積分解説01

2020/06/25に出題した問題です。
https://twitter.com/sounansya_29/status/1268788501606133760?s=21

[解説]
$\begin{array}{rcl}& & {\int }_0^{\mathrm{\infty }}\frac{\sin x}{\sinh x}dx\\ & =& 2{\int }_0^{\mathrm{\infty }}\frac{\sin x}{e^x-e^{-x}}dx\\ & =& 2{\int }_0^{\mathrm{\infty }}{e}^{-x}\sin x\sum _{k=0}^{\mathrm{\infty }}{e}^{-2kx}dx\\ & =& 2\sum _{k=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-(2k+1)x}\sin xdx\\ & =& 2\sum _{k=0}^{\mathrm{\infty }}\mathcal{L}[\sin x](2k+1)\\ & =& 2{\sum _{k=0}^{\mathrm{\infty }}\frac{1}{s^2+1}|}_{s=2k+1}\\ & =& \frac{1}{2}\sum _{k=0}^{\mathrm{\infty }}\frac{1}{(k+\frac{1}{2}{)}^2+\frac{1}{4}}\\ & =& \frac{\pi \tan \frac{\pi }{2}i}{2i}\\ & =& \frac{\pi }{2}\tanh \frac{\pi }{2}\end{array}$
よって、この問題の解答は${\displaystyle \frac{\pi }{2}\tanh \frac{\pi }{2}}$となります。