ABSTRACT

The series of Riemann functions and the series of hurwitz zeta functions are given by using integral basic identities. The series given is closed. Finally, the numerical series of the zeta function and hurwitz function series are given.

Keywords:Integral Basic Identity, Riemann Zeta Function, Hurwitz Zeta Function

一些涉及黎曼Zeta函数的无穷级数

马玮¹，及万会^2*

¹宁夏吴忠中学，宁夏 吴忠

²宁夏民族职业技术学院，宁夏 吴忠

收稿日期：2019年10月4日；录用日期：2019年10月23日；发布日期：2019年10月30日

摘 要

用积分基本恒等式给出涉及黎曼 $\zeta \left(2n\right)$ 函数级数与赫尔维茨zeta函数级数。所给出级数是封闭形的。最后给出关于 $\zeta \left(2n\right)$ 函数与赫尔维茨zeta函数级数的数值级数。

关键词 :积分基本恒等式，黎曼zeta函数，赫尔维茨zeta函数

Copyright © 2019 by author(s) and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY).

http://creativecommons.org/licenses/by/4.0/

1. 引言与准备

熟知，贝努里数 $B_{2n}={\left(-1\right)}^{n-1}\frac{2\left(2n\right)!\zeta \left(2n\right)}{2^{2n}{\pi }^{2n}}$，$\zeta \left(2n\right)$ 是黎曼zeta函数 $\zeta \left(s\right)$ ；黎曼zeta函数 $\zeta \left(s\right)$ 定义

$\zeta \left(s\right)=\{\begin{array}{l}{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{1}{n^s}=\frac{1}{1-2^{-s}}{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{1}{{\left(2n-1\right)}^s},R\left(s\right)>1}}\\ \frac{1}{1-2^{1-s}}{\displaystyle \underset{n=}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^{n-1}}{n^s},R\left(s\right)>0,s\ne 1}\end{array}$ (1.1)

赫尔维茨函数(黎曼zeta函数 $\zeta \left(s\right)$ )推广定义

$\zeta \left(s,a\right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{1}{{\left(n+a\right)}^s}}$，$\mathrm{Re}\left(s\right)>1$，$a\ne 0,-1,-2,\cdots$ (1.2)

黎曼zeta函数的极限值 [1] ：

$\zeta \left(-1\right)=-\frac{1}{12}$，$\zeta \left(0\right)=-\frac{1}{2}$，${\zeta }^{\prime }\left(0\right)=-\frac{1}{2}\ln \left(2\pi \right)$，$\underset{x\to 1}{\lim }\left(\zeta \left(s\right)-\frac{1}{s-1}\right)=\gamma$ (1.3)

欧拉常数 $\gamma$ 定义

$\gamma =\underset{n\to \infty }{\lim }\left({\displaystyle \underset{k=1}{\overset{n}{\sum }}\frac{1}{k}}-\ln n\right)\cong 0.577215664901532860606512\cdots$

赫尔维茨函数与黎曼zeta函数关系 [2] [3]， $\zeta \left(a,1\right)=\zeta \left(s\right)$，$\zeta \left(s\right)=\frac{1}{m^s-1}{\displaystyle \underset{j=1}{\overset{m-1}{\sum }}\zeta \left(s,\frac{i}{m}\right)}$ (1.4)

$\zeta \left(s,a+n\right)=\zeta \left(s,a\right)-{\displaystyle \underset{k=0}{\overset{n-1}{\sum }}\frac{1}{{\left(k+a\right)}^s}}$ (1.5)

升阶乘符号： ${\left(\lambda \right)}_n=\frac{\Gamma \left(\lambda +n\right)}{\Gamma \left(\lambda \right)}=\{\begin{array}{l}1,n=0\\ \lambda \left(\lambda +1\right)\cdots \left(\lambda +n-1\right),n\in N:1,2,\cdots \end{array}$ (1.6)

引理1 [4] 设 $p\left(x\right)$ 为可为函数，则积分基本恒等式成立

${\displaystyle \underset{a}{\overset{b}{\int }}p\left(x\right)\cot x\text{d}x}=2{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}{\displaystyle \underset{a}{\overset{b}{\int }}p\left(x\right)\sin 2nx\text{d}x}}$

引理2 [5] Clausen函数

$C{l}_{2n}\left(x\right)={\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\frac{\sin \left(kx\right)}{k^{2n}}}$，$n\ge 1$ ； $C{l}_{2n+1}\left(x\right)={\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\frac{\cos \left(kx\right)}{k^{2n+1}}}$，$n\ge 0$

2. 主要结果与证明

命题1 设 $\left|x\right|<\pi$，则黎曼zeta函数 $\zeta \left(2n\right)$ 的级数封闭型和式

1) ${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\zeta \left(2n\right)x^{2n-1}}{{\pi }^{2n}}}=\frac{1}{2x}-\frac{1}{2}\cot x$，$\left|x\right|<\pi$ (1)

2) ${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\zeta \left(2n\right)x^{2n}}{n{\pi }^{2n}}}=\ln x-\ln \sin x$ (2)

3) ${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\zeta \left(2n\right)x^{2n+1}}{n\left(2n+1\right){\pi }^{2n}}}=-x+x\ln \left|2\sin x\right|+\frac{1}{2}C{l}_2\left(2x\right)$ (3)

4) $\begin{array}{c}{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\zeta \left(2n\right)x^{2n+2}}{n\left(2n+2\right){\pi }^{2n}}}=\frac{1}{2}{x}^2\ln x-\frac{1}{4}{x}^2-\frac{1}{2}{x}^2\ln \sin x+x^2\ln \left|2\sin x\right|\\ +\frac{1}{2}xC{l}_2\left(2x\right)+\frac{1}{4}C{l}_3\left(2x\right)-\frac{1}{4}\zeta \left(3\right)\end{array}$ (4)

5) $\begin{array}{c}{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\zeta \left(2n\right)x^{2n+3}}{n\left(2n+3\right){\pi }^{2n}}}=\frac{1}{3}{x}^3\ln x-\frac{1}{9}{x}^3-\frac{1}{3}{x}^3\ln \sin x+\frac{1}{2}{x}^{2}C{l}_2\left(2x\right)\\ -\frac{1}{4}C{l}_4\left(2x\right)+\frac{1}{3}{x}^3\ln \left|2\sin x\right|+\frac{1}{2}C{l}_3\left(2x\right)\end{array}$ (5)

其中 $C{l}_j\left(t\right)\left(j=2,3,4\right)$ 为Clausen函数

6) ${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}{\left(-1\right)}^{n-1}\frac{\zeta \left(2n\right)x^{2n-1}}{{\pi }^{2n}}}=-\frac{1}{2x}+\frac{1}{2}\coth x$，$\left|x\right|\le \pi$ (6)

7) ${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}{\left(-1\right)}^{n-1}\frac{\zeta \left(2n\right)x^{2n}}{n{\pi }^{2n}}}=\ln \sinh x-\ln x$，$\left|x\right|\le \pi$ (7)

证明 1) 根据余切函数定义文 [6] ； $\cot x=\frac{1}{x}-2{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}{\left(-1\right)}^{n-1}}\frac{2^{2n}{B}_{2n}}{\left(2n\right)!}{x}^{2n-1}$，$B_{2n}$ 是贝努里数。

将贝努里数 $B_{2n}$ 代入余切表达式，得到关于系数为黎曼zeta 函数 $\zeta \left(2n\right)$ 的级数表达式(1)。

2) (1)两端关于x积分，给出(2)式。

3) (2)式两端关于x积分， ${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\zeta \left(2n\right)x^{2n+1}}{n\left(2n+1\right){\pi }^{2n}}}=x\ln x-x-x\ln x+{\displaystyle \underset{0}{\overset{x}{\int }}t\cot t\text{d}t}$

积分 ${\displaystyle \underset{0}{\overset{x}{\int }}t\cot t\text{d}t}$，使用分步积分法，利用积分基本恒等式，

，

由傅里叶级数知， ${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\cos 2nx}{n}}=-\ln \left|2\sin x\right|$，所以 ${\displaystyle \underset{0}{\overset{x}{\int }}t\cot t\text{d}t}=x\ln \left|2\sin x\right|+\frac{1}{2}C{l}_2\left(2x\right)$ (3)式成立。

4) (2)式两端乘以x再关于x积分

${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\zeta \left(2n\right)x^{2n+2}}{n\left(2n+2\right){\pi }^{2n}}}=\frac{1}{2}{x}^2\ln x-\frac{1}{4}{x}^2-\frac{1}{2}{x}^2\ln \sin x+\frac{1}{2}{\displaystyle \underset{0}{\overset{x}{\int }}{t}^2}\cot t\text{d}t$

${\displaystyle \underset{0}{\overset{x}{\int }}{t}^2}\cot t\text{d}t$ 使用分步积分法，利用积分基本恒等式

$\begin{array}{c}{\displaystyle \underset{0}{\overset{x}{\int }}{t}^2\cot t\text{d}t}=2{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}{\displaystyle \underset{0}{\overset{t}{\int }}{x}^2\sin 2nx\text{d}x}}={2{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\left(-\frac{t^2}{2n}\cos 2t+\frac{t}{2n^2}\sin 2t+\frac{1}{4}\frac{\cos 2nt}{n^3}\right)}|}_0^x\\ =-x^2{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\cos 2nx}{n}}+x{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\sin 2nx}{n^2}}+\frac{1}{2}{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\cos 2nx}{n^3}}-\frac{1}{2}{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{1}{n^3}}\\ =x^2\ln \left|2\sin x\right|+xC{l}_2\left(2x\right)+\frac{1}{2}C{l}_3\left(2x\right)-\frac{1}{2}\zeta (3)\end{array}$

将其代入上式，(4)成立。

5) (2) 式两端乘以，再关于x积分

${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\zeta \left(2n\right)x^{2n+3}}{n\left(2n+3\right){\pi }^{2n}}}={\displaystyle \underset{0}{\overset{x}{\int }}\left(t^2\ln t-t^2\ln \sin t\right)\text{d}t}=\frac{1}{3}{x}^3\ln x-\frac{1}{9}{x}^3-\frac{1}{3}{x}^3\ln \sin x+\frac{1}{3}{\displaystyle \underset{0}{\overset{x}{\int }}{t}^3}\cot t\text{d}t$

我们利用积分基本恒等式计算

$\begin{array}{c}{\displaystyle \underset{0}{\overset{x}{\int }}{t}^3}\cot t\text{d}t=2{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}{\displaystyle \underset{0}{\overset{x}{\int }}{t}^3}}\sin 2nt\text{d}t\\ ={\frac{3}{2}{t}^2{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\sin 2nt}{n^2}}-\frac{3}{4}{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\sin 2nt}{n^4}}-t^3{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\cos 2nt}{n}}+\frac{3}{2}t{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\cos 2nt}{n^3}}|}_0^x\\ =\frac{3}{2}{x}^2{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\sin 2nx}{n^2}}-\frac{3}{4}{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\sin 2nx}{n^4}}-x^3{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\cos 2nx}{n}}+\frac{3}{2}x{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\cos 2nx}{n^3}}\\ =\frac{3}{2}{x}^{2}C{l}_2\left(2x\right)-\frac{3}{4}C{l}_4\left(2x\right)+x^3\ln \left|2\sin x\right|+\frac{3}{2}C{l}_3(2x)\end{array}$

将积分 ${\displaystyle \underset{0}{\overset{x}{\int }}{t}^3}\cot t\text{d}t$ 表达式代入上式，得到(5)式。

我们讨论相关交错函数 $\zeta \left(2n\right)$ 级数

6) 根据双曲余切函数定义 [6] $\coth x=\frac{1}{x}+{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{2^{2n}{B}_{2n}{x}^{2n-1}}{\left(2n\right)!}}$，将贝努里数 $B_{2n}$ 代入双曲余切表达式得到(6)式。

7) (6)式两端关于x积分得(7)式。命题1证毕。

下面讨论赫尔维茨zeta函数级数封闭形和式。

命题2 设 $\left|t\right|<\left|a\right|$，则含有升阶乘 ${\left(s\right)}_k$ 的赫尔维茨zeta函数级数封闭形和式

1) ${\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\frac{{\left(s\right)}_k}{k!}\zeta \left(s+k,a\right)t^k=\zeta \left(s,a-t\right)}$，$\left(\left|t\right|<\left|a\right|\right)$ (8)

2) ${\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\frac{{\left(s\right)}_k}{\left(k+1\right)!}\zeta \left(s+k,a\right)t^{k+1}=\frac{1}{s-1}\zeta \left(s-1,a-t\right)}$ (9)

3) ${\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\frac{{\left(s\right)}_k}{\left(k+2\right)!}\zeta \left(s+k,a\right)t^{k+2}=\frac{1}{\left(s-1\right)\left(s-2\right)}\zeta \left(s-1,a-t\right)}$ (10)

4) ${\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\frac{{\left(s\right)}_k}{\left(k-1\right)!}\zeta \left(s+k,a\right)t^{k-1}=\left(s+1\right)\zeta \left(s+1,a-t\right)}$ (11)

5) ${\displaystyle \underset{k=2}{\overset{\infty }{\sum }}\frac{{\left(s\right)}_k}{\left(k-2\right)!}\zeta \left(s+k,a\right)t^{k-2}=\left(s+1\right)\left(s+2\right)\zeta \left(s+2,a-t\right)}$ (12)

证明 实指数s的二项式展开式为

$\begin{array}{c}{\left(1+x\right)}^s={\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\frac{s\left(s-1\right)\cdots \left(s-k+1\right)}{k!}}{x}^k\\ =1+\frac{s}{1!}x+\frac{s\left(s-1\right)}{2!}{x}^2+\cdots +\frac{s\left(s-1\right)\left(s-2\right)\cdots \left(s-k+1\right)}{k!}{x}^k+\cdots \end{array}$

当 $s<0$ 为负实数，即，利用升阶乘符号

$\begin{array}{c}{\left(1+x\right)}^{-s}=1+\frac{-s}{1!}x+\frac{-s\left(-s-1\right)}{2!}{x}^2+\cdots +\frac{-s\left(-s-1\right)\left(-s-2\right)\cdots \left(-s-k+1\right)}{k!}{x}^k+\cdots \\ =1+\frac{-s}{1!}x+\frac{s\left(s+1\right)}{2!}{x}^2+\cdots +\frac{{\left(-1\right)}^{k}s\left(s+1\right)\left(s+2\right)\cdots \left(s+k-1\right)}{k!}{x}^k+\cdots \\ ={\displaystyle \underset{k=0}{\overset{\infty }{\sum }}{\left(-1\right)}^k\frac{s\left(s+1\right)\cdots \left(s+k-1\right)}{k!}}{x}^k={\displaystyle \underset{k=0}{\overset{\infty }{\sum }}{\left(-1\right)}^k\frac{{\left(s\right)}_k}{k!}{x}^k}\end{array}$

同法可得 ${\left(1-x\right)}^{-s}={\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\frac{{\left(s\right)}_k}{k!}}{x}^k$

1) $\begin{array}{c}\zeta \left(s,a-t\right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{1}{{\left(n+a-t\right)}^s}}={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{1}{{\left(n+a\right)}^s}}\frac{{\left(n+a\right)}^s}{{\left(n+a-t\right)}^s}={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{1}{{\left(n+a\right)}^s}}\frac{1}{\frac{{\left(n+a-t\right)}^s}{{\left(n+a\right)}^s}}\\ ={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{1}{{\left(n+a\right)}^s}}\frac{1}{{\left(1-\frac{t}{n+a}\right)}^s}={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{1}{{\left(n+a\right)}^s}}{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\frac{{\left(s\right)}_k}{k!}}\frac{t^k}{{\left(n+a\right)}^k}\\ ={\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\frac{{\left(s\right)}_k}{k!}{t}^k{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{1}{{\left(n+a\right)}^{s+k}}}}={\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\frac{{\left(s\right)}_k}{k!}\zeta \left(s+k,a\right)t^k}\end{array}$ (1.1)成立。

2) 利用赫尔维茨zeta函数积分公式文 [7]， ${\displaystyle \int \zeta \left(s,q\right)\text{d}q}=\frac{1}{1-s}\zeta \left(s-1,q\right)$，对(8)式对 $a-t$ 积分1次，2次得到(9)，(10)式。

3) 利用赫尔维茨zeta函数微分公式文 [7]， $\frac{\text{d}}{\text{d}q}\zeta \left(s-1,q\right)=\left(1-s\right)\zeta \left(s,q\right)$，对(8)式对 $a-t$ 微分1次，2次得到(11)，(12)式。

命题3设 $\left|t\right|<\left|a\right|$，则含有升阶乘 ${\left(s\right)}_{2k}$ 赫尔维茨zeta函数级数封闭形和式

1) (13)

2) ${\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\frac{{\left(s\right)}_{2k}}{\left(2k+1\right)!}\zeta \left(s+2k,a\right)t^{2k+1}}=\frac{1}{2}\left[\frac{1}{s-1}\zeta \left(s-1,a-t\right)+\frac{1}{1-s}\zeta \left(s-1,a+t\right)\right]$ (14)

3) ${\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\frac{{\left(s\right)}_{2k}}{\left(2k+2\right)!}\zeta \left(s+2k,a\right)t^{2k+2}}=\frac{1}{2}\left[\frac{1}{\left(s-1\right)\left(s-2\right)}\zeta \left(s-2,a-t\right)+\frac{1}{\left(1-s\right)\left(2-s\right)}\zeta \left(s-2,a-t\right)\right]$ (15)

4) ${\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\frac{{\left(s\right)}_{2k}}{\left(2k-1\right)!}\zeta \left(s+2k,a\right)t^{2k-1}}=\frac{1}{2}\left[\left(s+1\right)\zeta \left(s+1,a-t\right)+\left(1-s\right)\zeta \left(s+1,a+t\right)\right]$ (16)

5)(17)

证明 1) $\begin{array}{l}\frac{1}{2}\left[\zeta \left(s,a-t\right)+\zeta \left(s,a+t\right)\right]\\ =\frac{1}{2}\left[{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{1}{{\left(n+a\right)}^s}}\frac{{\left(n+a\right)}^s}{{\left(n+a-t\right)}^s}+{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{1}{{\left(n+a\right)}^s}}\frac{{\left(n+a\right)}^s}{{\left(n+a+t\right)}^s}\right]\\ =\frac{1}{2}\left[{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{1}{{\left(n+a\right)}^s}}\frac{1}{\frac{{\left(n+a-t\right)}^s}{{\left(n+a\right)}^s}}+{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{1}{{\left(n+a\right)}^s}}\frac{1}{\frac{{\left(n+a+t\right)}^s}{{\left(n+a\right)}^s}}\right]\\ =\frac{1}{2}\left[{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{1}{{\left(n+a\right)}^s}}\frac{1}{{\left(1-\frac{t}{n+a}\right)}^s}+{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{1}{{\left(n+a\right)}^s}}\frac{1}{{\left(1+\frac{t}{n+a}\right)}^s}\right]\end{array}$

$\begin{array}{l}=\frac{1}{2}\left[{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{1}{{\left(n+a\right)}^s}}{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\frac{{\left(s\right)}_k}{k!}}\frac{t^k}{{\left(n+a\right)}^k}+{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{1}{{\left(n+a\right)}^s}}{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^k{\left(s\right)}_k}{k!}}\frac{t^k}{{\left(n+a\right)}^k}\right]\\ =\frac{1}{2}\left[{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\frac{{\left(s\right)}_k}{k!}{t}^k}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{1}{{\left(n+a\right)}^{s+k}}}+{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^k{\left(s\right)}_k}{k!}{t}^k}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{1}{{\left(n+a\right)}^{s+k}}}\right]\\ =\frac{1}{2}\left[{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{2}{{\left(n+a\right)}^s}}+\frac{2{\left(s\right)}_2{t}^2}{2!}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{1}{{\left(n+a\right)}^{s+2}}}+\cdots +\frac{2{\left(s\right)}_{2k}{t}^{2k}}{\left(2k\right)!}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{1}{{\left(n+a\right)}^{s+2k}}}+\cdots \right]\\ ={\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\frac{{\left(s\right)}_{2k}}{\left(2k\right)!}\zeta \left(s+2k,a\right)t^{2k}}\end{array}$ (13)式成立。

2) 利用赫尔维茨zeta函数积分公式 ${\displaystyle \int \zeta \left(s,q\right)\text{d}q}=\frac{1}{1-s}\zeta \left(s-1,q\right)$，对(13)式对 $a-t$ 积分1次，2次得到(14)，(15)式。

3) 利用广义黎曼zeta函数微分公式 $\frac{\text{d}}{\text{d}q}\zeta \left(s-1,q\right)=\left(1-s\right)\zeta \left(s,q\right)$，对(13)式对 $a-t$ 微分1次，2次得到(16)，(17)式。

类似命题2的方法可得如下

命题4设 $\left|t\right|<\left|a\right|$，则含有升阶乘 ${\left(s\right)}_{2k+1}$ 赫尔维茨zeta函数级数封闭形和式

1) ${\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\frac{{\left(s\right)}_{2k+1}}{\left(2k+1\right)!}\zeta \left(s+2k,a\right)t^{2k+1}}=\frac{1}{2}\left[\zeta \left(s,a-t\right)-\zeta \left(s,a+t\right)\right]$ (18)

2) ${\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\frac{{\left(s\right)}_{2k+1}}{\left(2k+2\right)!}\zeta \left(s+2k,a\right)t^{2k+2}}=\frac{1}{2}\left[\frac{1}{s-1}\zeta \left(s-1,a-t\right)-\frac{1}{1-s}\zeta \left(s-1,a+t\right)\right]$ (19)

3) ${\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\frac{{\left(s\right)}_{2k+1}}{\left(2k+3\right)!}\zeta \left(s+2k,a\right)t^{2k+3}}=\frac{1}{2}\left[\frac{1}{\left(s-1\right)\left(s-2\right)}\zeta \left(s-2,a-t\right)-\frac{1}{\left(1-s\right)\left(2-s\right)}\zeta \left(s-2,a-t\right)\right]$ (20)

4) ${\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\frac{{\left(s\right)}_{2k+1}}{\left(2k\right)!}\zeta \left(s+2k,a\right)t^{2k}}=\frac{1}{2}\left[\left(s+1\right)\zeta \left(s+1,a-t\right)-\left(1-s\right)\zeta \left(s+1,a+t\right)\right]$ (21)

5) ${\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\frac{{\left(s\right)}_{2k+1}}{\left(2k-1\right)!}\zeta \left(s+2k,a\right)t^{2k-1}}=\frac{1}{2}\left[\left(s+1\right)\left(s+2\right)\zeta \left(s+2,a-t\right)-\left(1-s\right)\left(2-s\right)\zeta \left(s+2,a+t\right)\right]$ (22)

3. 一些与黎曼zeta函数相关的数值级数

(1) 一些关于黎曼zeta函数 $\zeta \left(2n\right)$ 的数值级数

在(1)式，令 $x=\frac{\pi }{2},\frac{\pi }{3},\frac{\pi }{4}$，有关于 $\zeta \left(2n\right)$ 函数的数值级数

1) ${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\zeta \left(2n\right)}{2^{2n-1}\pi }}=\frac{1}{\pi }$ ；2) ${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\zeta \left(2n\right)}{3^{2n-1}\pi }}=\frac{3}{2\pi }-\frac{1}{2\sqrt{3}}$ ；3) ${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\zeta \left(2n\right)}{4^{2n-1}\pi }}=\frac{2}{\pi }-\frac{1}{2}$。

在(2)式，令 $x=\frac{\pi }{2},\frac{\pi }{3},\frac{\pi }{4}$，有关于 $\zeta \left(2n\right)$ 函数的数值级数

1) ${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\zeta \left(2n\right)}{2^{2n}n}}=\ln \frac{\pi }{2}$ ；2) ${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\zeta \left(2n\right)}{3^{2n}n}}=\ln \frac{\pi }{3}-\ln \frac{\sqrt{3}}{2}$ ；3) ${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\zeta \left(2n\right)}{4^{2n}n}}=\ln \frac{\pi }{4}-\ln \frac{\sqrt{2}}{2}$。

在(3)式令 $x=\frac{\pi }{2},\frac{\pi }{4},\frac{\pi }{6}$，有关于 $\zeta \left(2n\right)$ 函数的数值级数

1) ${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\zeta \left(2n\right)\pi }{2^{2n+1}n\left(2n+1\right)}}=-\frac{\pi }{2}+\frac{\pi }{2}\ln 2$ ；2) ${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\zeta \left(2n\right)\pi }{4^{2n+1}n\left(2n+1\right)}}=-\frac{\pi }{4}+\frac{\pi }{4}\ln \sqrt{2}+\frac{1}{2}G$ ；

3) ${\psi }^{\prime }\left(1/3\right){\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\zeta \left(2n\right)\pi }{6^{2n+1}n\left(2n+1\right)}}=-\frac{\pi }{6}+\frac{\sqrt{3}}{12}\left({\psi }^{\prime }\left(1/3\right)-\frac{2}{3}{\pi }^2\right)$。

这里 $G={\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^k}{{\left(2k+1\right)}^2}}=0.915965947\cdots$ 是卡大兰常数。为双伽马函数又称普西函数 ${\psi }^{\prime }\left(z\right)$ 是普西函数的导数。这里 ${\psi }^{\prime }\left(\frac{1}{3}\right)={\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\frac{1}{{\left(k+1/3\right)}^2}}$

在(4)式令 $x=\frac{\pi }{2},\frac{\pi }{3},\frac{\pi }{4},\frac{\pi }{6}$，有关于 $\zeta \left(2n\right)$ 函数的数值级数

1) $\begin{array}{l}{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\zeta \left(2n\right){\pi }^2}{2^{2n+3}n\left(n+1\right)}}\\ =\frac{1}{8}{\pi }^2\ln \frac{\pi }{2}-\frac{{\pi }^2}{16}-\frac{{\pi }^2}{8}\ln \sin \frac{\pi }{2}+\frac{{\pi }^2}{4}\ln 2+\frac{\pi }{4}C{l}_2\left(\pi \right)+\frac{1}{4}C{l}_3\left(\pi \right)-\frac{1}{4}\zeta \left(3\right)\\ =\frac{1}{8}{\pi }^2\ln \frac{\pi }{2}-\frac{{\pi }^2}{16}-0+\frac{{\pi }^2}{4}\ln 2+0+\frac{1}{4}\left(-\frac{3}{4}\zeta \left(3\right)\right)-\frac{1}{4}\zeta \left(3\right)\\ =\frac{1}{8}{\pi }^2\ln \frac{\pi }{2}-\frac{{\pi }^2}{16}+\frac{{\pi }^2}{4}\ln 2-\frac{7}{16}\zeta (3)\end{array}$

2) $\begin{array}{l}{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\zeta \left(2n\right){\pi }^2}{3^{2n+2}n\left(2n+2\right)}}\\ =\frac{1}{18}{\pi }^2\ln \frac{\pi }{3}-\frac{{\pi }^2}{36}-\frac{{\pi }^2}{18}\ln \sqrt{3}+\frac{\pi }{6}C{l}_2\left(2\pi /3\right)+\frac{1}{4}C{l}_3\left(2\pi /3\right)-\frac{1}{4}\zeta \left(3\right)\\ =\frac{1}{18}{\pi }^2\ln \frac{\pi }{3}-\frac{{\pi }^2}{36}-\frac{{\pi }^2}{36}\ln 3+\frac{\pi }{6}\frac{\sqrt{3}}{9}\left[{\psi }^{\prime }\left(1/3\right)-\frac{2}{3}{\pi }^2\right]+\frac{1}{4}\left(-\frac{4}{9}\zeta \left(3\right)\right)-\frac{1}{4}\zeta \left(3\right)\\ =\frac{1}{18}{\pi }^2\ln \frac{\pi }{3}-\frac{{\pi }^2}{36}-\frac{{\pi }^2}{36}\ln 3+\frac{\pi }{54}\sqrt{3}\left[{\psi }^{\prime }\left(1/3\right)-\frac{2}{3}{\pi }^2\right]-\frac{13}{36}\zeta (3)\end{array}$

3) $\begin{array}{l}{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\zeta \left(2n\right){\pi }^2}{4^{2n+2}n\left(2n+2\right)}}\\ =\frac{1}{36}{\pi }^2\ln \frac{\pi }{4}-\frac{{\pi }^2}{64}-\frac{{\pi }^2}{32}\ln \frac{\sqrt{2}}{2}+\frac{{\pi }^2}{16}\ln \sqrt{2}+\frac{\pi }{8}C{l}_2\left(\pi /2\right)+\frac{1}{4}C{l}_3\left(\pi /2\right)-\frac{1}{4}\zeta \left(3\right)\\ =\frac{1}{36}{\pi }^2\ln \frac{\pi }{4}-\frac{{\pi }^2}{64}-\frac{3{\pi }^2}{64}\ln 2+\frac{\pi }{8}G+\frac{1}{4}\left(-\frac{3}{32}\zeta \left(3\right)\right)-\frac{1}{4}\zeta \left(3\right)\\ =\frac{1}{36}{\pi }^2\ln \frac{\pi }{4}-\frac{{\pi }^2}{64}-\frac{3{\pi }^2}{64}\ln 2+\frac{\pi }{8}G-\frac{35}{128}\zeta (3)\end{array}$

4) $\begin{array}{l}{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\zeta \left(2n\right){\pi }^2}{6^{2n+2}n\left(2n+2\right)}}\\ =\frac{1}{72}{\pi }^2\ln \frac{\pi }{6}-\frac{{\pi }^2}{144}-\frac{{\pi }^2}{72}\ln \frac{1}{2}+\frac{{\pi }^2}{36}\ln \left(2\cdot \frac{1}{2}\right)+\frac{\pi }{12}C{l}_2\left(\pi /3\right)+\frac{1}{4}C{l}_3\left(\pi /3\right)-\frac{1}{4}\zeta \left(3\right)\\ =\frac{1}{72}{\pi }^2\ln \frac{\pi }{6}-\frac{{\pi }^2}{144}+\frac{{\pi }^2}{72}\ln 2+\frac{\pi }{12}\frac{\sqrt{3}}{6}\left[{\psi }^{\prime }\left(1/3\right)-\frac{2}{3}{\pi }^2\right]+\frac{1}{4}\left(\frac{1}{3}\zeta \left(3\right)\right)-\frac{1}{4}\zeta \left(3\right)\\ =\frac{1}{72}{\pi }^2\ln \frac{\pi }{6}-\frac{{\pi }^2}{144}+\frac{{\pi }^2}{72}\ln 2+\frac{\pi }{72}\sqrt{3}\left[{\psi }^{\prime }\left(1/3\right)-\frac{2}{3}{\pi }^2\right]-\frac{1}{6}\zeta (3)\end{array}$

在(5)式，令 $x=\frac{\pi }{6},\frac{\pi }{4},\frac{\pi }{3}$，关于 $\zeta \left(2n\right)$ 函数的数值级数如下

1) $\begin{array}{l}{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\zeta \left(2n\right){\pi }^3}{6^{2n+3}n\left(2n+3\right)}}\\ =\frac{1}{648}{\pi }^3\ln \frac{\pi }{6}-\frac{{\pi }^3}{1944}-\frac{{\pi }^3}{648}\ln \frac{1}{2}+\frac{1}{2}\frac{{\pi }^2}{36}C{l}_2\left(\pi /3\right)-\frac{1}{4}C{l}_4\left(\pi /3\right)+\frac{{\pi }^3}{648}\ln 1+\frac{1}{2}C{l}_3\left(\pi /3\right)\\ =\frac{1}{648}{\pi }^3\ln \frac{\pi }{6}-\frac{{\pi }^3}{1944}+\frac{{\pi }^3}{648}\ln 2+\frac{{\pi }^2}{72}\frac{\sqrt{3}}{6}\left[{\psi }^{\prime }\left(1/3\right)-\frac{2}{3}{\pi }^2\right]\\ -\frac{\sqrt{3}}{4}\left\{-\frac{40}{81}\zeta \left(4\right)-\frac{1}{6^4}\left[\zeta \left(4,\frac{1}{6}\right)+\zeta \left(4,\frac{1}{3}\right)\right]\right\}+\frac{1}{2}\left[\frac{1}{2}\left(1-2^{-2}\right)\left(1-3^{-2}\right)\zeta \left(3\right)\right]\\ =\frac{1}{648}{\pi }^3\ln \frac{\pi }{6}-\frac{{\pi }^3}{1944}+\frac{{\pi }^3}{648}\ln 2+\frac{{\pi }^2}{432}\sqrt{3}\left[{\psi }^{\prime }\left(1/3\right)-\frac{2}{3}{\pi }^2\right]\\ -\frac{\sqrt{3}}{4}\left\{\frac{3^{-4}-1}{2}\zeta \left(4\right)-\frac{1}{6^4}\left[\zeta \left(4,\frac{1}{6}\right)+\zeta \left(4,\frac{1}{3}\right)\right]\right\}+\frac{1}{6}\zeta (3)\end{array}$

2) $\begin{array}{l}{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\zeta \left(2n\right){\pi }^3}{4^{2n+3}n\left(2n+3\right)}}\\ =\frac{1}{192}{\pi }^3\ln \frac{\pi }{4}-\frac{{\pi }^3}{576}-\frac{{\pi }^3}{192}\ln \frac{1}{\sqrt{2}}+\frac{{\pi }^2}{32}C{l}_2\left(\pi /2\right)-\frac{1}{4}C{l}_4\left(\pi /2\right)+\frac{{\pi }^3}{192}\ln \sqrt{2}+\frac{1}{2}C{l}_3\left(\pi /2\right)\\ =\frac{1}{192}{\pi }^3\ln \frac{\pi }{4}-\frac{{\pi }^3}{576}+\frac{{\pi }^3}{192}\ln 2+\frac{{\pi }^2}{32}G-\frac{\sqrt{3}}{4}\left[\left(2^{-4}-1\right)\zeta \left(4\right)+2^{1-8}\zeta \left(4,\frac{1}{4}\right)\right]\\ +\frac{1}{2}\left[-2^{-3}\left(1-2^{-2}\right)\zeta \left(3\right)\right]\\ =\frac{1}{192}{\pi }^3\ln \frac{\pi }{4}-\frac{{\pi }^3}{576}+\frac{{\pi }^3}{192}\ln 2+\frac{{\pi }^2}{32}G-\frac{\sqrt{3}}{4}\left[-\frac{15}{16}\zeta \left(4\right)+\frac{1}{2^7}\zeta \left(4,\frac{1}{4}\right)\right]-\frac{3}{64}\zeta (3)\end{array}$

3) $\begin{array}{l}{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{\zeta \left(2n\right){\pi }^3}{3^{2n+3}n\left(2n+3\right)}}\\ =\frac{1}{81}{\pi }^3\ln \frac{\pi }{3}-\frac{{\pi }^3}{243}-\frac{{\pi }^3}{81}\ln \frac{\sqrt{3}}{2}+\frac{{\pi }^2}{18}C{l}_2\left(2\pi /3\right)\\ -\frac{1}{4}C{l}_4\left(2\pi /3\right)+\frac{{\pi }^3}{81}\ln \sqrt{3}+\frac{1}{2}C{l}_3\left(2\pi /3\right)\\ =\frac{1}{81}{\pi }^3\ln \frac{\pi }{3}-\frac{{\pi }^3}{243}+\frac{{\pi }^3}{81}\ln 2+\frac{{\pi }^2}{18}\frac{\sqrt{3}}{9}\left[{\psi }^{\prime }\left(1/3\right)-\frac{2}{3}{\pi }^2\right]\\ -\frac{\sqrt{3}}{4}\left[\frac{3^{-4}-1}{2}\zeta \left(4\right)+3^{-4}\zeta \left(4,\frac{1}{3}\right)\right]+\frac{1}{2}\frac{1}{2}\left(3^{1-3}-1\right)\zeta \left(3\right)\\ =\frac{1}{81}{\pi }^3\ln \frac{\pi }{3}-\frac{{\pi }^3}{243}+\frac{{\pi }^3}{81}\ln 2+\frac{{\pi }^2}{162}\sqrt{3}\left[{\psi }^{\prime }\left(1/3\right)-\frac{2}{3}{\pi }^2\right]\\ -\frac{\sqrt{3}}{4}\left[-\frac{40}{81}\zeta \left(4\right)+\frac{1}{81}\zeta \left(4,\frac{1}{3}\right)\right]-\frac{2}{9}\zeta (3)\end{array}$