積分解説04

2020/08/06に出題した問題です。
https://twitter.com/sounansya_29/status/1291334440119984129?s=21

[解説]
$\begin{array}{rcl}& & {\int }_0^{\mathrm{\infty }}\frac{e^{-x}{\sin }^{2}x}{x}dx\\ & =& {\int }_0^{\mathrm{\infty }}\frac{e^{-x}}{x}{\int }_0^1\frac{\partial }{\partial t}{\sin }^{2}txdtdx\\ & =& {\int }_0^{\mathrm{\infty }}\frac{e^{-x}}{x}{\int }_0^{1}2x\sin tx\cos txdtdx\\ & =& {\int }_0^{\mathrm{\infty }}{e}^{-x}{\int }_0^1\sin 2txdtdx\\ & =& {\int }_0^1{\int }_0^{\mathrm{\infty }}{e}^{-x}\sin 2txdxdt\\ & =& {\int }_0^1\mathcal{L}[\sin 2tx](1)dt\\ & =& {{\int }_0^1\frac{2t}{s^2+4t^2}dt|}_{s=1}\\ & =& {\int }_0^1\frac{2t}{1+4t^2}dt\\ & =& \frac{1}{2}{\int }_0^{\arctan 2}\tan \theta d\theta & (t=\frac{1}{2}\tan \theta )\\ & =& -\frac{1}{2}{[\log \cos \theta ]}_0^{\arctan 2}\\ & =& -\frac{1}{2}\log \cos \arctan 2\\ & =& -\frac{1}{2}\log \frac{1}{\sqrt{1+2^2}}\\ & =& \frac{1}{4}\log 5\end{array}$
よって、この問題の解答は${\displaystyle \frac{1}{4}\log 5}$となります。