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337

今回解いたのはこれです。

リンク
$ζ (3)$ の無理数倍というなかなか途中過程が想像しにくい値ですが、なんとか自分の力だけで解けたので、(わざわざ記事にする意味があるか謎ですが)つらつら解法を書きます。
$I = \int_{0}^{1} \frac{x}{\tanh^{- 1} (x)} d x$
とする。

$I = \frac{1}{2} \int_{0}^{\infty} \tanh^{2} (1 / x) d x$

$\begin{aligned} I & = \int_{0}^{1} \frac{x}{\tanh^{- 1} (x)} d x \\ = \int_{0}^{\infty} \frac{\tanh (u)}{u \cosh^{2} (u)} d u (x = \tanh u と置換) \\ = \frac{1}{2} \int_{0}^{\infty} \frac{1}{u} {(1 - \frac{1}{\cosh^{2} (u)})}^{'} d u \\ = \frac{1}{2} {[\frac{1}{u} (1 - \frac{1}{\cosh^{2} (u)})]}_{0}^{\infty} + \frac{1}{2} \int_{0}^{\infty} \frac{1}{u^{2}} (1 - \frac{1}{\cosh^{2} (u)}) d u \\ = \frac{1}{2} \int_{0}^{\infty} {(\frac{\tanh (u)}{u})}^{2} d u \\ = \frac{1}{2} \int_{0}^{\infty} \tanh^{2} (1 / x) d x (u = 1 / x と置換) \end{aligned}$

$\tanh^{2} (1 / x) = \sum_{n, m \geq 0} \frac{4 x^{2}}{(1 + x^{2} π^{2} {(n + 1 / 2)}^{2}) (1 + x^{2} π^{2} {(m + 1 / 2)}^{2})}$

三角関数の部分分数展開より、
$π \tan π x = - \sum_{n = 0}^{\infty} \frac{2 x}{x^{2} - {(n + \frac{1}{2})}^{2}}$
$π x \to i / x$ として整理すると、 $\tan (i x) = i \tanh (x)$ より、
$\tanh (1 / x) = \sum_{n = 0}^{\infty} \frac{2 x}{1 + x^{2} π^{2} {(n + 1 / 2)}^{2}}$
となる。よって、
$\begin{aligned} \tanh^{2} (1 / x) & = \tanh (1 / x) \cdot \tanh (1 / x) \\ = \sum_{n, m \geq 0} \frac{4 x^{2}}{(1 + x^{2} π^{2} {(n + 1 / 2)}^{2}) (1 + x^{2} π^{2} {(m + 1 / 2)}^{2})} \end{aligned}$

$\int_{0}^{\infty} \frac{d x}{(x^{2} + a^{2}) (x^{2} + b^{2})} = \frac{π}{2} \cdot \frac{1}{a b (a + b)} (a, b > 0)$

$a \neq b$ のとき、
$\begin{aligned} \int_{0}^{\infty} \frac{d x}{(x^{2} + a^{2}) (x^{2} + b^{2})} & = \frac{1}{b^{2} - a^{2}} \int_{0}^{\infty} (\frac{1}{x^{2} + a^{2}} - \frac{1}{x^{2} + b^{2}}) d x \\ = \frac{1}{b^{2} - a^{2}} {[\frac{1}{a} \arctan (\frac{x}{a}) - \frac{1}{b} \arctan (\frac{x}{b})]}_{0}^{\infty} \\ = \frac{1}{b^{2} - a^{2}} \cdot \frac{b - a}{a b} \cdot \frac{π}{2} \\ = \frac{π}{2} \cdot \frac{1}{a b (a + b)} \end{aligned}$
また、 $a = b$ のときは、
$\begin{aligned} \int_{0}^{\infty} \frac{d x}{(x^{2} + a^{2}) (x^{2} + b^{2})} & = \int_{0}^{\infty} \frac{1}{{(x^{2} + a^{2})}^{2}} d x \\ = \int_{0}^{\frac{π}{2}} \frac{\cos^{4} (θ)}{a^{4}} \cdot \frac{a}{\cos^{2} (θ)} d θ (x = a \tan θ と置換) \\ = \frac{1}{a^{3}} \int_{0}^{\frac{π}{2}} \frac{1 + \cos (2 θ)}{2} d θ \\ = \frac{π}{4 a^{3}} = \frac{π}{2} \cdot \frac{1}{a \cdot a (a + a)} \end{aligned}$
より、 $a = b$ でも成立する。

$\sum_{n, m \geq 0}^{} \frac{1}{(n + \frac{1}{2}) (m + \frac{1}{2}) (n + m + 1)} = 7 ζ (3)$

この等式は示すのに多くのステップを要したので、それぞれ補題に分けて証明します。(MZVをしっかりと学んでいる方はもっと簡単に(級数の変形だけで？)示せると思います。私は学んでいないので面倒に導出しました。)

$\sum_{n, m \geq 0}^{} \frac{1}{(n + \frac{1}{2}) (m + \frac{1}{2}) (n + m + 1)} = \sum_{m = 0}^{\infty} \frac{8}{(2 m + 1)^{2}} (\sum_{n = 0}^{\infty} (\frac{1}{2 n + 1} - \frac{1}{2 n + 2 m + 2}))$

$\begin{aligned} \sum_{n, m \geq 0}^{} \frac{1}{(n + \frac{1}{2}) (m + \frac{1}{2}) (n + m + 1)} & = \sum_{n, m \geq 0}^{} \frac{8}{(2 n + 1) (2 m + 1) (2 n + 2 m + 2)} \\ = \sum_{m = 0}^{\infty} (\frac{8}{2 m + 1} \sum_{n = 0}^{\infty} \frac{1}{(2 n + 1) (2 n + 2 m + 2)}) \\ = \sum_{m = 0}^{\infty} (\frac{8}{{(2 m + 1)}^{2}} \sum_{n = 0}^{\infty} (\frac{1}{2 n + 1} - \frac{1}{2 n + 2 m + 2})) \end{aligned}$

$\sum_{m = 0}^{\infty} (\frac{8}{{(2 m + 1)}^{2}} \sum_{n = 0}^{\infty} (\frac{1}{2 n + 1} - \frac{1}{2 n + 2 m + 2})) = 2 \int_{0}^{1} \frac{\ln^{2} (\frac{1 + u}{1 - u})}{u} d u$

$\begin{aligned} \sum_{m = 0}^{\infty} (\frac{8}{{(2 m + 1)}^{2}} \sum_{n = 0}^{\infty} (\frac{1}{2 n + 1} - \frac{1}{2 n + 2 m + 2})) \\ = \sum_{m = 0}^{\infty} \frac{8}{{(2 m + 1)}^{2}} \sum_{n = 0}^{\infty} (\int_{0}^{1} (u^{2 n} - u^{2 n + 2 m + 1}) d u) \\ = \sum_{m = 0}^{\infty} \frac{8}{{(2 m + 1)}^{2}} \int_{0}^{1} \frac{1 - u^{2 m + 1}}{1 - u^{2}} d u \\ = \int_{0}^{1} \frac{8 \sum_{m = 0}^{\infty} \frac{1}{{(2 m + 1)}^{2}} - 8 \sum_{m = 0}^{\infty} \frac{u^{2 m + 1}}{{(2 m + 1)}^{2}}}{1 - u^{2}} d u \\ = \int_{0}^{1} (π^{2} - 8 \sum_{m = 0}^{\infty} \frac{u^{2 m + 1}}{{(2 m + 1)}^{2}}) {(\frac{1}{2} \ln (\frac{1 + u}{1 - u}))}^{'} d u \\ = 4 \int_{0}^{1} \ln (\frac{1 + u}{1 - u}) \sum_{m = 0}^{\infty} \frac{u^{2 m}}{2 m + 1} d u \\ = 4 \int_{0}^{1} \ln (\frac{1 + u}{1 - u}) \cdot \frac{1}{u} \int_{0}^{u} \frac{1}{1 - t^{2}} d t d u \\ = 2 \int_{0}^{1} \frac{\ln^{2} (\frac{1 + u}{1 - u})}{u} d u \end{aligned}$

$2 \int_{0}^{1} \frac{\ln^{2} (\frac{1 + u}{1 - u})}{u} d u = 7 ζ (3)$

$2 \int_{0}^{1} \frac{\ln^{2} (\frac{1 + u}{1 - u})}{u} d u$
について、 $\frac{1 - u}{1 + u} = w$ と置換すると、
$u = \frac{1 - w}{1 + w} ⟹ d u = \frac{- 2}{(1 + w)^{2}} d w, u : 0 \to 1, w : 1 \to 0$
より、
$\begin{aligned} 2 \int_{0}^{1} \frac{\ln^{2} (\frac{1 + u}{1 - u})}{u} d u \\ = 2 \int_{1}^{0} \frac{\ln^{2} (1 / w)}{\frac{1 - w}{1 + w}} \frac{- 2}{(1 + w)^{2}} d w \\ = 4 \int_{0}^{1} \frac{\ln^{2} (w)}{1 - w^{2}} d w \\ = 4 \sum_{n = 0}^{\infty} \int_{0}^{1} \ln^{2} (w) w^{2 n} d w \\ = 4 \sum_{n = 0}^{\infty} \frac{2}{(2 n + 1)^{3}} \\ = 8 (\sum_{n = 1}^{\infty} \frac{1}{n^{3}} - \sum_{n = 1}^{\infty} \frac{1}{(2 n)^{3}}) \\ = 8 \cdot \frac{7}{8} \sum_{n = 1}^{\infty} \frac{1}{n^{3}} = 7 ζ (3) \end{aligned}$

ここまでの定理、補題から表題の等式を証明します。

$\begin{aligned} I & = \int_{0}^{1} \frac{x}{\tanh^{- 1} (x)} d x \\ = \frac{1}{2} \int_{0}^{\infty} \tanh^{2} (1 / x) d x 補題1より \\ = \int_{0}^{\infty} \sum_{n, m \geq 0} \frac{2 x^{2}}{(1 + x^{2} π^{2} {(n + 1 / 2)}^{2}) (1 + x^{2} π^{2} {(m + 1 / 2)}^{2})} 補題2より \\ = \sum_{n, m \geq 0} \int_{0}^{\infty} \frac{2 x^{2}}{(1 + x^{2} π^{2} {(n + 1 / 2)}^{2}) (1 + x^{2} π^{2} {(m + 1 / 2)}^{2})} d x \\ = 2 \sum_{n, m \geq 0} \int_{0}^{\infty} \frac{d u}{(u^{2} + π^{2} (n + 1 / 2)^{2}) (u^{2} + π^{2} (m + 1 / 2)^{2})} (1 / x = u と置換) \\ = 2 \sum_{n, m \geq 0} \frac{π}{2} \frac{1}{π (n + 1 / 2) \cdot π (m + 1 / 2) π (n + m + 1)} (補題3において、 a = π (n + 1 / 2), b = π (m + 1 / 2) とする) \\ = \frac{1}{π^{2}} \sum_{n, m \geq 0}^{} \frac{1}{(n + \frac{1}{2}) (m + \frac{1}{2}) (n + m + 1)} \\ = \frac{1}{π^{2}} 7 ζ (3) (補題4より) \\ = \frac{7 ζ (3)}{π^{2}} \end{aligned}$