the sum \(C\), the \(r=m\) terms are separated and all of the values \(\zeta '(0)=-\ln (2\pi )/2\), (2.3a), and (2.3b) are used. Combining terms then yields (1.4). \(\square \)

Remark

The expression for \(C\) in the proof may be written in terms of \(\gamma _1(j/m)\) by using (2.5). For the evaluation of general \(\gamma _k(j/m)\), there will always be a sum \(\sum _{r=1}^{m-1} \cos {{2\pi jr} \over m}\zeta ^{(k+1)}\left( 0,{r \over m}\right) \), as according to the highest derivative of \(\zeta (s,a)\) present, there will be no derivative of the sine factor in (2.1). The evaluation will then also contain a contribution of \(\zeta ^{(k+1)}(0,1)=\zeta ^{(k+1)}(0)\).

Corollary 2. The proof uses the discrete orthogonality of sine and cosine functions, implying

$$\begin{aligned} \sum _{j=1}^{m-1}\cot {{\pi j} \over m}\cos {{2\pi j\ell } \over m}=0 \end{aligned}$$

and

$$\begin{aligned} \sum _{j=1}^{m-1}\cot {{\pi j} \over m}\sin {{2\pi j\ell } \over m}=m-2\ell \end{aligned}$$

along with

$$\begin{aligned} \sum _{j=1}^{m-1}\psi \left( {j \over m}\right) \cos {{2\pi j \ell } \over m}=m\ln \left( 2\sin {{\pi \ell } \over m}\right) +\gamma \end{aligned}$$

and

$$\begin{aligned} \sum _{j=1}^{m-1}\psi \left( {j \over m}\right) \sin {{2\pi j \ell } \over m}={\pi \over 2} (2\ell -m). \end{aligned}$$

The three sums following part (b) may be determined by successively differentiating the relation \(\sum _{\ell =1}^{m-1}\zeta \left( s,{\ell \over m}\right) =(m^s-1)\zeta (s)\) and putting \(s\) to \(0\). \(\square \)

Proposition 2. We will indicate four proofs. For the first, we may use a standard integral representation for the polygamma function [16], p. 943] to write

$$\begin{aligned} \sum _{n=0}^\infty (k-1)^n {z^n \over {n!}}\psi ^{(n)}(z)&=\psi (z)+\sum _{n=1}^\infty (k-1)^n {z^n \over {n!}}\int _0^1 {{t^{z-1}\ln ^n t} \over {t-1}}\mathrm{{d}}t \nonumber \\&=\psi (z)+\int _0^1 {{[1-t^{(k-1)z}]} \over {t-1}}t^{z-1}\mathrm{{d}}t \nonumber \\&=\psi (z)-\psi (z)+\psi (kz)=\psi (kz). \end{aligned}$$

\(\square \)

For the second, we may start with the expansion [16], p. 944]

$$\begin{aligned} \psi (x+1)=\psi (x)+{1 \over x}=-\gamma +\sum _{j=2}^\infty (-1)^j \zeta (j)x^{j-1}, \quad |x|<1. \end{aligned}$$

Then

$$\begin{aligned} \psi ^{(n)}(z)={{(-1)^{n+1}n!} \over z^{n+1}}+\sum _{j=2}^\infty (-1)^j \zeta (j){{(j-1)!} \over {(j-n-1)!}}z^{j-n-1}, \end{aligned}$$

so that

$$\begin{aligned}&\sum _{n=0}^\infty (k-1)^n {z^n \over {n!}}\psi ^{(n)}(z) \nonumber \\&\quad =\psi (z)+\sum _{n=1}^\infty (k-1)^n {z^n \over {n!}}\left[ {{(-1)^{n+1}n!} \over z^{n+1}}+\sum _{j=2}^\infty (-1)^j\zeta (j){{(j-1)!} \over {(j-n-1)!}}z^{j-n-1}\right] \nonumber \\&\quad =\psi (z)-{1 \over {kz}}+{1 \over z}+\sum _{j=2}^\infty (-1)^j \zeta (j)z^{j-1}\sum _{n=1}^{j-1}(k-1)^n {{j-1} \atopwithdelims ()n} \nonumber \\&\quad =\psi (z)-{1 \over {kz}}+{1 \over z}+\sum _{j=2}^\infty (-1)^j \zeta (j)z^{j-1}(k^{j-1}-1)\nonumber \\&\quad =\psi (z)-{1 \over {kz}}+{1 \over z}+\psi (kz)+{1 \over {kz}}-\psi (z)-{1 \over z}=\psi (kz). \end{aligned}$$

\(\square \)

A third method of proof follows from the representations

$$\begin{aligned} \gamma _0(a)=-\psi (a)=-\ln a-\sum _{k=1}^\infty {1 \over {k+1}}\sum _{\ell =0}^k (-1)^\ell {k \atopwithdelims ()\ell } \ln (\ell +a) \end{aligned}$$

and

$$\begin{aligned} \psi ^{(j)}(a)=(-1)^{j-1}{{(j-1)!} \over a^j}+(j-1)!\sum _{k=1}^\infty {1 \over {k+1}}\sum _{\ell =0}^k (-1)^\ell {k \atopwithdelims ()\ell } {{(-1)^{j-1}} \over {(\ell +a)^j}}. \end{aligned}$$

\(\square \)

The fourth method of proof follows the first proof of Proposition 3 so we omit it.

Proposition 3. We have the multiplication formula

$$\begin{aligned} \zeta (s,kz)=\sum _{n=0}^\infty {{s+n-1} \atopwithdelims ()n}(1-k)^n z^n \zeta (s+n,z), \end{aligned}$$

being a case of a more general result of Truesdell [30]. We then expand both sides about \(s=1\). Equating the coefficients of \((s-1)^0\) on both sides gives another means of demonstrating Proposition 2. Equating the coefficients of \((s-1)^1\) and using Proposition 2 gives part (a). Equating the coefficients of \((s-1)^2\) gives part (b). \(\square \)

For a second method of proof of part (a), we may use the integral representation for Re \(s>1\) and Re \(a>0\),

$$\begin{aligned} \zeta (s,a)={1 \over {\Gamma (s)}}\int _0^\infty {{t^{s-1}e^{-(a-1)t}} \over {e^t-1}}\mathrm{{d}}t, \end{aligned}$$

so that

$$\begin{aligned} {{\partial \zeta (s,a)} \over {\partial s}}=-\psi (s)\zeta (s,a)+{1 \over {\Gamma (s)}}\int _0^\infty {{t^{s-1}e^{-(a-1)t}} \over {e^t-1}}\ln t ~\mathrm{{d}}t. \end{aligned}$$

Then

$$\begin{aligned}&\sum _{n=1}^\infty (1-k)^n z^n \left. {{\partial \zeta (s,z)} \over {\partial s}}\right| _{s=n+1} \nonumber \\&\quad =\sum _{n=1}^\infty (1-k)^n z^n \left[ -\psi (n+1)\zeta (n+1,z)+{1 \over {n!}}\int _0^\infty {{t^n e^{-(a-1)t}} \over {e^t-1}}\ln t ~\mathrm{{d}}t\right] \nonumber \\&\quad =-\sum _{n=1}^\infty (1-k)^n z^n \psi (n+1)\zeta (n+1,z)+\int _0^\infty e^{-(z-1)t}{{[e^{-(k-1)tz} -1]} \over {e^t-1}}\ln t ~\mathrm{{d}}t. \end{aligned}$$

Now

$$\begin{aligned} \int _0^\infty t^{s-1}e^{-(z-1)t}{{[e^{-(k-1)tz}-1]} \over {e^t-1}}\mathrm{{d}}t=\Gamma (s)[\zeta (s,kz)-\zeta (s,z)], \end{aligned}$$

giving

$$\begin{aligned}&\int _0^\infty t^{s-1}e^{-(z-1)t}{{[e^{-(k-1)tz}-1]} \over {e^t-1}}\ln t ~\mathrm{{d}}t \nonumber \\&\quad =\Gamma (s)\psi (s)[\zeta (s,kz)-\zeta (s,z)] +\Gamma (s) \left[ {{\partial \zeta } \over {\partial s}}(s,kz)-{{\partial \zeta }\over {\partial s}}(s,z)\right] . \end{aligned}$$

Since by (1.1)

$$\begin{aligned} {{\partial \zeta }\over {\partial s}}(s,a)&=-{1 \over {(s-1)^2}}+\sum _{n=0}^\infty {{(-1)^{n+1}} \over {n!}}\gamma _{n+1}(a)(s-1)^n \nonumber \\&=-{1 \over {(s-1)^2}}-\gamma _1(a)(s-1)^0+{{\gamma _2(a)} \over 2}(s-1)^1 +O[(s-1)^2], \end{aligned}$$

we find

$$\begin{aligned} \int _0^\infty e^{-(z-1)t}{{[e^{-(k-1)tz}-1]} \over {e^t-1}}\ln t ~\mathrm{{d}}t=\gamma [\psi (kz)-\psi (z)] +\gamma _1(kz)-\gamma _1(z), \end{aligned}$$

and the Proposition again follows. \(\square \)

Remarks

Obviously we may evaluate the integrals

$$\begin{aligned} \int _0^\infty e^{-(z-1)t}{{[e^{-(k-1)tz}-1]} \over {e^t-1}}\ln ^j t ~\mathrm{{d}}t \end{aligned}$$

in terms of the difference \(\gamma _j(kz)-\gamma _j(z)\).

The harmonic numbers \(H_n=\sum _{k=1}^n 1/k=\psi (n)+\gamma \) and generalized harmonic numbers

$$\begin{aligned} H_n^{(r)}=&\sum _{k=1}^n {1 \over k^r} ={{(-1)^{r-1}} \over {(r-1)!}}\left[ \psi ^{(r-1)}(n+1) -\psi ^{(r-1)}(1)\right] \nonumber \\ =&{{(-1)^{r-1}} \over {(r-1)!}}\int _0^1{{(t^n-1)} \over {t-1}}\ln ^{r-1}t ~\mathrm{{d}}t \end{aligned}$$

enter the representations of Proposition 3 and for the higher Stieltjes constants. This is part of the elaboration of the following discussion section.

Proposition 4. We first write, for \(0<p<q\),

$$\begin{aligned} I^k_{pq}\equiv q\int _0^1 \left( {{x^{q-1}-x^{p-1}} \over {1-x^q}}\right) \ln ^k(-\ln x)\mathrm{{d}}x =\sum _{k=1}^{q-1} \int _0^1 \left( {{\omega _k^p-1} \over {x-\omega _k}}\right) \ln ^k(-\ln x)\mathrm{{d}}x. \end{aligned}$$

By performing logarithmic differentiation on the integral

$$\begin{aligned} \int _0^1 {{(-\ln x)^a} \over {x-\omega }}\mathrm{{d}}x=-\Gamma (a)\text{ Li }_a(\omega ^{-1}), \end{aligned}$$

one then has the following expressions:

$$\begin{aligned} I_{pq}^1=-\sum _{k=1}^{q-1}(\omega _k^p-1)\left[ \gamma \ln \left( {{\omega _k-1} \over \omega _k} \right) +\left. {{\partial \text{ Li }_s} \over {\partial s}}(\omega _k^{-1})\right| _{s=1}\right] \end{aligned}$$

and

$$\begin{aligned} I_{pq}^2&=\sum _{k=1}^{q-1}(\omega _k^p-1)\left[ \left( \gamma ^2+{\pi ^2 \over 6}\right) \ln \left( {{\omega _k-1} \over \omega _k}\right) \nonumber \right. \\&\quad \left. +2\gamma \left. {{\partial \text{ Li }_s} \over {\partial s}}(\omega _k^{-1})\right| _{s=1}-\left. {{\partial ^2 \text{ Li }_s} \over {\partial s^2}}(\omega _k^{-1})\right| _{s=1}\right] . \end{aligned}$$

We next present the partial derivatives of the polylogarithm function. These result from expansion of the following expression in powers of \(s-1\) [14]:

$$\begin{aligned} \text{ Li }_s(z)=\Gamma (1-s)\ln ^{s-1}\left( {1 \over z}\right) +\sum _{n=0}^\infty \zeta (s-n) {{\ln ^n z} \over {n!}}, \quad |\ln z|<2\pi , \end{aligned}$$

wherein the polar part of the first term on the right is cancelled by the pole \(1/(s-1)\) of the \(n=0\) term of the sum. We obtain

$$\begin{aligned}&\left. {{\partial \text{ Li }_s} \over {\partial s}}(z)\right| _{s=1}=-\gamma _1-{\gamma ^2 \over 2}-{\pi ^2 \over {12}}-\gamma \ln (-\ln z)-{1 \over 2}\ln ^2(-\ln z) \nonumber \\&\qquad \qquad \qquad \qquad +\sum _{n=1}^\infty \zeta '(1-n){{\ln ^n z} \over {n!}} \end{aligned}$$
(2.6)

and

$$\begin{aligned}&\left. {{\partial ^2 \text{ Li }_s} \over {\partial s^2}}(z)\right| _{s=1}={1 \over 6}\left[ -2\gamma ^3-\gamma \pi ^2-6\gamma ^2\ln (-\ln z)-\pi ^2\ln (-\ln z)-6\gamma \ln ^2(-\ln z) \right. \nonumber \\&\qquad \qquad \qquad \qquad \,\left. -2\ln ^3(-\ln z) -4\zeta (3)\right] +\gamma _2+\sum _{n=1}^\infty \zeta ''(1-n){{\ln ^n z} \over {n!}}. \end{aligned}$$
(2.7)

Part (a) then makes use of the first derivative and the sum \(\sum _{k=1}^{q-1}(\omega _k^p-1)=-q\).

For the second evaluation of \(I_{pq}^k\), we use

$$\begin{aligned} I_{pq}^k&=q\sum _{m=0}^\infty \int _0^1 (x^{q-1}-x^{p-1})x^{qm}\ln ^k(-\ln x)\mathrm{{d}}x \nonumber \\&=q\sum _{m=0}^\infty \int _0^\infty \left( e^{-(q-1)u}-e^{-(p-1)u}\right) e^{-(qm+1)u}\ln ^k u ~\mathrm{{d}}u. \end{aligned}$$

By using logarithmic differentiation of the Gamma function integral,

$$\begin{aligned} I_{pq}^1&=q\sum _{m=0}^\infty \left[ {{\gamma +\ln (p+mq)} \over {p+mq}}-{{\gamma +\ln (m+1)q} \over {(m+1)q}}\right] \nonumber \\&=-(\gamma +\ln q)\left[ \gamma +\psi \left( {p \over q}\right) \right] +\sum _{m=0}^\infty \left[ {{\ln (m+p/q)} \over {m+p/q}}-{{\ln (m+1)} \over {m+1}}\right] \nonumber \\&=-(\gamma +\ln q)\left[ \gamma +\psi \left( {p \over q}\right) \right] +\gamma _1\left( {p \over q} \right) -\gamma _ 1, \end{aligned}$$

wherein we applied (1.2) and the well-known summation (e.g., [16], p. 943])

$$\begin{aligned} \psi (x)=-\gamma -\sum _{k=0}^\infty \left( {1 \over {x+k}}-{1 \over {k+1}}\right) . \end{aligned}$$

The other evaluation of (b) goes similarly, with

$$\begin{aligned} I_{pq}^2&=q\sum _{m=0}^\infty \int _0^\infty \left( e^{-(q-1)u}-e^{-(p-1)u}\right) e^{-(qm+1)u} \ln ^2 u ~\mathrm{{d}}u \nonumber \\&=\sum _{m=0}^\infty \left\{ {1 \over {6(m+1)}}\left[ 6\gamma ^2+\pi ^2+12\gamma \ln (m+1)q+6\ln ^2 (m+1)q\right] \right. \nonumber \\&\quad \left. -{1 \over {6(m+p/q)}}\left[ 6\gamma ^2+\pi ^2+12\gamma \ln (mq+p)q+6\ln ^2 (mq+p)\right] \right\} \nonumber \\&=\left( \gamma ^2+{\pi ^2 \over 6}\right) \left[ \gamma +\psi \left( {p \over q}\right) \right] +2\gamma \sum _{m=0}^\infty \left[ {{\ln (m+1)q} \over {m+1}}-{{\ln (mq+p)} \over {m+p/q}}\right] \nonumber \\&\quad +\sum _{m=0}^\infty \left[ {{\ln ^2(m+1)q} \over {m+1}}-{{\ln ^2(mq+p)} \over {m+p/q}}\right] \nonumber \\&=\left( \gamma ^2+{\pi ^2 \over 6}+2\gamma \ln q\right) \left[ \gamma +\psi \left( {p \over q}\right) \right] -2\gamma \left[ \gamma _1\left( {p \over q}\right) -\gamma _1\right] \nonumber \\&\quad +\sum _{m=0}^\infty \left[ {{\ln ^2(m+1)} \over {m+1}}-{{\ln ^2(m+p/q)} \over {m+p/q}} +2\ln q\left( {{\ln (m+1)} \over {m+1}}-{{\ln (m+p/q)} \over {m+p/q}}\right) \right. \nonumber \\&\quad \left. +\ln ^2q \left( {1 \over {m+1}}-{1 \over {m+p/q}}\right) \right] \nonumber \\&=\left( \gamma ^2+{\pi ^2 \over 6}+2\gamma \ln q+\ln ^2 q\right) \left[ \gamma +\psi \left( {p \over q}\right) \right] \nonumber \\&\quad -2(\gamma +\ln q)\left[ \gamma _1\left( {p \over q}\right) -\gamma _1\right] +\gamma _2- \gamma _2\left( {p \over q}\right) . \end{aligned}$$

\(\square \)

Remark

For applications or computation with Proposition 4, it is important that the values of \(\ln (\pm \omega _k^{\pm 1})\) are kept to the principal branch, e.g., with \(-\pi <\text{ Im } ~\ln z \le \pi \). This requirement maintains a real-valued result for \(I_{pq}^k\).

Elaboration of the partial derivative (2.6).

The partial derivative (2.6) may also be represented as

$$\begin{aligned} -\left. {{\partial \text{ Li }_s(z)} \over {\partial s}}\right| _{s=1}=\int _1^\infty {z^x \over x}\ln x ~\mathrm{{d}}x +\int _1^\infty z^x\left( {{(\ln z)\ln x} \over x}+{1 \over x^2}-{{\ln x} \over x^2} \right) P_1(x)\mathrm{{d}}x,\nonumber \\ \end{aligned}$$
(2.8)

following from [8]

$$\begin{aligned} -{{\partial \text{ Li }_s(z)} \over {\partial s}}=\int _1^\infty {z^x \over x^s} \ln x ~\mathrm{{d}}x +\int _1^\infty z^x\left( {{(\ln z)\ln x} \over x^s}+{1 \over x^{s+1}}-s {{\ln x} \over x^{s+1}}\right) P_1(x)\mathrm{{d}}x, \end{aligned}$$

wherein \(P_1(x) \equiv B_1(x-[x])\). We relate (2.6) and (2.8). By using the expansion \(z^x=\sum _{j=0}^\infty \ln ^j z{x^j \over {j!}}\), we have

$$\begin{aligned} \int _1^a {z^x \over x}\ln x ~\mathrm{{d}}x&=\sum _{j=0}^\infty {{\ln ^j z} \over {j! j^2}}[1+a^j(j\ln a-1)] \nonumber \\&=-\gamma \ln a-\Gamma (0,-a\ln z)\ln a +{{\ln ^2 a} \over 2} \nonumber \\&\quad \, +\ln z~_3F_3(1,1,1;2,2,2;\ln z) \nonumber \\&\quad \, -a\ln z ~_3F_3(1,1,1;2,2,2;a\ln z)-\ln a \ln (-a\ln z), \end{aligned}$$

where \(_pF_q\) is the generalized hypergeometric function and the incomplete Gamma function \(\Gamma (\alpha ,x)=\Gamma (\alpha )-\sum _{n=0}^\infty {{(-1)^nx^{n+\alpha }} \over {n!(n+\alpha )}}\). The asymptotic form of the \(_3F_3\) function as \(a \rightarrow \infty \) may be considered as in [11] and the result is

$$\begin{aligned} \int _1^\infty {z^x \over x}\ln x ~\mathrm{{d}}x&={\gamma ^2 \over 2}+{\pi ^2 \over {12}}+\ln z ~_3F_3(1,1,1;2,2,2;\ln z) \nonumber \\&\quad +\gamma \ln (-\ln z)+{1 \over 2}\ln ^2(-\ln z), \quad |z|<1. \end{aligned}$$
(2.9)

This result (2.9) may also be obtained as a reduction of a Meijer-\(G\) function. However, we omit details of this evaluation.

By using respectively the partial derivative of (1.1) with respect to \(s\) and then (2.8) and (2.9), we have these additional expressions for the partial derivative (2.6):

$$\begin{aligned} \left. {{\partial \text{ Li }_s} \over {\partial s}}(z)\right| _{s=1}&=-\gamma _1-{\gamma ^2 \over 2}-{\pi ^2 \over {12}}-\gamma \ln (-\ln z)-{1 \over 2}\ln ^2(-\ln z) \nonumber \\&\quad -\ln z ~_3F_3(1,1,1;2,2,2;\ln z)-\sum _{j=0}^\infty {\gamma _{j+1} \over {j!}}\sum _{n=1}^\infty {{\ln ^n z} \over {n!}}n^j \nonumber \\&=-{\gamma ^2 \over 2}-{\pi ^2 \over {12}}-\gamma \ln (-\ln z)-{1 \over 2}\ln ^2(-\ln z) \nonumber \\&\quad -\ln z ~_3F_3(1,1,1;2,2,2;\ln z) \nonumber \\&\quad -\int _1^\infty z^x\left( {{(\ln z)\ln x} \over x}+{1 \over x^2}-{{\ln x} \over x^2}\right) P_1(x)\mathrm{{d}}x \nonumber \\&=-{\gamma ^2 \over 2}-{\pi ^2 \over {12}}-\gamma \ln (-\ln z)-{1 \over 2}\ln ^2(-\ln z) \nonumber \\&\quad -\ln z ~_3F_3(1,1,1;2,2,2;\ln z) \nonumber \\&\quad -\sum _{j=1}^\infty \int _0^1 z^{y+j}\left[ {{(\ln z)\ln (y\!+\!j)} \over {y\!+\!j}}\!+\!{1 \over {(y\!+\!j)^2}} \!-\!{{\ln (y\!+\!j)} \over {(y\!+\!j)^2}}\right] \left( y\!-\!{1 \over 2}\right) \mathrm{{d}}y. \end{aligned}$$

By comparing (2.6) with the second expression above, we conclude that

$$\begin{aligned}&\gamma _1-\sum _{n=1}^\infty \zeta '(1-n){{\ln ^n z} \over {n!}} \nonumber \\&=\ln z ~_3F_3(1,1,1;2,2,2;\ln z) +\int _1^\infty z^x\left( {{(\ln z)\ln x} \over x}+{1 \over x^2}-{{\ln x} \over x^2}\right) P_1(x)\mathrm{{d}}x. \end{aligned}$$

Proposition 5. Let \(C_k(a) \equiv \gamma _k(a) -(\ln ^k a)/a\). With \(B_n(x)\) being the Bernoulli polynomials, their periodic extension is denoted as \(P_n(x) \equiv B_n(x-[x])\) and we have the representation [32]

$$\begin{aligned} C_n(a) = (-1)^{n-1} n!\sum _{k=0}^{n+1} {{s(n+1,n+1-k)} \over {k!}} \int _1^\infty P_n(x-a) {{\ln ^k x} \over x^{n+1}} \mathrm{{d}}x, \quad n \ge 1,\nonumber \\ \end{aligned}$$
(2.10)

with \(s(n,k)\) as the Stirling numbers of the first kind. We recall the Fourier expansions of \(P_n(x)\) [1], p. 805],

$$\begin{aligned} P_{2n}(x-a)=(-1)^{n-1} {{2(2n)!} \over {(2\pi )^{2n}}}\sum _{k=1}^\infty {{\cos 2\pi k(x-a)} \over {k^{2n}}}, \end{aligned}$$

and

$$\begin{aligned} P_{2n-1}(x-a)=(-1)^n {{2(2n-1)!} \over {(2\pi )^{2n-1}}}\sum _{k=1}^\infty {{\sin 2\pi k(x-a)} \over {k^{2n-1}}}. \end{aligned}$$

We therefore obtain

$$\begin{aligned} P_{2n}(x-a)=(-1)^{n-1} {{2(2n)!} \over {(2\pi )^{2n}}}\sum _{k=1}^\infty {1 \over {k^{2n}}}[\cos 2\pi k x+2\pi ka\sin 2\pi kx +O(a^2)] \end{aligned}$$

and

$$\begin{aligned} P_{2n-1}(x-a)=(-1)^n {{2(2n-1)!} \over {(2\pi )^{2n-1}}}\sum _{k=1}^\infty {1 \over {k^{2n-1}}}[\sin 2\pi kx-2\pi k a\cos 2\pi kx +O(a^2)]. \end{aligned}$$

These forms are then inserted into (2.10). Noting that \(C_k(1)=\gamma _k\), the \(a^0\) term produces \(\gamma _k\), \(C_k(a) \rightarrow \gamma _k\) as \(a \rightarrow 0\) and hence the result. \(\square \)

3 Discussion

Here we first discuss the equivalence of Proposition 3 as a case of an addition formula which we have previously presented [12] (Proposition 1). We then show applications of differences of Stieltjes constants to some classic integrals of analytic number theory. We exhibit a new proof technique for certain log–log integrals.

As regards the Truesdell representation of \(\zeta (s,kz)\), we note

$$\begin{aligned} {{s+n-1} \atopwithdelims ()n}&={{(-1)^n} \over {n!}}(1-s-n)_n={{(-1)^n} \over {n!}}{{\Gamma (1-s)} \over {\Gamma (1-s-n)}} \nonumber \\&\quad ={1 \over {n!}}{{\Gamma (s+n)} \over {\Gamma (s)}}={1 \over {n!}}(s)_n, \end{aligned}$$

so that

$$\begin{aligned} \zeta (s,kz)=\sum _{n=0}^\infty {{(s)_n} \over {n!}}(1-k)^n z^n \zeta (s+n,z). \end{aligned}$$

On the other hand, an old formula of Wilton [31] may be written as

$$\begin{aligned} \zeta (s,a+b)=\sum _{j=0}^\infty {{(-b)^j} \over {j!}}(s)_j\zeta (s+j,a), ~~|b|<|a|, ~~ \text{ Re } ~a>0. \end{aligned}$$

Thus the two formulas correspond with \(b=-(1-k)z\) and \(a=z\). Lemma 1 of [12] provides the derivative values

$$\begin{aligned} \left. \left( {d \over {ds}}\right) ^\ell (s)_j \right| _{s=1}=(-1)^{j+\ell } \ell ! s(j+1,\ell +1). \end{aligned}$$

Therefore, from Proposition 1 of [12], we know the general form of the multiplication formula for the Stieltjes constants,

$$\begin{aligned} \gamma _\ell (kz)=\gamma _\ell (z)+(-1)^\ell \sum _{j=2}^\infty {(k-1)^{j-1}z^{j-1} \over {(j-1)!}} \sum _{k=0}^\ell (-1)^k {\ell \atopwithdelims ()k} s(j,k+1) k!\zeta ^{(\ell -k)}(j,z). \end{aligned}$$

The Stirling numbers of the first kind may indeed be written with the generalized harmonic numbers, and the first few are given by \(s(n+1,1)=(-1)^n n!\), \(s(n+1,2)=(-1)^{n+1}n!H_n\), \(s(n+1,3)=(-1)^n {{n!} \over 2}[H_n^2-H_n^{(2)}]\), and \(s(n+1,4)=(-1)^{n+1}{{n!} \over 6} [H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}]\).

We demonstrate how differences of Stieltjes constants may be used to efficiently evaluate some example log–log integrals, including

$$\begin{aligned} I_{\pm }\equiv \int _0^1 {{\ln (-\ln x)} \over {1\pm x+x^2}}\mathrm{{d}}x. \end{aligned}$$

For \(I_-\),

$$\begin{aligned} I_-=&\int _0^1 \left( {{1+x} \over {1+x^3}}\right) \ln (-\ln x)\mathrm{{d}}x \nonumber \\ =&\sum _{m=0}^\infty (-1)^m\int _0^1 (1+x)x^{3m}\ln (-\ln x)\mathrm{{d}}x \nonumber \\ =&\sum _{m=0}^\infty (-1)^m \int _0^\infty (1+e^{-u})e^{-(3m+1)u}\ln u ~\mathrm{{d}}u \nonumber \\ =&-\sum _{m=0}^\infty (-1)^m\left[ \gamma \left( {1 \over {3m+2}}+{1 \over {3m+1}}\right) +{{\ln (3m+2)} \over {3m+2}}+{{\ln (3m+1)} \over {3m+1}}\right] \nonumber \\ =&-\gamma {{2\pi } \over {3\sqrt{3}}}-{1 \over 6}\sum _{m=0}^\infty \left[ \ln 6\left( {1 \over {m+1/3}}+{1 \over {m+1/6}}-{1 \over {m+5/6}}-{1 \over {m+2/3}}\right) \right. \nonumber \\&\left. +{{\ln (m+1/3)} \over {m+1/3}}+{{\ln (m+1/6)} \over {m+1/6}}-{{\ln (m+5/6)} \over {m+5/6}}-{{\ln (m+2/3)} \over {m+2/3}}\right] \nonumber \\ =&-(\gamma +\ln 6){{2\pi } \over {3\sqrt{3}}}+{1 \over 6}\left[ \gamma _1\left( {2 \over 3} \right) -\gamma _1\left( {1 \over 3}\right) +\gamma _1\left( {5 \over 6}\right) -\gamma _1\left( {1 \over 6}\right) \right] . \end{aligned}$$

We used polygamma function values and (1.2).

For \(I_+\),

$$\begin{aligned} I_+=&\int _0^1 \left( {{1-x} \over {1-x^3}}\right) \ln (-\ln x)\mathrm{{d}}x \nonumber \\ =&\sum _{m=0}^\infty \int _0^1 (1-x)x^{3m}\ln (-\ln x)\mathrm{{d}}x \nonumber \\ =&\sum _{m=0}^\infty \int _0^\infty (1-e^{-u})e^{-(3m+1)u}\ln u ~\mathrm{{d}}u \nonumber \\ =&\sum _{m=0}^\infty \left[ \gamma \left( {1 \over {3m+2}}-{1 \over {3m+1}}\right) +{{\ln (3m+2)} \over {3m+2}}-{{\ln (3m+1)} \over {3m+1}}\right] \nonumber \\ =&-{{\gamma \pi } \over {3\sqrt{3}}}+{1 \over 3}\sum _{m=0}^\infty \left[ {{\ln 3(m+2/3)} \over {m+2/3}}-{{\ln 3(m+1/3)}\over {m+1/3}}\right] \nonumber \\ =&-(\gamma +\ln 3){\pi \over {3\sqrt{3}}}+{1 \over 3}\left[ \gamma _1\left( {2 \over 3} \right) -\gamma _1\left( {1 \over 3}\right) \right] . \end{aligned}$$

More generally,

$$\begin{aligned} I_{+n}=&\int _0^1 {{\ln (-\ln x)} \over {x^{n-1}+x^{n-2}+\cdots +x+1}}\mathrm{{d}}x \nonumber \\ =&{{(\gamma +\ln n)} \over n}\left[ \psi \left( {1 \over n}\right) -\psi \left( {2 \over n}\right) \right] +{1 \over n}\left[ \gamma _1\left( {2 \over n} \right) -\gamma _1\left( {1 \over n}\right) \right] , \nonumber \\ I_{+n}^q=&\int _0^1 {{x^q\ln (-\ln x)} \over {x^{n-1}+x^{n-2}+\cdots +x+1}}\mathrm{{d}}x \nonumber \\ =&{{(\gamma +\ln n)} \over n}\left[ \psi \left( {{q+1} \over n}\right) -\psi \left( {{q+2} \over n}\right) \right] \nonumber \\&+{1 \over n}\left[ \gamma _1\left( {{q+2} \over n} \right) -\gamma _1\left( {{q+1} \over n}\right) \right] , ~~\text{ Re } ~q>-1, \end{aligned}$$

and for \(n\) odd,

$$\begin{aligned} I_{-n}^q =&\int _0^1 {{x^q \ln (-\ln x)} \over {x^{n-1}-x^{n-2}+\cdots -x+1}}\mathrm{{d}}x \nonumber \\ =&{{(\gamma +\ln n)} \over {2n}}\left[ \psi \left( {{q+2} \over {2n}}\right) +\psi \left( {{q+1} \over {2n}}\right) \right. \nonumber \\&\quad \left. -\psi \left( {{n+q+2} \over {2n}}\right) -\psi \left( {{n+q+1} \over {2n}}\right) \right] \nonumber \\&\quad +{1 \over {2n}}\left[ \gamma _1\left( {{n+q+2} \over {2n}}\right) +\gamma _1\left( {{n+q+1} \over {2n}}\right) \right. \nonumber \\&\quad \left. -\gamma _1\left( {{q+2} \over {2n}}\right) -\gamma _1\left( {{q+1} \over {2n}}\right) \right] , ~~\text{ Re } ~q>-1. \end{aligned}$$

Similarly,

$$\begin{aligned} \int _0^1 {{\ln (-\ln x)} \over {1+x^2}}\mathrm{{d}}x=-(\gamma +\ln 4){\pi \over 4}+{1 \over 4} \left[ \gamma _1\left( {3 \over 4}\right) -\gamma _1\left( {1 \over 4}\right) \right] , \end{aligned}$$
$$\begin{aligned} J_p&\equiv \int _0^1 {{\ln (-\ln x)} \over {1+x^p}}\mathrm{{d}}x={1 \over 2}\int _0^\infty {{e^{-(1-p/2)u} \ln u} \over {\cosh (pu/2)}}\mathrm{{d}}u\nonumber \\&={{(\gamma +\ln 2p)} \over {2p}}\left[ \psi \left( {1 \over {2p}}\right) -\psi \left( {{p+1} \over {2p}} \right) \right] \nonumber \\&\quad +{1 \over {2p}}\left[ \gamma _1\left( {{p+1} \over {2p}}\right) -\gamma _1\left( {1 \over {2p}} \right) \right] , ~~\text{ Re } ~p>0, \end{aligned}$$

and

$$\begin{aligned} J_p^2&\equiv \int _0^1 {{\ln ^2(-\ln x)} \over {1+x^p}}\mathrm{{d}}x\nonumber \\&={1 \over {2p}}\left[ \gamma ^2+{\pi ^2 \over 6}+2\gamma \ln 2p+\ln ^2 2p\right] \left[ \psi \left( {{p+1} \over {2p}}\right) -\psi \left( {1 \over {2p}}\right) \right] \nonumber \\&\quad +{1 \over p}(\gamma +\ln 2p)\left[ \gamma _1\left( {1 \over {2p}}\right) -\gamma _1\left( {{p+1} \over {2p}}\right) \right] \nonumber \\&\quad +{1 \over {2p}}\left[ \gamma _2\left( {1 \over {2p}}\right) -\gamma _2\left( {{p+1} \over {2p}}\right) \right] , \end{aligned}$$

with the limits

$$\begin{aligned} \lim _{p \rightarrow \infty }J_p=-\gamma , \quad \lim _{p \rightarrow \infty }J_p^2=\gamma ^2+\zeta (2). \end{aligned}$$

As they should be, these limits are consistent with Corollary 3 and the more general Proposition 5.

From the \(J_p\) evaluation follows the integral identity

$$\begin{aligned}&\int _0^1 \left\{ {{(\gamma \!+\!\ln 2p)} \over {2p}}\left[ \psi \left( {1 \over {2p}}\right) \!-\!\psi \left( {{p+1} \over {2p}}\right) \right] \!+\!{1 \over {2p}}\left[ \gamma _1\left( {{p+1} \over {2p}}\right) \!-\!\gamma _1\left( {1 \over {2p}}\right) \right] \right\} \mathrm{{d}}p\nonumber \\&\quad =\int _0^1 {{[\ln (2x)\!-\!\ln (x+1)]} \over {\ln x}}\ln (-\ln x)\mathrm{{d}}x. \end{aligned}$$

An analogous result applies for

$$\begin{aligned} \int _0^1 J_p^2 \mathrm{{d}}p=\int _0^1 {{[\ln (2x)-\ln (x+1)]} \over {\ln x}}\ln ^2(-\ln x)\mathrm{{d}}x. \end{aligned}$$

As an extension of \(J_p\), for Re \(p>0\) and Re \(q>-1\), we have

$$\begin{aligned} J_p^q&\equiv \int _0^1 {x^q \over {1+x^p}}\ln (-\ln x)\mathrm{{d}}x\nonumber \\&={{(\gamma +\ln 2p)} \over {2p}}\left[ \psi \left( {{q+1} \over {2p}}\right) -\psi \left( {{p+q+1} \over {2p}}\right) \right] \nonumber \\&\quad +{1 \over {2p}}\left[ \gamma _1\left( {{p+q+1} \over {2p}}\right) -\gamma _1\left( {{q+1} \over {2p}}\right) \right] , \end{aligned}$$

with

$$\begin{aligned} \lim _{p \rightarrow \infty }J_p^q =-{{\gamma +\ln (q+1)} \over {q+1}}. \end{aligned}$$

Propositions 1 and 4 apply to all of these integrals. As a brief example, one finds

$$\begin{aligned} I_2\equiv \int _0^1 {{\ln (-\ln x)} \over {1+x^2}}\mathrm{{d}}x={\pi \over 4}\left[ \ln {{8\pi \Gamma ^2(3/4)} \over {\Gamma ^2(1/4)}}-\ln 4\right] ={\pi \over 2}\ln {{\sqrt{2\pi } \Gamma (3/4)} \over {\Gamma (1/4)}}. \end{aligned}$$

The value of \(I_2\) has been known for a long time, and it may of course be written in many equivalent forms. However, the following method of evaluation may be new.

Demonstration 1

$$\begin{aligned} I_2\equiv \int _0^1 {{\ln (-\ln x)} \over {1+x^2}}\mathrm{{d}}x=\int _{\pi /4}^{\pi /2} \ln (\ln (\tan x)) \mathrm{{d}}x ={\pi \over 4}\left[ \ln \left( {\pi \over 8} \right) +2\ln {{\Gamma (3/4)} \over {\Gamma (5/4)}}\right] . \end{aligned}$$

The method below applies to a large variety of integrals, enabling another determination of differences of Stieltjes constants at rational arguments. A key feature of these integrals is integrands with polynomial denominators with zeros at roots of unity.

Proof

Write

$$\begin{aligned} I_2={1 \over {2i}}\int _0^1 \left( {1 \over {x-i}}-{1 \over {x+i}}\right) \ln (-\ln x) ~\mathrm{{d}}x, \end{aligned}$$

and then apply

$$\begin{aligned} \int _0^1 {{\ln (-\ln x)} \over {x-a}}\mathrm{{d}}x=-\gamma \ln \left( {{a-1} \over a}\right) -\left. {{\partial \text{ Li }_s} \over {\partial s}}\right| _{s=1}(a^{-1}), \end{aligned}$$
(3.1)

and (2.6) for the partial derivative to find

$$\begin{aligned} I_2={{\gamma \pi } \over 4}+{\pi \over 2}\ln \left( {\pi \over 2}\right) +{1 \over {2i}} \sum _{n=1}^\infty {{\zeta '(1-n)} \over {n!}}\left( {\pi \over 2}\right) ^n i^n[1-(-1)^n] \end{aligned}$$
$$\begin{aligned} =\pi \left[ {{\gamma } \over 4}+{1 \over 2}\ln \left( {\pi \over 2}\right) \right] +\sum _{m=0}^\infty {{(-1)^m \zeta '(-2m)} \over {(2m+1)!}}\left( {\pi \over 2}\right) ^{2m+1}. \end{aligned}$$

Next separate the \(m=0\) term of the sum and use the functional equation of the zeta function, \(\pi ^{1-z}\zeta (z)=2^z\Gamma (1-z)\zeta (1-z)\sin {{\pi z} \over 2}\), along with \(\zeta (-2m)=0\) for \(m\ge 1\), to determine that for \(m \ge 1\), \(2(-1)^m \zeta '(-2m)=(2m)!\zeta (2m+1)/(2\pi )^{2m}\). There results

$$\begin{aligned} I_2={\pi \over 4}\left[ \gamma +\ln \left( {\pi \over 8}\right) +\sum _{m=1}^\infty {{\zeta (2m+1)} \over {16^m(2m+1)}}\right] . \end{aligned}$$

Using (e.g., [16], p. 939])

$$\begin{aligned} \sum _{n=1}^\infty {x^{2n+1} \over {2n+1}}\zeta (2n+1)=-\gamma x+{1 \over 2}\left[ \ln \Gamma (1-x)-\ln \Gamma (x+1)\right] , \quad |x|<1,\nonumber \\ \end{aligned}$$
(3.2)

completes the evaluation.Footnote 1 \(\square \)

As a further indication of the applicability of this method, we mention

Lemma 1

For \(-\pi <\delta \le \pi \),

$$\begin{aligned} I_\omega \equiv \int _0^1 {{\ln (-\ln x)} \over {x^2-2x\cos \delta +1}}\mathrm{{d}}x =-{\pi \over {2\sin \delta }}\left[ {\delta \over \pi }\ln (2\pi )-\ln \delta + \ln {{\Gamma \left( 1+{\delta \over {2p}}\right) } \over {\Gamma \left( 1-{\delta \over {2p}}\right) }}\right] . \end{aligned}$$

We only sketch the proof as this is a known integral.

Proof

We let \(\omega =e^{i\delta }\) and use the factorization \((x-\omega )(x-\omega ^*)=x^2-(\omega +\omega ^*)x+|\omega |^2=x^2-2x\cos \delta +1\), giving

$$\begin{aligned} I_\omega ={1 \over {\omega -\omega ^*}}\int _0^1\left( {1 \over {x-\omega }}-{1 \over {x-\omega ^*}} \right) \ln (-\ln x) \mathrm{{d}}x. \end{aligned}$$

We employ the integral (3.1), the partial derivative (2.6), and finally the summation (3.2). Along the way we use elementary relations such as \(1/\omega ^*=\omega \) and

$$\begin{aligned} \ln \left[ {{(\omega ^*-1)\omega } \over {\omega ^*(\omega -1)}}\right] =\ln \left( {{1-e^{i\delta }} \over {1-e^{-i\delta }}}\right) =\ln [e^{i(\delta +\pi )}]=i(\delta +\pi ). \end{aligned}$$

\(\square \)

Similarly, integrals of the form

$$\begin{aligned} \int _0^1 {{\ln ^2(-\ln x)} \over {p(x)}}\mathrm{{d}}x, \end{aligned}$$

with \(p\) a polynomial having as zeros roots of unity, may be evaluated with the aid of

$$\begin{aligned}&\int _0^1 {{\ln ^2(-\ln x)} \over {x-a}}\mathrm{{d}}x=\left( \gamma ^2+{\pi ^2 \over 6}\right) \ln \left( {{a-1} \over a}\right) \\&\quad +2\gamma \left. {{\partial \text{ Li }_s} \over {\partial s}}\right| _{s=1} (a^{-1})-\left. {{\partial ^2 \text{ Li }_s} \over {\partial s^2}}\right| _{s=1} (a^{-1}), \end{aligned}$$

and the partial derivative (2.7). In summary, this method comprises the use of partial fractions, logarithmic differentiation of a polylogarithm integral, application of the partial derivatives of Li\(_s\) at \(s=1\), application of the functional equation of the Riemann zeta function, and summation to \(\ln \Gamma \) constants where pertinent.

Many other integrals follow from the results of this paper. For instance, for \(|a|=1\) but \(a\ne 1\) or Re \(a>1\), we have

$$\begin{aligned} \int _0^1 {{\ln (-\ln x)} \over {(x-a)^{m+1}}}\mathrm{{d}}x={\gamma \over m}(-1)^m\left[ {1 \over {(a-1)^m}} -{1 \over a^m}\right] -{1 \over {m!}}\left( {\partial \over {\partial a}}\right) ^m \left. {{\partial \text{ Li }_s} \over {\partial s}}\right| _{s=1} \left( {1 \over a}\right) . \end{aligned}$$

This follows from (3.1) and (2.6). In particular,

$$\begin{aligned} \int _0^1 {{\ln (-\ln x)} \over {(x-a)^2}}\mathrm{{d}}x={1 \over a}\left\{ \gamma \left[ {1 \over {\ln a}} -{1 \over {(a-1)}}\right] +{{\ln (\ln a)} \over {\ln a}}+\sum _{n=0}^\infty {{\zeta '(-n)} \over {n!}} \ln ^n(a^{-1}). \right\} \end{aligned}$$

Such a result may be combined with the use of partial fractions to yield yet other integrals.