Partial Sums of the Hurwitz and Allied Functions and Their Special Values

Abstract

We supplement the formulas for partial sums of the Hurwitz zeta-function and its derivatives, producing more integral representations and generic definitions of important constants. Then, these are used, coupled with the functional equation for the completed zeta-function to clarify the results of Choudhury, giving rise to closed expressions for the Riemann zeta-function and its derivatives.

1. Introduction and Some Examples

Let $s=\sigma +it$ be the complex variable, and let

$$\zeta \left(s\right)=\prod _p{\left(1-p^{-s}\right)}^{-1}=\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n^s}},$$

denote the Riemann zeta-function, where $\sigma >1$ in the first instance. $\zeta \left(s\right)$ is analytically continued to a meromorphic function over the whole plane with a simple pole at $s=1$ with residue 1. Cf. [1].

Let $X\left(s\right):={\displaystyle \frac{1}{\zeta \left(s\right)}}$, A be the Glaisher–Kinkelin constant defined by (5), and let $A_2$ be the second Glaisher-Kinkelin constant defined by (6), respectively. We have some interesting special values as

$$\begin{array}{cc}\hfill & X\left(2\right)={\displaystyle \frac{6}{{\pi }^2}},X^{\prime }\left(2\right)=-{\displaystyle \frac{6}{{\pi }^2}}(\gamma +log\left(2\pi \right)-12logA),\hfill \\ \hfill & X^{″}\left(2\right)={\displaystyle \frac{-{\zeta }^{″}\left(2\right){\zeta }^2\left(2\right)+2\zeta \left(2\right){\left({\zeta }^{\prime }\left(2\right)\right)}^2}{{\zeta }^4\left(2\right)}}=-{\displaystyle \frac{{\zeta }^{″}\left(2\right)}{{\zeta }^2\left(2\right)}}+2{\displaystyle \frac{{\zeta }^{\prime }\left(2\right)}{\zeta \left(2\right)}}{\displaystyle \frac{{\zeta }^{\prime }\left(2\right)}{{\zeta }^2\left(2\right)}}.\hfill \end{array}$$

(1)

By the functional equation of the logarithmic derivative of $\zeta \left(s\right)$, we also have

$$\begin{array}{c}\hfill {\zeta }^{″}\left(2\right)={\displaystyle \frac{{\pi }^2}{6}}\left({(\gamma +log2\pi )}^2-24(\gamma +log2\pi -1)logA-12A_2+{\displaystyle \frac{{\pi }^2}{12}}-{\displaystyle \frac{3}{4}}\right).\end{array}$$

(2)

Substituting

$$\begin{array}{c}\hfill {\displaystyle \frac{{\zeta }^{\prime }}{\zeta }}\left(2\right)=\gamma +log2\pi -12logA,\end{array}$$

and (2) in (1), we deduce the closed form

$$\begin{array}{cc}\hfill & X^{″}\left(2\right)=-{\displaystyle \frac{6}{{\pi }^2}}\left({(\gamma +log2\pi )}^2-24(\gamma +log2\pi -1)logA-12A_2+{\displaystyle \frac{{\pi }^2}{12}}-{\displaystyle \frac{3}{4}}\right)\hfill \\ \hfill & +2\left({\displaystyle \frac{6}{{\pi }^2}}\right){(\gamma +log2\pi -12logA)}^2.\hfill \end{array}$$

Our method consists of considering

$$\begin{array}{c}\hfill Z\left(s\right)={\displaystyle \frac{{\zeta }^{\prime }}{\zeta }}\left(s\right)+{\displaystyle \frac{1}{s-1}},\end{array}$$

and using the functional equation

$$\begin{array}{c}\hfill Z\left(s\right)=-Z(1-s)+log\pi -{\displaystyle \frac{1}{2}}\left(\psi \left({\displaystyle \frac{s}{2}}+1\right)+\psi \left({\displaystyle \frac{3}{2}}-{\displaystyle \frac{s}{2}}\right)\right),\end{array}$$

(3)

where $\psi$ is the Euler digamma-function defined by (22) in the following Lemma 2. The introduction of the function $Z\left(s\right)$ is only for dealing with the most essential case of $s\to 1$; for other arguments, there is no need to use it, and one can just compute normally. It can be seen that (3) with s, a bigger value (and $1-s$ a small value) is the most appropriate to ensure the convergence of the arising integral. We illustrate the case $s=2$ of (3).

$$\begin{array}{c}\hfill Z\left(2\right)=-\left({\displaystyle \frac{{\zeta }^{\prime }}{\zeta }}(-1)-{\displaystyle \frac{1}{2}}\right)+log\pi -{\displaystyle \frac{1}{2}}\left(\psi \left(1\right)+1+\psi \left({\displaystyle \frac{1}{2}}\right)\right).\end{array}$$

(4)

Substituting (5) in (4), and noting (23), we deduce the third equality in (1).

As has been shown above, we often omit the steps regarding the digamma-function hereafter. We gather some basic formulas in Lemma 2.

It is interesting to see how the Glaisher–Kinkelin constant comes in the formula for $X^{\prime }\left(2\right)$. For instance, see (28) with $u=1$, $l=3$, and $a=1$, and $x\in \mathbb{N}$ (for simplicity) reads

$$\begin{array}{cc}\hfill & -{\zeta }^{\prime }(-1)+{\displaystyle \frac{1}{12}}\hfill \\ \hfill & =\underset{x\to \infty }{lim}\left(\sum _{n\le x}nlogn-{\displaystyle \frac{1}{2}}\left(x^2+x+{\displaystyle \frac{1}{6}}\right)logx+{\displaystyle \frac{x^2}{4}}\right)\hfill \\ \hfill & =:logA.\hfill \end{array}$$

(5)

Cf. (31) for a generalization.

This suggests the following. To find a closed form for ${\zeta }^{″}\left(2\right)$, we may proceed as above to introduce a constant through a counterpart of (5) and apply the above procedure of using the functional equation.

In parallel to (5), we may prove a counterpart of (28), whose special case with $u=1$, $l=3$, and $a=1$, and $x\in \mathbb{N}$ (for simplicity) reads

$$\begin{array}{cc}\hfill & {\zeta }^{″}(-1)\hfill \\ \hfill & =\underset{x\to \infty }{lim}\left(\sum _{n\le x}n{log}^{2}n-{\displaystyle \frac{1}{2}}{x}^2\left({log}^{2}x-logx+{\displaystyle \frac{1}{4}}\right)-{\displaystyle \frac{1}{2}}x({log}^{2}x+x)-{\displaystyle \frac{1}{12}}({log}^{2}x+2logx)\right)\hfill \\ \hfill & =:A_2,\hfill \end{array}$$

(6)

the second Glaisher–Kinkelin constant.

Then, we apply the differentiated form of (3)

$$\begin{array}{cc}\hfill & Z^{\prime }\left(s\right)={\left({\displaystyle \frac{{\zeta }^{\prime }}{\zeta }}\left(s\right)\right)}^{\prime }-{\displaystyle \frac{1}{{(s-1)}^2}}\hfill \\ \hfill & =Z^{\prime }(1-s)-{\displaystyle \frac{1}{4}}\left({\psi }^{\prime }\left({\displaystyle \frac{s}{2}}+1\right)-{\psi }^{\prime }\left({\displaystyle \frac{3}{2}}-{\displaystyle \frac{s}{2}}\right)\right),\hfill \end{array}$$

(7)

to deduce (2).

2. Partial Sums for Higher Derivatives of the Hurwitz Zeta-Function

Let

$$\zeta (s,a)=\sum _{n=0}^{\infty }{(n+a)}^{-s},\sigma >1,0<a\le 1,$$

be the Hurwitz zeta-function, where we write $s=\sigma +it$ throughout. We consider the asymptotic formula as well as the integral representation for the partial sum

$$L_u(x,a)=\sum _{0\le n\le x}{(n+a)}^u,0<a\le 1,$$

and its derivatives. The following results can be found in [2] (Chapter 3). Let

$$\mathsf{\Gamma }\left(s\right)={\int }_0^{\infty }{e}^{-s}{x}^{s-1}\mathit{d}x,\sigma >0,$$

be the gamma function, let $\psi \left(z\right)={\displaystyle \frac{{\mathsf{\Gamma }}^{\prime }}{\mathsf{\Gamma }}}\left(z\right)$ be the Euler digamma function in Lemma 2, and let ${\overline{B}}_r\left(x\right)=B_r\left(x-\left[x\right]\right)$ denote the rth periodic Bernoulli polynomial, where the rth Bernoulli polynomial is defined by

$${\displaystyle \frac{t{e}^{xt}}{e^t-1}}=\sum _{r=0}^{\infty }{\displaystyle \frac{B_r\left(x\right)}{r!}}{t}^r\left(\right|t|<2\pi ).$$

Subsequent results depend on the Euler–Maclaurin sum formula [2] (p. 201, Theorem B.5),

$$\begin{array}{cc}\hfill \sum _{\alpha <n\le x}f\left(n\right)=& {\int }_{\alpha }^{x}f\left(t\right)\mathrm{d}t+\sum _{r=1}^l{\displaystyle \frac{{(-1)}^r}{r!}}{\left[{\overline{B}}_r\left(t\right)f^{(r-1)}\left(t\right)\right]}_{\alpha }^x\hfill \\ \hfill & +{\displaystyle \frac{{(-1)}^{l+1}}{l!}}{\int }_{\alpha }^x{\overline{B}}_l\left(t\right)f^{\left(l\right)}\left(t\right)\mathrm{d}t,\hfill \end{array}$$

(8)

which follows by repeated application of integration by parts to Euler’s summation formula in view of

$${\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}}{\overline{B}}_r\left(t\right)=r{\overline{B}}_{r-1}\left(t\right).$$

The Euler–Maclaurin ‘constant’ $C=C\left(f\right)$ (EM constant for short) is defined by ($\alpha \ge 0$)

$$C\sim \sum _{\alpha \le n\le x}f\left(n\right)-{\int }^{x}f\left(t\right)\mathrm{d}t-\sum _{r=1}^l{\displaystyle \frac{{(-1)}^r}{r!}}{\overline{B}}_r\left(x\right)f^{(r-1)}\left(x\right),$$

where ${\int }^x$ means the lower limit is removed, and l may be ∞ as the case may be, cf. [3] (8). In the case of the sums of powers, this corresponds to the constant term in (10). The following theorem and its corollaries ([2] (pp. 55–61)) are far-reaching, but, in the literature, it went unnoticed. A prototype can be found in [4]. We will show its effect by interpreting the results in [3].

Lemma 1

(Integral Representations [2]). For every $u\in \mathbb{C}$ and any $l\in \mathbb{N}$ with $l>\mathrm{Re}u+1$, we have the integral representation (for any $x\ge 0$)

$$\begin{array}{cc}\hfill L_u(x,a)& =\sum _{r=1}^l{\displaystyle \frac{\mathsf{\Gamma }(u+1)}{\mathsf{\Gamma }(u+2-r)}}{\displaystyle \frac{{(-1)}^r}{r!}}\overline{B_r}\left(x\right){(x+a)}^{u-r+1}\hfill \\ \hfill & +{\displaystyle \frac{{(-1)}^l}{l!}}{\displaystyle \frac{\mathsf{\Gamma }(u+1)}{\mathsf{\Gamma }(u+1-l)}}{\int }_x^{\infty }\overline{B_l}\left(t\right){(t+a)}^{u-l}\mathrm{d}t\hfill \\ \hfill & +\left\{\begin{array}{cc}{\displaystyle \frac{1}{u+1}}{(x+a)}^{u+1}+\zeta (-u,a),\hfill & u\ne -1\hfill \\ log(x+a)-\psi \left(a\right),\hfill & u=-1\hfill \end{array}\right..\hfill \end{array}$$

(9)

Also, the asymptotic formula

$$\begin{array}{cc}\hfill L_u(x,a)& =\sum _{r=1}^l{\displaystyle \frac{{(-1)}^r}{r}}\left({\displaystyle \genfrac{}{}{0pt}{}{u}{r-1}}\right)\overline{B_r}\left(x\right){(x+a)}^{u-r+1}+O\left(x^{\mathrm{Re}u-l}\right)\hfill \\ \hfill & +\left\{\begin{array}{cc}{\displaystyle \frac{1}{u+1}}{(x+a)}^{u+1}+\zeta (-u,a),\hfill & u\ne -1\hfill \\ log(x+a)-\psi \left(a\right),\hfill & u=-1\hfill \end{array}\right..\hfill \end{array}$$

(10)

holds true as $x\to \infty$, where $\zeta (-u,a)$ is to be interpreted as (11).

Furthermore, the integral representation

$$\begin{array}{cc}\hfill \zeta (-u,a)& =a^u-{\displaystyle \frac{1}{u+1}}{a}^{u+1}-\sum _{r=1}^l{\displaystyle \frac{{(-1)}^r}{r}}\left({\displaystyle \genfrac{}{}{0pt}{}{u}{r-1}}\right)B_r{a}^{u-r+1}\hfill \\ \hfill & +{(-1)}^{l+1}\left({\displaystyle \genfrac{}{}{0pt}{}{u}{l}}\right){\int }_0^{\infty }\overline{B_l}\left(t\right){(t+a)}^{u-l}\mathrm{d}t,\hfill \end{array}$$

(11)

holds true for all complex $u\ne -1$, $\mathrm{Re}u<l-1$, which follows from (9) by putting $x=0$. For $u=-1$, we have the integral representation for the digamma function

$$\psi \left(a\right)=loga-{\displaystyle \frac{1}{2}}{a}^{-1}+{\int }_0^{\infty }{\overline{B}}_1\left(t\right){(t+a)}^{-2}\mathrm{d}t.$$

Proof. 

By the Euler–Maclaurin summation Formula (8), we have

$$\begin{array}{cc}\hfill & L_u(x,a)={(-1)}^{l+1}\left({\displaystyle \genfrac{}{}{0pt}{}{u}{l}}\right){\int }_0^{\infty }{\overline{B}}_l\left(t\right){(t+a)}^{u-l}\mathrm{d}t\hfill \\ \hfill & -\sum _{r=1}^l\left({\displaystyle \genfrac{}{}{0pt}{}{u}{r-1}}\right){\displaystyle \frac{1}{r}}{(-1)}^r{B}_r{a}^{u-r+1}+a^u+{\int }_0^x{(t+a)}^u\mathrm{d}t\hfill \\ \hfill & +\sum _{r=1}^l\left({\displaystyle \genfrac{}{}{0pt}{}{u}{r-1}}\right){\displaystyle \frac{1}{r}}{(-1)}^r{\overline{B}}_r\left(x\right){(x+a)}^{u-r+1}+{(-1)}^l\left({\displaystyle \genfrac{}{}{0pt}{}{u}{l}}\right){\int }_x^{\infty }{\overline{B}}_l\left(t\right){(t+a)}^{u-l}\mathrm{d}t,\hfill \end{array}$$

(12)

where

$${\int }_0^x{(t+a)}^u\mathrm{d}t=\left\{\begin{array}{cc}{\displaystyle \frac{1}{u+1}}\left({(x+a)}^{u+1}-a^{u+1}\right),\hfill & u\ne -1\hfill \\ log(x+a)-loga,\hfill & u=-1\hfill \end{array}\right..$$

In the case $\mathrm{Re}u<-1$, we let $x\to \infty$ in (12) to see that it amounts to (11) for $\mathrm{Re}u<-1$. Since the integral in (11) is absolutely convergent for $\mathrm{Re}u<\ell -1$ and ℓ can be as large as we wish, (12) gives a meromorphic continuation of $\zeta (s,a)$ over the whole plane with a simple pole at $u=-1$. In the case $u=-1$, (12) reads for x large

$$L_{-1}(x,a)-log(x+a)=\sum _{0\le n\le x}{\displaystyle \frac{1}{n+a}}-log(x+a)=f\left(a\right)+O\left(x^{-1}\right),$$

where $f\left(a\right)$ is a function in a only. Since this is the generic definition (22) of $\psi \left(a\right)$, we must have $f\left(a\right)=-\psi \left(a\right)$, which determines the constant term in the asymptotic Formula (10). □

For $u=n\in \mathbb{N}\cup \left\{0\right\}$ and $a=1$, (11) gives

$$\zeta (-n)=-{\displaystyle \frac{1}{n+1}}\sum _{r=0}^{n+1}{(-1)}^r\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{r}}\right)B_r=-{\displaystyle \frac{B_{n+1}}{n+1}},\zeta \left(0\right)=B_1,$$

and

$$\zeta (-2n)=0(n>0),\zeta \left(0\right)=-{\displaystyle \frac{1}{2}}.$$

Since the integrals appearing in Theorem 1 are analytic in the region $\Re u<1-l$, we can differentiate (9) and (11) in u. The results can be found in [2] (pp. 54–60). The case $u=-1$ needs to be treated separately as in the proof of Theorem 1 above. In the following, we give some illustrating examples of derivatives.

Theorem 1.

Suppose $\mathrm{Re}u<1$. Then, we have asymptotic formulas

$$\begin{array}{cc}\hfill & {\displaystyle \frac{{\partial }^k}{{\partial }^{k}u}}{L}_u(x,a)=-\overline{B_1}\left(x\right){(x+a)}^u{log}^k(x+a)+O\left(x^{\mathrm{Re}u-1}{log}^k(x+a)\right)\hfill \\ \hfill & +\left\{\begin{array}{cc}{(x+a)}^{u+1}\sum _{r=0}^k{(-1)}^{r}r!\left({\displaystyle \genfrac{}{}{0pt}{}{k}{r}}\right){(u+1)}^{-1-r}{log}^{k-r}(x+a)+{(-1)}^k{\zeta }^{\left(k\right)}(-u,a),\hfill & u\ne -1\hfill \\ {\displaystyle \frac{1}{k+1}}{log}^{k+1}(x+a)+{(-1)}^k{\gamma }_k\left(a\right),\hfill & u=-1\hfill \end{array}\right..\hfill \end{array}$$

(13)

for $\mathrm{Re}u<0$ and

$$\begin{array}{cc}\hfill & {\displaystyle \frac{{\partial }^k}{{\partial }^{k}u}}{L}_u(x,a)=-{\overline{B}}_1\left(x\right){(x+a)}^u{log}^k(x+a)+{\displaystyle \frac{{\overline{B}}_2\left(x\right)}{2}}\sum _{r=0}^1\left({\displaystyle \genfrac{}{}{0pt}{}{k}{r}}\right)u^{\left(r\right)}{(x+a)}^{u-1}{log}^{k-r}(x+a)\hfill \\ \hfill & +{(x+a)}^{u+1}\sum _{r=0}^k{(-1)}^{r}r!\left({\displaystyle \genfrac{}{}{0pt}{}{k}{r}}\right){(u+1)}^{-1-r}{log}^{k-r}(x+a)+{(-1)}^k{\zeta }^{\left(k\right)}(-u,a)\hfill \\ \hfill & +O\left(x^{\mathrm{Re}u-2}{log}^k(x+a)\right),\hfill \end{array}$$

(14)

for $0\le \mathrm{Re}u<1$.

The EM constant for the function $f\left(t\right)={(t+a)}^u{log}^k(t+a)$ (for $\mathrm{Re}u<1$) is ${(-1)}^k{\zeta }^{\left(k\right)}(-u,a)$.

Proof. 

We distinguish two cases $\mathrm{Re}u<0$ and $0\le \mathrm{Re}u<1$ and consider the former case in which we may take $l=1$. Then, (9) reads

$$\begin{array}{cc}\hfill L_u(x,a)& =-\overline{B_1}\left(x\right){(x+a)}^u-u{\int }_x^{\infty }\overline{B_1}\left(t\right){(t+a)}^{u-1}\mathrm{d}t\hfill \\ \hfill & +\left\{\begin{array}{cc}{\displaystyle \frac{1}{u+1}}{(x+a)}^{u+1}+\zeta (-u,a),\hfill & u\ne -1\hfill \\ log(x+a)-\psi \left(a\right),\hfill & u=-1\hfill \end{array}\right..\hfill \end{array}$$

For $0\le \mathrm{Re}u<1$, we may choose $l=2$, and (9) reads

$$\begin{array}{cc}\hfill L_u(x,a)=& -\overline{B_1}\left(x\right){(x+a)}^u+{\displaystyle \frac{u}{2}}\overline{B_2}\left(x\right){(x+a)}^{u-1}+{\displaystyle \frac{1}{u+1}}{(x+a)}^{u+1}+\zeta (-u,a)\hfill \\ \hfill & +{\displaystyle \frac{u(u-1)}{2}}{\int }_x^{\infty }\overline{B_2}\left(t\right){(t+a)}^{u-2}\mathrm{d}t.\hfill \end{array}$$

□

Example 1.

(i) We state the case $k=1$ of (14). For a generalization of (14), cf. (28) whose special case is (30), we state (14).

The case $k=1$ of (14) leads to the asymptotic formula

$$\begin{array}{cc}\hfill & -{\zeta }^{\prime }(-u,a)\hfill \\ \hfill & =\sum _{0\le n\le x}{(n+a)}^{u}log(n+a)-{(u+1)}^{-1}{(x+a)}^{u+1}log(x+a)+{(u+1)}^{-2}{(x+a)}^{u+1}\hfill \\ \hfill & +{\overline{B}}_1\left(x\right){(x+a)}^{u}log(x+a)-{\displaystyle \frac{1}{2}}{\overline{B}}_2\left(x\right){(x+a)}^{u-1}(1+ulog(x+a))\hfill \\ \hfill & +O\left(x^{\mathrm{Re}u-2}\right).\hfill \end{array}$$

Hence, as a generic definition ($N\in \mathbb{N}$)

$$\begin{array}{cc}\hfill & -{\zeta }^{\prime }(0,a)\hfill \\ \hfill & =\underset{N\to \infty }{lim}\left(\sum _{0\le n\le N}log(n+a)-(N+a)log(N+a)+N+a+B_{1}log(N+a)\right).\hfill \end{array}$$

(15)

The generic definition (15) plays an essential role in the proof of Lerch’s formula [2] (p. 82, Theorem 5.1)

$${\zeta }^{\prime }(0,a)=log{\displaystyle \frac{\mathsf{\Gamma }\left(a\right)}{\sqrt{2\pi }}},$$

(16)

and the Weierstrass product for the gamma function [2] (p. 88, Exercise 5.2). Hence, in particular,

$${\zeta }^{\prime }\left(0\right)=-log\sqrt{2\pi }.$$

(17)

Substituting this and data in Lemma 2 in (3), we conclude that

$$\underset{s\to 1}{lim}\left({\displaystyle \frac{{\zeta }^{\prime }}{\zeta }}+{\displaystyle \frac{1}{s-1}}\right)=\gamma .$$

(18)

Applying (18), (17) was deduced by substituting (36) in (3).

(ii) We state the case $k=2$ of (14).

Equation (14) leads to the asymptotic formula

$$\begin{array}{cc}\hfill & {\zeta }^{″}(-u,a)=\sum _{0\le n\le x}{(n+a)}^u{log}^2(n+a)\hfill \\ \hfill & -{(x+a)}^{u+1}({(u+1)}^{-1}{log}^2(x+a)-2{(u+1)}^{-2}log(x+a)+{(u+1)}^{-3})\hfill \\ \hfill & +{\overline{B}}_1\left(x\right){(x+a)}^u{log}^2(x+a)-{\displaystyle \frac{1}{2}}{\overline{B}}_2\left(x\right){(x+a)}^{u-1}(u{log}^2(x+a)+2log(x+a))\hfill \\ \hfill & +O(x^{\mathrm{Re}u-2}logx).\hfill \end{array}$$

Hence, we have an asymptotic formula ($N\in \mathbb{N}$)

$$\begin{array}{cc}\hfill {\zeta }^{″}(0,a)=\sum _{n=0}^N{log}^2(n+a)& -(N+a)\left\{{log}^2(N+a)-2log(N+a)+1\right\}-{\displaystyle \frac{1}{2}}{log}^2(N+a)\hfill \\ \hfill & -{\int }_N^{\infty }{\displaystyle \frac{{\overline{B}}_1\left(t\right)}{t+a}}{log}^2(t+a)\mathrm{d}t,\hfill \end{array}$$

and the generic definition

$$\begin{array}{cc}\hfill & -{\zeta }^{″}(0,a)\hfill \\ \hfill & =\underset{N\to \infty }{lim}\left(\sum _{n=0}^N{log}^2(n+a)-(N+a)\left\{{log}^2(N+a)-2log(N+a)+1\right\}-{\displaystyle \frac{1}{2}}{log}^2(N+a)\right).\hfill \end{array}$$

(19)

See that (19) plays an essential role in the theory of Deninger’s R-function [2] (p. 92, (5.36))

$$R\left(a\right)=-{\zeta }^{″}(0,a).$$

(20)

In particular, (19) contains [5] (Theorem 2.3) involving the Gaussian representation and the Weierstrass representation [2] (pp. 89–91).

Lerch’s Formula (16) and Deninger’s Formula (20) are principal solutions to the difference equation furnished by the Dufrenoy–Pisot theorem; cf. [6]. As (17) is proved on the ground of Stirling’s formula, the closed-form expression

$$\begin{array}{c}\hfill -R\left(1\right)={\zeta }^{″}\left(0\right)=-{\displaystyle \frac{1}{2}}{(log2\pi )}^2+{\gamma }_1+{\displaystyle \frac{1}{2}}{\gamma }^2-{\displaystyle \frac{\zeta \left(2\right)}{4}}.\end{array}$$

(21)

depends on an extraneous factor of the logarithmic differentiation, cf. Section 3 below.

For a systematic study on the values ${\zeta }^{\left(k\right)}\left(0\right)$, we refer to [7].

Lemma 2.

Under the definition of the Euler constant

$$\gamma =\underset{N\to \infty }{lim}\left(\sum _{n=1}^N{\displaystyle \frac{1}{n}}-log(N+z)\right).$$

for any z other than negative integers, the two definitions for the Euler digamma function ψ are equivalent: the generic definition

$$\psi \left(z\right)=-\underset{N\to \infty }{lim}\left(\sum _{n=0}^N{\displaystyle \frac{1}{n+z}}-log(N+z)\right),$$

(22)

where z is not a non-positive integer, and the Gaussian representation

$$\psi \left(z\right)+\gamma =\sum _{n=1}^{\infty }\left({\displaystyle \frac{1}{n}}-{\displaystyle \frac{1}{z+n-1}}\right),$$

We assemble here some well-known formulas for ψ:

$$\psi (z+1)=\psi \left(z\right)+{\displaystyle \frac{1}{z}}{\psi }^{\prime }(z+1)-{\psi }^{\prime }\left(z\right)=-{\displaystyle \frac{1}{z^2}},$$

$$\psi \left({\displaystyle \frac{1}{2}}\right)=-\gamma -2log2,{\psi }^{\prime }\left(1\right)=\zeta \left(2\right),{\psi }^{\prime }\left({\displaystyle \frac{1}{2}}\right)=3\zeta \left(2\right).$$

(23)

Example 2.

The kth Laurent coefficient of the Hurwitz zeta-function (at $s=1$) is given by ${\displaystyle \frac{{(-1)}^k}{k!}}{\gamma }_k\left(a\right)$, where

$${\gamma }_k\left(a\right)=\underset{x\to \infty }{lim}\left(\sum _{0\le n\le x}{\displaystyle \frac{{log}^k(n+a)}{n+a}}-{\displaystyle \frac{{log}^{k+1}(x+a)}{k+1}}\right),$$

(24)

is sometimes called the Stieltjes or generalized Euler constant. ${\gamma }_k\left(a\right)$ admits the integral representation

$$\begin{array}{cc}\hfill {\gamma }_k\left(a\right)=& {\displaystyle \frac{1}{2a}}{log}^{k}a-{\displaystyle \frac{1}{k+1}}{log}^{k+1}a\hfill \\ \hfill & -{\int }_0^{\infty }{\displaystyle \frac{{\overline{B}}_1\left(t\right)}{{(t+a)}^2}}\left({log}^k(t+a)-k{log}^{k-1}(t+a)\right)\mathrm{d}t.\hfill \end{array}$$

(25)

In particular, ${\gamma }_0\left(a\right)=-\psi \left(a\right)$.

Proof. 

We prove the integral representation (27) corresponding to (9), which also establishes (13) with $u=-1$. Theorem 1 with $l=1$ and $-s$ ($s\ne 1$, $\sigma >0$) for u

$$L_{-s}(x,a)={\displaystyle \frac{{(x+a)}^{1-s}}{1-s}}+\zeta (s,a)-{\displaystyle \frac{{\overline{B}}_1\left(x\right)}{{(x+a)}^s}}+s{\int }_x^{\infty }{\displaystyle \frac{{\overline{B}}_1\left(t\right)}{{(t+a)}^{s+1}}}\mathrm{d}t.$$

(26)

Since both sides of (26) are analytic in $\sigma >0$, we may compute the kth Taylor coefficients around $s=1$. Equating them, we conclude that

$$\begin{array}{cc}\hfill {\gamma }_k\left(a\right)=& \sum _{0\le n\le x}{(n+a)}^{-1}{log}^k(n+a)\hfill \\ \hfill & -{\displaystyle \frac{{log}^{k+1}(x+a)}{k+1}}+{\displaystyle \frac{{\overline{B}}_1\left(x\right)}{x+a}}{log}^k(x+a)\hfill \\ \hfill & -{\int }_x^{\infty }{\displaystyle \frac{{\overline{B}}_1\left(t\right)}{{(t+a)}^2}}\left({log}^k(t+a)-k{log}^{k-1}(t+a)\right)\mathrm{d}t.\hfill \end{array}$$

(27)

As in Theorem 1, letting $x\to \infty$ and $x=0$ implies (24) and (25), respectively. □

Example 3.

As Corollary 1 establishes [3] Theorems 2 and 3, (28) gives [3] Theorem 1 to the effect that the EM constant is $-{\zeta }^{\prime }(u,a)$ in general and in the special case of m being an even positive integer that

$${\zeta }^{\prime }(-m)={\displaystyle \frac{B_{m+1}}{m+1}}\sum _{j=1}^m{\displaystyle \frac{1}{j}}.$$

We state special cases of the Lemma 1. In the case of $u=m$, $m\in \mathbb{N}\cup \left\{0\right\}$. For $\mathbb{N}\ni l>m+1$, we have the generic definition

$$\begin{array}{cc}\hfill -{\zeta }^{\prime }(-m,a)& =\underset{N\to \infty }{lim}(\sum _{n=0}^N{(n+a)}^{m}log(n+a)\hfill \\ \hfill & -{\displaystyle \frac{1}{m+1}}{(N+a)}^{m+1}log(N+a)+{\displaystyle \frac{1}{{(m+1)}^2}}{(N+a)}^{m+1}\hfill \\ \hfill & -{\displaystyle \frac{1}{2}}{(N+a)}^{m}log(N+a)-\sum _{r=2}^{m+1}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{r-1}}\right){\displaystyle \frac{B_r}{r!}}\hfill \\ \hfill & ·\left({\displaystyle \frac{1}{m}}+\cdots +{\displaystyle \frac{1}{m-r+2}}+log(N+a)\right){(N+a)}^{m-r+1}),\hfill \end{array}$$

(28)

where an empty sum is to be interpreted as 0 and the integral representation

$$\begin{array}{cc}\hfill & {\zeta }^{\prime }(-m,a)\hfill \\ \hfill & ={\displaystyle \frac{1}{m+1}}{a}^{m+1}loga-{\displaystyle \frac{1}{{(m+1)}^2}}{a}^{m+1}-{\displaystyle \frac{1}{2}}{a}^{m}loga+{\displaystyle \frac{1}{12}}{a}^{m-1}loga\hfill \\ \hfill & +\sum _{r=4}^{m+1}{\displaystyle \frac{B_r}{r}}\left(\sum _{j=0}^{r-2}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{m}{j}}\right){\displaystyle \frac{1}{r-1-j}}+\left({\displaystyle \genfrac{}{}{0pt}{}{m}{r-1}}\right)loga\right)a^{m-r+1}\hfill \\ \hfill & +{\displaystyle \frac{1}{m+1}}\sum _{r=m+2}^l{B}_r\left(\sum _{j=0}^{r-1}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{r-m-2}{j}}\right){\displaystyle \frac{1}{r-j}}\right)a^{m-r+1}\hfill \\ \hfill & +{(-1)}^{l+1}{\int }_0^{\infty }\left(\sum _{j=0}^{l-1}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{l-m-1}{j}}\right){\displaystyle \frac{1}{l-j}}{\overline{B}}_l\left(t\right){(t+a)}^{m-l}\right)\mathrm{d}t,\hfill \end{array}$$

(29)

where (28) and (29) correspond to (10) and (11), respectively. Additionally, (28) implies Example 3.

So, (28) with $m=1$ leads to a generalization of (5):

$$\begin{array}{c}\hfill -{\zeta }^{\prime }(-1,a)=logA\left(a\right)-{\displaystyle \frac{B_2}{2}},\end{array}$$

(30)

where

$$\begin{array}{cc}\hfill & logA\left(a\right)\hfill \\ \hfill & =\underset{N\to \infty }{lim}(\sum _{n=0}^N(n+a)log(n+a)-{\displaystyle \frac{1}{2}}{(N+a)}^{2}log(N+a)+{\displaystyle \frac{1}{2^2}}{(N+a)}^2\hfill \\ \hfill & -{\displaystyle \frac{1}{2}}(N+a)log(N+a)-{\displaystyle \frac{B_2}{2}}log(N+a)).\hfill \end{array}$$

(31)

3. Special Values of the Hurwitz Zeta-Function and Its Derivatives

As we have shown in Example 1 above, it is the relations between the derivatives of the Hurwitz zeta-function and the allied gamma-functions ($log\mathsf{\Gamma }\left(a\right)$, $R\left(x\right)$). However, the special values of the zeta- and gamma-allied functions are of interest. Equation (21) is a typical example in which one knows the closed form for the value $R\left(1\right)$, which is of importance as with the values $\mathsf{\Gamma }\left(1\right)=1$.

The other value that can be readily obtained is

$$R\left({\displaystyle \frac{1}{2}}\right)=-{\zeta }^{″}\left(0,{\displaystyle \frac{1}{2}}\right)={\displaystyle \frac{1}{2}}{log}^{2}2-2log2log\sqrt{2\pi }.$$

This follows by differentiating

$$\zeta \left(s,{\displaystyle \frac{1}{2}}\right)=(2^s-1)\zeta \left(s\right),$$

(32)

twice

$${\zeta }^{″}\left(0,{\displaystyle \frac{1}{2}}\right)=2^s{(log2)}^2\zeta \left(s\right)+2^{s+1}log2{\zeta }^{\prime }\left(s\right)+(2^s-1){\zeta }^{″}\left(s\right).$$

Example 4

([8], Part I). For the alternating series, we have, similar to (32),

$${\ell }_s\left({\displaystyle \frac{1}{2}}\right)=(2^{1-s}-1)\zeta \left(s\right),$$

(33)

where

$${\ell }_s\left(x\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{e^{2\pi inx}}{n^s}},$$

indicates the Lerch zeta-function. The method of logarithmic differentiation seems to be first used by de la Vallée-Poussin. Applying it to (33), we deduce that

$${\displaystyle \frac{{\zeta }^{\prime }}{\zeta }}\left(s\right)={\displaystyle \frac{{\ell }_s^{\prime }}{{\ell }_s}}\left({\displaystyle \frac{1}{2}}\right)+{\displaystyle \frac{log2}{2^{1-s}-1}}.$$

(34)

It suffices to find the Laurent constant at $s=1$, and so below all asymptotic equations mean the ones near $a=1$. Since ${\displaystyle \frac{log2}{2^{1-s}-1}}=-{\displaystyle \frac{1}{s-1}}+{\displaystyle \frac{1}{2}}log2+O(s-1)$, (34) implies that

$${\displaystyle \frac{{\zeta }^{\prime }}{\zeta }}\left(s\right)+{\displaystyle \frac{1}{s-1}}={\displaystyle \frac{{\ell }_s^{\prime }}{{\ell }_s}}\left({\displaystyle \frac{1}{2}}\right)+{\displaystyle \frac{1}{2}}{log}^{2}2+O(s-1).$$

Substituting the Laurent expansion

$${\ell }_s^{\prime }\left({\displaystyle \frac{1}{2}}\right)=-\gamma log2+{\displaystyle \frac{1}{2}}log2+O(s-1),$$

(35)

we conclude

$$\underset{s\to 1}{lim}\left({\displaystyle \frac{{\zeta }^{\prime }}{\zeta }}\left(s\right)+{\displaystyle \frac{1}{s-1}}\right)=\gamma .$$

(36)

To prove (35), we differentiate both sides of (33) to obtain

$${\ell }_s^{\prime }\left({\displaystyle \frac{1}{2}}\right)=(2^{1-s}-1){\zeta }^{\prime }\left(s\right)-2^{1-s}(log2)\zeta \left(s\right).$$

Substituting the Laurent expansions

$$\zeta \left(s\right)={\displaystyle \frac{1}{s-1}}+\gamma +O(s-1),{\zeta }^{\prime }\left(s\right)=-{\displaystyle \frac{1}{{(s-1)}^2}}+{\gamma }_1+O(s-1),$$

and

$$2^{1-s}=1-(log2)(s-1)+{\displaystyle \frac{{log}^{2}2}{2}}{(s-1)}^2+O\left({(s-1)}^2\right),$$

we deduce (35) after some manipulations. Combining

$${\ell }_1\left({\displaystyle \frac{1}{2}}\right)=-log2,$$

and (35), we obtain

$${\displaystyle \frac{{\ell }_1^{\prime }}{{\ell }_1}}\left({\displaystyle \frac{1}{2}}\right)=\gamma -{\displaystyle \frac{1}{2}}log2.$$

Remark 1.

Example 4 is stated by de la Vallée-Poussin [8], available as [9] (pp. 63–64) and reproduced in [10] (pp. 216–217). de la Vallée-Poussin uses the alternating property of ${\ell }_1\left({\displaystyle \frac{1}{2}}\right)$ and appeals to the generic definition of γ without using the integral representation.

Now, we briefly describe a modification of Song’s method [11] of logarithmic differentiation and non-trivial zeros. As described in Section 1, our method depends on the functional equation and integral representation furnished by Theorem 1. We rewrite (11) with $u=-s$, $l=1$, and $a=1$ as

$$(s-1)\zeta \left(s\right)=1+J\left(s\right),J\left(s\right)=(s-1)\left({\displaystyle \frac{1}{2}}-s{\int }_1^{\infty }{\overline{B}}_1\left(t\right)t^{-s-1}\mathrm{d}t\right),$$

where $\sigma >0$.

Now, apply the logarithmic differentiation to obtain

$$Z\left(s\right)={\displaystyle \frac{{\zeta }^{\prime }}{\zeta }}\left(s\right)+{\displaystyle \frac{1}{s-1}}={\displaystyle \frac{J^{\prime }\left(s\right)}{1+J\left(s\right)}},$$

(37)

where

$$\begin{array}{c}\hfill J^{\prime }\left(s\right)={\displaystyle \frac{1}{2}}+(1-2s){\int }_1^{\infty }{\overline{B}}_1\left(t\right)t^{-s-1}\mathrm{d}t+s(s-1){\int }_1^{\infty }{\overline{B}}_1\left(t\right)t^{-s-1}logt\mathrm{d}t.\end{array}$$

(38)

Hence, $J^{\prime }\left(s\right)={\displaystyle \frac{1}{2}}-s{\int }_1^{\infty }{\overline{B}}_1\left(t\right)t^{-s-1}\mathrm{d}t+O(s-1)$ and (36) followed by (25).

One more differentiation of (37) gives us

$$\begin{array}{cc}\hfill & {\displaystyle \frac{{\zeta }^{″}}{\zeta }}\left(s\right)-{\left({\displaystyle \frac{{\zeta }^{\prime }}{\zeta }}\left(s\right)\right)}^2-{\displaystyle \frac{1}{{(s-1)}^2}}\hfill \\ \hfill & =Z^{\prime }\left(s\right)={\left({\displaystyle \frac{{\zeta }^{\prime }}{\zeta }}\left(s\right)+{\displaystyle \frac{1}{s-1}}\right)}^{\prime }={\displaystyle \frac{J^{″}\left(s\right)(1+J\left(s\right))-{\left(J^{\prime }\left(s\right)\right)}^2}{{(1+J\left(s\right))}^2}}.\hfill \end{array}$$

(39)

From (38), we have

$$\begin{array}{cc}\hfill J^{″}\left(s\right)& =-2{\int }_1^{\infty }{\overline{B}}_1\left(t\right)t^{-s-1}\mathrm{d}t+2(2s-1){\int }_1^{\infty }{\overline{B}}_1\left(t\right)t^{-s-1}logt\mathrm{d}t\hfill \\ \hfill & -s(s-1){\int }_1^{\infty }{\overline{B}}_1\left(t\right)t^{-s-1}{log}^{2}t\mathrm{d}t.\hfill \end{array}$$

(40)

We take the limit as $s\to 1$ of $Z^{\prime }(1-s)$ for which we put $s=1$ in (40).

We have

$$\begin{array}{c}\hfill J^{″}\left(1\right)=-2{\int }_1^{\infty }{\overline{B}}_1\left(t\right)t^{-2}\mathrm{d}t+2{\int }_1^{\infty }{\overline{B}}_1\left(t\right)t^{-2}logt\mathrm{d}t.\end{array}$$

(41)

Substituting (25) in (41), we conclude that

$$\begin{array}{c}\hfill J^{″}\left(1\right)=-2{\gamma }_1.\end{array}$$

Substituting the data (23) in the far-right side of (39), we obtain

$$\begin{array}{c}\hfill \underset{s\to 1}{lim}{\left({\displaystyle \frac{{\zeta }^{\prime }}{\zeta }}\left(s\right)+{\displaystyle \frac{1}{s-1}}\right)}^{\prime }={\displaystyle \frac{J^{″}\left(1\right)(1+J\left(1\right))-{\left(J^{\prime }\left(1\right)\right)}^2}{{(1+J\left(1\right))}^2}}=-2{\gamma }_1-{\gamma }^2.\end{array}$$

Substituting the above data and ${\psi }^{\prime }\left({\displaystyle \frac{3}{2}}\right)=3\zeta \left(2\right)-4$ into (7), we deduce (21).