級数•積分botの積分を示してみる①

示す積分をIとします。
$$\begin{array}{rl}I& ={\displaystyle {\int }_0^{\mathrm{\infty }}x(\log (1-e^{-2x})-\log (1+e^{-2x}))dx}\\ & =-{\displaystyle {\int }_0^{\mathrm{\infty }}x(\sum _{n=1}^{\mathrm{\infty }}\frac{e^{-2nx}}{n}-\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n{e}^{-2nx}}{n})dx}\\ & =-2{\displaystyle {\int }_0^{\mathrm{\infty }}x{\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{e^{-2(2n+1)x}}{2n+1}dx}}\\ & =-2{\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{1}{2n+1}{\displaystyle {\int }_0^{\mathrm{\infty }}x{e}^{-2(2n+1)x}dx}}\\ & =-{\displaystyle \frac{1}{2}}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1{)}^3}}\\ & =-{\displaystyle \frac{1}{2}}\left({\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^3}-{\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{(2n{)}^3}}}\right)\\ & =-{\displaystyle \frac{7}{16}}\zeta (3)\end{array}$$