積分解説09

2020/11/08に出題した問題です。
https://mathlog.info/articles/297

[解説]
$\begin{array}{rcl}& & {\int }_0^1\frac{\log \left(\frac{1+x}{1-x}\right)}{x}dx\\ & =& 2{\int }_0^1\frac{\text{arctanh}x}{x}dx\\ & =& 2{\int }_0^1\sum _{n=0}^{\mathrm{\infty }}\frac{1}{2n+1}{x}^{2n}dx\\ & =& 2\sum _{n=0}^{\mathrm{\infty }}\frac{1}{2n+1}{\left[\frac{1}{2n+1}{x}^{2n+1}\right]}_0^1\\ & =& 2\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1{)}^2}\\ & =& 2\sum _{n=0}^{\mathrm{\infty }}(\frac{1}{n^2}-\frac{1}{4n^2})\\ & =& \frac{3}{2}\zeta (2)\\ & =& \frac{{\pi }^2}{4}\end{array}$
よって、この問題の解答は${\displaystyle \frac{{\pi }^2}{4}}$となります。