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
$$ \displaystyle \int_0^\infty \frac{\arctan x^2}{x^2} dx=\frac\pi{\sqrt2} $$
$$ \displaystyle \int_0^\infty \frac{\arctan^2x}{x^2} dx=\pi\log2 $$
[解説]
$ \begin{eqnarray*} &&\int_0^\infty \frac{\arctan x^2}{x^2}dx\\ &=&\frac12\int_0^\infty \frac{\arctan t}{t\sqrt t}dt~~~~~(t=x^2)\\ &=&-\left[\frac{\arctan t}{\sqrt t} \right]_0^\infty+\int_0^\infty \frac1{\sqrt t(1+t^2)}dt\\ &=&\int_0^{\frac\pi2} \frac1{\sqrt{\tan\theta}}d\theta~~~~~(t=\tan\theta)\\ &=&\int_0^{\frac\pi2}\sin^{-\frac12}\theta\cos^{\frac12}\theta d\theta\\ &=&\frac12B\left(\frac14,\frac34 \right)\\ &=&\frac12\Gamma\left(\frac14\right)\Gamma\left(\frac34\right)\\ &=&\frac\pi{2\sin\frac\pi4}\\ &=&\frac\pi{\sqrt2}\qed \end{eqnarray*} $
$ \begin{eqnarray*} &&\int_0^\infty \frac{\arctan^2x}{x^2}dx\\ &=&-\left[\frac{\arctan^2x}x \right]_0^\infty+2\int_0^\infty\frac{\arctan x}{x(1+x^2)}dx\\ &=&2\int_0^{\frac\pi2}\frac x{\tan x}dx\\ &=&2\left[x\log\sin x\right]_0^{\frac\pi2}-2\int_0^{\frac\pi2}\log\sin xdx\\ &=&-\int_0^{\frac\pi2}\log\sin xdx-\int_0^{\frac\pi2}\log\cos xdx\\ &=&-\int_0^{\frac\pi2}(\log\sin2x-\log2)dx\\ &=&\frac\pi2\log2-\frac12\int_0^\pi\log\sin xdx\\ &=&\frac\pi2\log2-\int_0^{\frac\pi2}\log\sin xdx\\ &=&\pi\log2\qed \end{eqnarray*} $
以上で2つの積分が証明されました。
