指数関数を分母と分子に含む対数の定積分

\begin{align*} \int_{0}^{\infty}\log\left(\frac{e^{x}+1}{e^{x}-1}\right)dx & =\int_{0}^{\infty}\log\left(\frac{1+e^{-x}}{1-e^{-x}}\right)dx\\ & =\int_{0}^{\infty}\log\left(1+e^{-x}\right)dx-\int_{0}^{\infty}\log\left(1-e^{-x}\right)dx\\ & =\int_{0}^{\infty}\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}\left(e^{-x}\right)^{k}}{k}dx-\int_{0}^{\infty}\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}\left(-e^{-x}\right)^{k}}{k}dx\cmt{\because-1<x\leq1\rightarrow\log\left(1+x\right)=\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}x^{k}}{k}}\\ & =\int_{0}^{\infty}\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}e^{-kx}}{k}dx+\int_{0}^{\infty}\sum_{k=1}^{\infty}\frac{e^{-kx}}{k}dx\\ & =\int_{0}^{\infty}\sum_{k=1}^{\infty}\frac{\left(\left(-1\right)^{k+1}+1\right)e^{-kx}}{k}dx\\ & =\sum_{k=1}^{\infty}\frac{\left(\left(-1\right)^{k+1}+1\right)}{k}\int_{0}^{\infty}e^{-kx}dx\\ & =\sum_{k=1}^{\infty}\frac{\left(\left(-1\right)^{k+1}+1\right)}{k}\left[\frac{e^{-kx}}{-k}\right]_{0}^{\infty}\\ & =\sum_{k=1}^{\infty}\frac{\left(\left(-1\right)^{k+1}+1\right)}{k}\cdot\frac{1}{k}\\ & =\sum_{k=1}^{\infty}\frac{\left(\left(-1\right)^{k+1}+1\right)}{k^{2}}\\ & =\sum_{k=1}^{\infty}\frac{\left(\left(-1\right)^{\left(2k-1\right)+1}+1\right)}{\left(2k-1\right)^{2}}+\sum_{k=1}^{\infty}\frac{\left(\left(-1\right)^{2k+1}+1\right)}{\left(2k\right)^{2}}\\ & =2\sum_{k=1}^{\infty}\frac{1}{\left(2k-1\right)^{2}}\\ & =2\left(\zeta\left(2\right)-\frac{1}{2^{2}}\zeta\left(2\right)\right)\\ & =2\left(1-\frac{1}{2^{2}}\right)\zeta\left(2\right)\\ & =\frac{3}{2}\zeta\left(2\right)\\ & =\frac{\pi^{2}}{4} \end{align*}