$\mathcal{Riemann}\,\zeta$ 函数

伯努利数

先给出伯努利数的生成函数定义

\frac{x}{e^{x} - 1} = \sum_{k = 0}^{\infty} \frac{B_{k}}{k!} x^{k}

可以用麦克劳林公式计算前几项伯努利数

\begin{aligned} B_{0} = 1 \\ B_{1} = - 1 / 2 \\ B_{2} = 1 / 6 \\ B_{3} = 0 \\ B_{4} = - 1 / 30 \\ B_{5} = 0 \\ \dots \end{aligned}

可以发现几个显然的规律，接下来依次将证明

一.

\sum_{k = 0}^{n} (\binom{n + 1}{k}) B_{k} = 0

$Proof$ :

根据伯努利数的定义，可以得到

\begin{aligned} x & = (e^{x} - 1) \sum_{k = 0}^{\infty} \frac{B_{k}}{k!} x^{k} \\ = \sum_{i = 1}^{\infty} \frac{x^{n}}{n!} \cdot \sum_{k = 0}^{\infty} \frac{B_{k}}{k!} x^{k} \\ = \sum_{i = 0}^{\infty} \frac{x^{i + 1}}{(i + 1)!} \cdot \sum_{k = 0}^{\infty} \frac{B_{k}}{k!} x^{k} \end{aligned}

根据两级数的柯西乘积

\sum_{n = 0}^{\infty} a_{n} \cdot \sum_{n = 0}^{\infty} b_{n} = \sum_{n = 0}^{\infty} \sum_{k = 0}^{n} a_{n} b_{n - k}

可以得到

\begin{aligned} x & = \sum_{i = 0}^{\infty} \frac{x^{i + 1}}{(i + 1)!} \cdot \sum_{k = 0}^{\infty} \frac{B_{k}}{k!} x^{k} \\ = \sum_{n = 0}^{\infty} \sum_{k = 0}^{n} \frac{x^{n + 1 - k}}{(n + 1 - k)!} \cdot \frac{B_{k}}{k!} x^{k} \\ = \sum_{n = 1}^{\infty} \sum_{k = 0}^{n} \frac{(n + 1)!}{(n + 1 - k)! k!} B_{k} \cdot \frac{x^{n + 1}}{(n + 1)!} \\ = \sum_{n = 1}^{\infty} \sum_{k = 0}^{n} (\binom{n + 1}{k}) B_{k} \frac{x^{n + 1}}{(n + 1)!} \end{aligned}

对照等式两边的系数即可得到

B_{0} = 1, \sum_{k = 0}^{n} (\binom{n + 1}{k}) B_{k} = 0

二.

B_{2 k + 1} = 0 (k ⩾ 1)

$Proof$ :

构造函数

\begin{aligned} g (x) & = \frac{x}{e^{x} - 1} - B_{1} x \\ = \frac{x}{e^{x} - 1} + \frac{1}{2} x \\ = \frac{2 x + x (e^{x} - 1)}{2 (e^{x} - 1)} \\ = \frac{x (e^{x} + 1)}{2 (e^{x} - 1)} \\ = \frac{x (e^{x} + 1)}{2 (e^{x} - 1)} \cdot \frac{e^{- x / 2}}{e^{- x / 2}} \\ = \frac{x (e^{x / 2} + e^{- x / 2})}{2 (e^{x / 2} - e^{- x / 2})} \end{aligned}

$g(x)$ 使偶函数，所以麦克劳林展开式中的奇次项系数都为零，也就是

B_{2 k + 1} = 0 (k ⩾ 1)

$\mathcal{Riemann}\,\zeta$ 函数

$Riemann\,\zeta$ 函数前，我们需要用刚才伯努利数的结论推一些式子.

$g(x)$ .

\begin{aligned} g (x) & = \sum_{n = 0}^{\infty} \frac{B_{2 n}}{(2 n)!} x^{2 n} \\ = \frac{x}{2} \coth \frac{x}{2} \end{aligned}

这也就推出了

\coth x = \sum_{n = 0}^{\infty} \frac{4^{n} B_{2 n}}{(2 n)!} x^{2 n - 1}

$x=ix$ .

\cot x = \sum_{n = 0}^{\infty} \frac{(- 1)^{n} 4^{n} B_{2 n}}{(2 n)!} x^{2 n - 1}

根据恒等式

2 \coth 2 x - \coth x = \tanh x

可以得到

\tanh x = \sum_{n = 1}^{\infty} \frac{4^{n} (4^{n} - 1) B_{2 n}}{(2 n)!} x^{2 n - 1}

$x=ix$ ，可得

\tan x = \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1} 4^{n} (4^{n} - 1) B_{2 n}}{(2 n)!} x^{2 n - 1}

$Riemann\,\zeta$ 函数的定义

ζ (z) = \sum_{n = 1}^{\infty} \frac{1}{n^{z}} = \frac{1}{1^{z}} + \frac{1}{2^{z}} + \frac{1}{3^{z}} + \dots (z \in C)

$\zeta(2n)$ 的公式进行推导.

根据这篇文章，我们知道

\frac{\sin π x}{π x} = \prod_{n = 1}^{\infty} (1 - \frac{x^{2}}{n^{2}})

两边取对数

\ln \frac{\sin π x}{π x} = \sum_{n = 1}^{\infty} \ln (1 - \frac{x^{2}}{n^{2}})

两边求导（过程省略）

\begin{aligned} π \cot π x - \frac{1}{x} = \sum_{n = 1}^{\infty} (\frac{1}{x - n} + \frac{1}{x + n}) \\ \cot x = \frac{1}{x} + \sum_{n = 1}^{\infty} (\frac{1}{x - n π} + \frac{1}{x + n π}) \end{aligned}

将分式展开

\begin{aligned} \cot x & = \frac{1}{x} + \sum_{n = 1}^{\infty} \frac{1}{π n} (\frac{1}{1 + \frac{x}{π n}} - \frac{1}{1 - \frac{x}{π n}}) \\ = \frac{1}{x} + \sum_{n = 1}^{\infty} \frac{1}{π n} [\sum_{k = 0}^{\infty} (- 1)^{n} {(\frac{x}{π n})}^{k} - \sum_{k = 0}^{\infty} {(\frac{x}{π n})}^{k}] \\ = \frac{1}{x} - 2 \sum_{n = 1}^{\infty} \frac{1}{π n} \sum_{k = 1}^{\infty} {(\frac{x}{π n})}^{2 k - 1} \\ = \frac{1}{x} - 2 \sum_{n = 1}^{\infty} \sum_{k = 1}^{\infty} \frac{x^{2 k - 1}}{(π n)^{2 k}} \\ = \frac{1}{x} - 2 \sum_{n = 1}^{\infty} \sum_{k = 1}^{\infty} \frac{x^{2 k - 1}}{π^{2 k}} \cdot \frac{1}{n^{2 k}} \\ = \frac{1}{x} - 2 \sum_{k = 1}^{\infty} \frac{x^{2 k - 1}}{π^{2 k}} \cdot ζ (2 k) \end{aligned}

我们又知道

\begin{aligned} \cot x & = \sum_{k = 0}^{\infty} \frac{(- 1)^{k} 4^{k} B_{2 k}}{(2 k)!} x^{2 k - 1} \\ = \frac{1}{x} + \sum_{k = 1}^{\infty} \frac{(- 1)^{k} 4^{k} B_{2 k}}{(2 k)!} x^{2 k - 1} \end{aligned}

也就是说

\frac{1}{x} + \sum_{k = 1}^{\infty} \frac{(- 1)^{k} 4^{k} B_{2 k}}{(2 k)!} x^{2 k - 1} = \frac{1}{x} - 2 \sum_{k = 1}^{\infty} \frac{x^{2 k - 1}}{π^{2 k}} \cdot ζ (2 k)

$x^{2k-1}$ 的系数即得

\begin{aligned} ζ (2 k) & = \frac{(- 1)^{k} 4^{k} B_{2 k} π^{2 k}}{2 (2 k)!} \\ = \frac{2^{2 k - 1} | B_{2 k} | π^{2 k}}{(2 k)!} \end{aligned}

$\zeta(2k)$ 的通项

ζ (2 k) = \frac{2^{2 k - 1} | B_{2 k} | π^{2 k}}{(2 k)!}

$k$ 的前几个值

\begin{aligned} ζ (2) = \frac{π^{2}}{6} ζ (4) = \frac{π^{4}}{90} \\ ζ (6) = \frac{π^{6}}{945} ζ (8) = \frac{π^{8}}{9450} \\ ζ (10) = \frac{π^{10}}{93555} ζ (12) = \frac{691 π^{12}}{638512875} \end{aligned}