積分解説24

2020/12/14にツイートした積分です。

https://twitter.com/sounansya_29/status/1338416268429250560?s=21

[解説]

$\begin{array}{rcl}& & {\int }_0^{\mathrm{\infty }}\frac{\arctan x^2}{x^2}dx\\ & =& \frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{\arctan t}{t\sqrt{t}}dt(t=x^2)\\ & =& -{\left[\frac{\arctan t}{\sqrt{t}}\right]}_0^{\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }}\frac{1}{\sqrt{t}(1+t^2)}dt\\ & =& {\int }_0^{\frac{\pi }{2}}\frac{1}{\sqrt{\tan \theta }}d\theta (t=\tan \theta )\\ & =& {\int }_0^{\frac{\pi }{2}}{\sin }^{-\frac{1}{2}}\theta {\cos }^{\frac{1}{2}}\theta d\theta \\ & =& \frac{1}{2}B(\frac{1}{4},\frac{3}{4})\\ & =& \frac{1}{2}\mathrm{\Gamma }\left(\frac{1}{4}\right)\mathrm{\Gamma }\left(\frac{3}{4}\right)\\ & =& \frac{\pi }{2\sin \frac{\pi }{4}}\\ & =& \frac{\pi }{\sqrt{2}}◻\end{array}$

$\begin{array}{rcl}& & {\int }_0^{\mathrm{\infty }}\frac{{\arctan }^{2}x}{x^2}dx\\ & =& -{\left[\frac{{\arctan }^{2}x}{x}\right]}_0^{\mathrm{\infty }}+2{\int }_0^{\mathrm{\infty }}\frac{\arctan x}{x(1+x^2)}dx\\ & =& 2{\int }_0^{\frac{\pi }{2}}\frac{x}{\tan x}dx\\ & =& 2{[x\log \sin x]}_0^{\frac{\pi }{2}}-2{\int }_0^{\frac{\pi }{2}}\log \sin xdx\\ & =& -{\int }_0^{\frac{\pi }{2}}\log \sin xdx-{\int }_0^{\frac{\pi }{2}}\log \cos xdx\\ & =& -{\int }_0^{\frac{\pi }{2}}(\log \sin 2x-\log 2)dx\\ & =& \frac{\pi }{2}\log 2-\frac{1}{2}{\int }_0^{\pi }\log \sin xdx\\ & =& \frac{\pi }{2}\log 2-{\int }_0^{\frac{\pi }{2}}\log \sin xdx\\ & =& \pi \log 2◻\end{array}$

以上で2つの積分が証明されました。