積分bot を解いてみた

\begin{align} I&=\int_{0}^{1}\frac {x}{\tanh^{-1}(x)}dx \newline &=\int_{0}^{\infty}\frac {\tanh(u)}{u\cosh^{2}(u)}du\qquad (x=\tanh u \text{と置換}) \newline &=\frac {1}{2}\int_{0}^{\infty}\frac 1 u \left(1-\frac 1 {\cosh^2 (u)} \right)'du \newline &=\frac 1{2}\left[\frac 1 u\left(1- \frac 1 {\cosh^2 (u)} \right)\right]^{\infty}_{0}+\frac {1}{2}\int_{0}^{\infty}\frac {1} {u^2}\left(1- \frac 1 {\cosh^2 (u)} \right)du \newline &=\frac 12 \int_{0}^{\infty}\left(\frac {\tanh (u)} u\right)^2du \newline &=\frac 12 \int_{0}^{\infty}\tanh^2 (1/x)dx\qquad (u=1/x\text{と置換}) \end{align}

三角関数の部分分数展開 より、
$$ \displaystyle \pi \tan {{\pi }x}=-\sum _{n=0}^{\infty }{\frac {2x}{x^{2}-\left(n+\textstyle {\frac {1}{2}}\right)^{2}}}$$
$\pi x\to i/x$として整理すると、$\tan(ix)=i\tanh(x)$より、
$$\tanh (1/x)=\sum_{n=0}^{\infty}\frac {2x}{1+x^2\pi^2\left(n+1/2\right)^2}$$
となる。よって、
\begin{align} \tanh ^2(1/x)&=\tanh(1/x)\cdot \tanh(1/x) \newline &=\sum_{n,m\ge 0}\frac {4x^2}{\left(1+x^2\pi^2\left(n+1/2\right)^2\right)\left(1+x^2\pi^2\left(m+1/2\right)^2\right)}\end{align}

$a\ne b$のとき、
\begin{align} \int_{0}^{\infty}\frac{dx}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}&=\frac{1}{b^{2}-a^{2}}\int_{0}^{\infty}\left(\frac{1}{x^{2}+a^{2}}-\frac{1}{x^{2}+b^{2}}\right)dx \newline &=\frac{1}{b^{2}-a^{2}}\left[\frac{1}{a}\arctan\left(\frac{x}{a}\right)-\frac{1}{b}\arctan\left(\frac{x}{b}\right)\right]_{0}^{\infty} \newline &=\frac{1}{b^{2}-a^{2}}\cdot\frac {b-a}{ab}\cdot \frac {\pi}{2} \newline &=\frac {\pi}{2}\cdot \frac {1}{ab(a+b)} \end{align}
また、$a=b$のときは、
\begin{align} \int_{0}^{\infty}\frac{dx}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}&=\int_{0}^{\infty}\frac{1}{\left(x^{2}+a^{2}\right)^{2}}dx \newline &=\int_{0}^{\frac{\pi}{2}}\frac{\cos^{4}\left(\theta\right)}{a^{4}}\cdot\frac{a}{\cos^{2}\left(\theta\right)}d\theta\quad (x=a\tan\theta \text{と置換} ) \newline &=\frac{1}{a^{3}}\int_{0}^{\frac{\pi}{2}}\frac{1+\cos\left(2\theta\right)}{2}d\theta\newline &=\frac{\pi}{4a^{3}}=\frac {\pi}{2}\cdot \frac {1}{a\cdot a(a+a)} \end{align}
より、$a=b$でも成立する。

\begin{align} \sum_{n,m\ge0}^{ }\frac{1}{\left(n+\frac{1}{2}\right)\left(m+\frac{1}{2}\right)\left(n+m+1\right)}&=\sum_{n,m\ge0}^{ }\frac{8}{\left(2n+1\right)\left(2m+1\right)\left(2n+2m+2\right)} \newline &=\sum_{m=0}^{\infty}\left(\frac{8}{2m+1}\sum_{n=0}^{\infty}\frac{1}{\left(2n+1\right)\left(2n+2m+2\right)}\right) \newline &=\sum_{m=0}^{\infty}\left(\frac{8}{\left(2m+1\right)^{2}}\sum_{n=0}^{\infty}\left(\frac{1}{2n+1}-\frac{1}{2n+2m+2}\right)\right) \end{align}

\begin{align} &\sum_{m=0}^{\infty}\left(\frac{8}{\left(2m+1\right)^{2}}\sum_{n=0}^{\infty}\left(\frac{1}{2n+1}-\frac{1}{2n+2m+2}\right)\right) \newline &=\sum_{m=0}^{\infty}\frac{8}{\left(2m+1\right)^{2}}\sum_{n=0}^{\infty}\left(\int_{0}^{1}\left(u^{2n}-u^{2n+2m+1}\right)du\right) \newline &=\sum_{m=0}^{\infty}\frac{8}{\left(2m+1\right)^{2}}\int_{0}^{1}\frac{1-u^{2m+1}}{1-u^{2}}du \newline &=\int_{0}^{1}\frac{8\sum_{m=0}^{\infty}\frac{1}{\left(2m+1\right)^{2}}-8\sum_{m=0}^{\infty}\frac{u^{2m+1}}{\left(2m+1\right)^{2}}}{1-u^{2}}du \newline &=\int_{0}^{1}\left(\pi^{2}-8\sum_{m=0}^{\infty}\frac{u^{2m+1}}{\left(2m+1\right)^{2}}\right)\left(\frac{1}{2}\ln\left(\frac{1+u}{1-u}\right)\right)'du \newline &=4\int_{0}^{1}\ln\left(\frac{1+u}{1-u}\right)\sum_{m=0}^{\infty}\frac{u^{2m}}{2m+1}du \newline &=4\int_{0}^{1}\ln\left(\frac{1+u}{1-u}\right)\cdot\frac{1}{u}\int_{0}^{u}\frac{1}{1-t^{2}}dtdu \newline &=2\int_{0}^{1}\frac{\ln^{2}\left(\frac{1+u}{1-u}\right)}{u}du \end{align}

$$2\int_{0}^{1}\frac{\ln^{2}\left(\frac{1+u}{1-u}\right)}{u}du$$
について、$\frac {1-u}{1+u}=w$と置換すると、
$$u=\frac {1-w}{1+w} \Longrightarrow du=\frac {-2}{(1+w)^2}dw,\quad u:0\to1,w:1\to 0 $$
より、
\begin{align} &2\int_{0}^{1}\frac{\ln^{2}\left(\frac{1+u}{1-u}\right)}{u}du \newline &=2\int_{1}^{0}\frac {\ln^2{(1/w)}}{\frac {1-w}{1+w}}\frac {-2}{(1+w)^2}dw \newline &=4\int_{0}^{1}\frac {\ln^2{(w)}}{1-w^2}dw \newline &=4\sum_{n=0}^{\infty}\int_{0}^{1}\ln^2{(w)}w^{2n}dw \newline &=4\sum_{n=0}^{\infty}\frac {2}{(2n+1)^3} \newline &=8\left(\sum_{n=1}^{\infty}\frac 1 {n^3}-\sum_{n=1}^{\infty}\frac 1 {(2n)^3}\right) \newline &=8\cdot \frac {7}{8}\sum_{n=1}^{\infty}\frac 1 {n^3}=7\zeta(3) \end{align}

\begin{align} I&=\int_{0}^{1}\frac {x}{\tanh^{-1}(x)}dx \newline &= \frac 12 \int_{0}^{\infty}\tanh^{2}(1/x)dx\qquad \text{補題1より} \newline &= \int_{0}^{\infty}\sum_{n,m\ge 0}\frac {2x^2}{\left(1+x^2\pi^2\left(n+1/2\right)^2\right)\left(1+x^2\pi^2\left(m+1/2\right)^2\right)} \qquad \text{補題2より} \newline &=\sum_{n,m\ge 0}\int_{0}^{\infty}\frac {2x^2}{\left(1+x^2\pi^2\left(n+1/2\right)^2\right)\left(1+x^2\pi^2\left(m+1/2\right)^2\right)}dx \newline &=2\sum_{n,m\ge 0}\int_{0}^{\infty}\frac {du}{\left(u^2+\pi^2(n+1/2)^2\right)\left(u^2+\pi^2(m+1/2)^2\right)} \qquad (1/x=u\text {と置換}) \newline &=2\sum_{n,m\ge 0}\frac {\pi}{2}\frac {1}{\pi(n+1/2)\cdot \pi(m+1/2) \pi(n+m+1)} \quad (\text{補題3において、}a=\pi(n+1/2),b=\pi(m+1/2)\text{とする}) \newline &=\frac 1 {\pi^2}\sum_{n,m\ge0}^{ }\frac{1}{\left(n+\frac{1}{2}\right)\left(m+\frac{1}{2}\right)\left(n+m+1\right)} \newline &=\frac {1}{\pi^2}7\zeta(3)\qquad (\text{補題4より}) \newline &=\frac {7\zeta(3)}{\pi^2} \end{align}