大学数学基礎解説

ζ(1,1,･･･,1,2)について

多重ゼータ値,リーマンゼータ関数

425

1.はじめに

多重ゼータ値関連の内容を書きます。
今回やるやつ↓↓↓

$r$ は2以上の正整数
$ζ ({1}^{r - 2}, 2) = ζ (r)$

例
$ζ (3) = ζ (1, 2)$
$ζ (4) = ζ (1, 1, 2)$
$ζ (5) = ζ (1, 1, 1, 2)$
$⋮$

2.補題

$ζ (s) = \frac{1}{Γ (s)} \int_{0}^{\infty} \frac{x^{s - 1}}{e^{x} - 1} d x$

$\begin{aligned} Γ (s) & = \int_{0}^{\infty} x^{s - 1} e^{- x} d x \\ = n^{s} \int_{0}^{\infty} t^{s - 1} e^{- n t} d t (x = n t) \end{aligned}$
$∴ \frac{1}{n^{s}} = \frac{1}{Γ (s)} \int_{0}^{\infty} t^{s - 1} e^{- n t} d t$
両辺 $1 \leq n$ の範囲で和をとると
$\begin{aligned} ζ (s) & = \sum_{n = 1}^{\infty} \frac{1}{Γ (s)} \int_{0}^{\infty} t^{s - 1} e^{- n t} d t \\ = \frac{1}{Γ (s)} \int_{0}^{\infty} t^{s - 1} \sum_{n = 1}^{\infty} e^{- n t} d t \\ = \frac{1}{Γ (s)} \int_{0}^{\infty} \frac{t^{s - 1}}{e^{t} - 1} d t ◼ \end{aligned}$

$r$ は正整数
$\frac{1}{r!} (- \log (1 - x))^{r} = \sum_{0 < n_{1} < n_{2} < \dots < n_{r}} \frac{x^{n_{r}}}{n_{1} n_{2} \dots n_{r}}$

シャッフル積を用いる
$\begin{aligned} \frac{1}{r!} (- \log (1 - x))^{r} & = \frac{1}{r!} (\int_{0}^{x} \frac{1}{1 - t} d t)^{r} \\ = \int_{0 < t_{1} < t_{2} < \dots < t_{r} < x} \frac{d t_{1}}{1 - t_{1}} \frac{d t_{2}}{1 - t_{2}} \dots \frac{d t_{r}}{1 - t_{r}} \\ = \int_{0}^{x} \frac{d t_{r}}{1 - t_{r}} \dots \int_{0}^{t_{4}} \frac{d t_{3}}{1 - t_{3}} \int_{0}^{t_{3}} \frac{d t_{2}}{1 - t_{2}} \int_{0}^{t_{2}} \frac{d t_{1}}{1 - t_{1}} \\ = \int_{0}^{x} \frac{d t_{r}}{1 - t_{r}} \dots \int_{0}^{t_{4}} \frac{d t_{3}}{1 - t_{3}} \int_{0}^{t_{3}} \frac{d t_{2}}{1 - t_{2}} \int_{0}^{t_{2}} \sum_{1 \leq n_{1}} {t_{1}}^{n_{1} - 1} \\ = \sum_{1 \leq n_{1}} \int_{0}^{x} \frac{d t_{r}}{1 - t_{r}} \dots \int_{0}^{t_{4}} \frac{d t_{3}}{1 - t_{3}} \int_{0}^{t_{3}} \frac{d t_{2}}{1 - t_{2}} \int_{0}^{t_{2}} {t_{1}}^{n_{1} - 1} \\ = \sum_{1 \leq n_{1}} \frac{1}{n_{1}} \int_{0}^{x} \frac{d t_{r}}{1 - t_{r}} \dots \int_{0}^{t_{4}} \frac{d t_{3}}{1 - t_{3}} \int_{0}^{t_{3}} \frac{{t_{2}}^{n_{1}}}{1 - t_{2}} d t_{2} \\ = \sum_{1 \leq n_{1}, n_{2}} \frac{1}{n_{1}} \int_{0}^{x} \frac{d t_{r}}{1 - t_{r}} \dots \int_{0}^{t_{4}} \frac{d t_{3}}{1 - t_{3}} \int_{0}^{t_{3}} {t_{2}}^{n_{1} + n_{2} - 1} d t_{2} \\ = \sum_{1 \leq n_{1}, n_{2}} \frac{1}{n_{1} (n_{1} + n_{2})} \int_{0}^{x} \frac{d t_{r}}{1 - t_{r}} \dots \int_{0}^{t_{4}} \frac{{t_{3}}^{n_{1} + n_{2}}}{1 - t_{3}} d t_{3} \\ ⋮ \\ = \sum_{1 \leq n_{1}, n_{2} \dots n_{r}} \frac{x^{n_{1} + n_{2} + \dots + n_{r}}}{n_{1} (n_{1} + n_{2}) \dots (n_{1} + n_{2} + \dots + n_{r})} \\ = \sum_{0 < n_{1} < n_{2} < \dots < n_{r}} \frac{x^{n_{r}}}{n_{1} n_{2} \dots n_{r}} ◼ \end{aligned}$

3.本題

(再掲)

$r$ は2以上の正整数
$ζ ({1}^{r - 2}, 2) = ζ (r)$

$\begin{aligned} \frac{1}{(r - 1)!} \int_{0}^{1} \frac{(- \log (1 - x))^{r - 1}}{x} d x & = \int_{0}^{1} \sum_{0 < n_{1} < n_{2} < \dots < n_{r - 1}} \frac{x^{n_{r - 1} - 1}}{n_{1} n_{2} \dots n_{r - 1}} d x (∵ 補題 2) \\ = \sum_{0 < n_{1} < n_{2} < \dots < n_{r - 1}} \frac{1}{n_{1} n_{2} \dots n_{r - 1}^{2}} \\ = ζ ({1}^{r - 2}, 2) \end{aligned}$
一方で
$\begin{aligned} \frac{1}{(r - 1)!} \int_{0}^{1} \frac{(- \log (1 - x))^{r - 1}}{x} d x & = \frac{1}{(r - 1)!} \int_{0}^{\infty} \frac{t^{r - 1}}{1 - e^{- t}} e^{- t} d t (x = 1 - e^{- t}) \\ = \frac{1}{(r - 1)!} \int_{0}^{\infty} \frac{t^{r - 1}}{e^{t} - 1} d t \\ = \frac{1}{(r - 1)!} Γ (r) ζ (r) (補題 1) \\ = ζ (r) \end{aligned}$
したがって
$ζ ({1}^{r - 2}, 2) = ζ (r) ◼$