級数解説07

2020/12/03に出題した問題です。

https://twitter.com/sounansya_29/status/1314883373463031808?s=21

[解説]

$\begin{array}{rcl}& & \sum _{0\le n,k}\frac{(-1{)}^{n+k}}{(2n+1)(2k+1)(2n+2k+1)}\\ & =& \sum _{0\le n,k}\frac{(-1{)}^{n+k}}{(2n+1)(2k+1)}{\int }_0^1{x}^{2n+2k}dx\\ & =& {\int }_0^1\frac{1}{x^2}\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{2n+1}{x}^{2n+1}\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k}{2k+1}{x}^{2k+1}dx\\ & =& {\int }_0^1\frac{{\arctan }^{2}x}{x^2}dx\\ & =& {\int }_0^{\frac{\pi }{4}}\frac{t^2}{{\tan }^{2}t{\cos }^{2}t}dt(x=\tan t)\\ & =& {\int }_0^{\frac{\pi }{4}}\frac{t^2}{{\sin }^{2}t}dt\\ & =& {[-\frac{t^2}{\tan t}]}_0^{\frac{\pi }{4}}+2{\int }_0^{\frac{\pi }{4}}\frac{t}{\tan t}dt\\ & =& -\frac{{\pi }^2}{16}+2{[t\log \sin x]}_0^{\frac{\pi }{4}}-2{\int }_0^{\frac{\pi }{4}}\log \sin xdx\\ & =& -\frac{{\pi }^2}{16}-\frac{\pi }{4}\log 2+2{\int }_0^{\frac{\pi }{4}}\log \frac{1}{2\sin x}dx+\frac{\pi }{2}\log 2\\ & =& \frac{\pi }{4}\log 2-\frac{{\pi }^2}{16}+2{\int }_0^{\frac{\pi }{4}}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}\cos 2nxdx\\ & =& \frac{\pi }{4}\log 2-\frac{{\pi }^2}{16}+2\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}{\int }_0^{\frac{\pi }{4}}\cos 2nxdx\\ & =& \frac{\pi }{4}\log 2-\frac{{\pi }^2}{16}+\sum _{n=1}^{\mathrm{\infty }}\frac{\sin \frac{\pi n}{2}}{n^2}\\ & =& \frac{\pi }{4}\log 2-\frac{{\pi }^2}{16}+\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n+1{)}^2}\\ & =& \frac{\pi }{4}\log 2-\frac{{\pi }^2}{16}+G\end{array}$

よって、この問題の解答は${\displaystyle \frac{\pi }{4}\log 2-\frac{{\pi }^2}{16}+G}$となります。