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
$$ \displaystyle \sum_{0\le n,k}\frac{(-1)^{n+k}}{(2n+1)(2k+1)(2n+2k+1)} $$
[解説]
$ \begin{eqnarray*} &&\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)}\i1x^{2n+2k}dx\\ &=&\i1\frac1{x^2}\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}x^{2n+1}\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}x^{2k+1}dx\\ &=&\i1\frac{\arctan^2x}{x^2}dx\\ &=&\i{\frac\pi4}\frac{t^2}{\tan^2t\cos^2t}dt~~~~~~~~~~(x=\tan t)\\ &=&\i{\frac\pi4}\frac{t^2}{\sin^2t}dt\\ &=&\left[-\frac{t^2}{\tan t} \right]_0^{\frac\pi4}+2\int_0^{\frac\pi4}\frac t{\tan t}dt\\ &=&-\frac{\pi^2}{16}+2\left[t\log\sin x \right]_0^{\frac\pi4}-2\int_0^{\frac\pi4}\log\sin xdx\\ &=&-\frac{\pi^2}{16}-\frac\pi4\log2+2\int_0^{\frac\pi4}\log\frac1{2\sin x}dx+\frac\pi2\log2\\ &=&\frac\pi4\log2-\frac{\pi^2}{16}+2\int_0^{\frac\pi4}\sum_{n=1}^\infty\frac1n\cos2nxdx\\ &=&\frac\pi4\log2-\frac{\pi^2}{16}+2\sum_{n=1}^\infty\frac1n\i{\frac\pi4}\cos2nxdx\\ &=&\frac\pi4\log2-\frac{\pi^2}{16}+\sum_{n=1}^\infty \frac{\sin\frac{\pi n}2}{n^2}\\ &=&\frac\pi4\log2-\frac{\pi^2}{16}+\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}\\ &=&\frac\pi4\log2-\frac{\pi^2}{16}+G \end{eqnarray*} $
よって、この問題の解答は$\d\frac\pi4\log2-\frac{\pi^2}{16}+G$となります。