複素積分10

今回はこちらの積分botさんの積分を解説します。
https://twitter.com/integralsbot/status/1444190581983301635?s=21

$\displaystyle I(a):=\int_0^{\frac{\pi}{2}}\tan^{-1}\frac{a}{\sqrt{\tan x}}dx$
$=\displaystyle\Im\int_0^{\frac{\pi}{2}}\log\big(\sqrt{\tan x}+ia\big)dx$
ここで、$I(0)=0$
(これは余談ですが、実部も0になりますね。)
$\displaystyle\frac{\partial I(a)}{\partial a}=\Im\int_0^{\frac{\pi}{2}}\frac{i}{\sqrt{\tan x}+ia}dx$
$=\displaystyle\int_0^{\frac{\pi}{2}}\frac{\sqrt{\tan x}}{\tan x+a^2}dx$
$=\displaystyle\int_{-∞}^∞\frac{1}{x^2+a^2}\frac{x^2}{x^4+1}dx$
$=\displaystyle\Re\int_{-∞}^∞\frac{1}{x^2+a^2}\frac{1}{x^2+i}dx$
$=\displaystyle\Re2\pi i\Big(\left.\frac{1}{(z+ai)(z^2+i)}\right|_{z=ai}+\left.\frac{1}{(z^2+a^2)(z-e^{-\frac{\pi}{4}i})}\right|_{z=e^{\frac{3}{4}\pi i}}\Big)$
$=\displaystyle\pi\Re\big(\frac{1}{a}\frac{1}{i-a^2}+\frac{1}{a^2-i}e^{-\frac{\pi}{4}i}\big)$
$I(a)=\displaystyle\pi\Re\int\big(\frac{1}{a}\frac{1}{i-a^2}+\frac{1}{a^2-i}e^{-\frac{\pi}{4}i}\big)da$
$=\displaystyle\pi\Re-i\big(\log a-\log(a+e^{\frac{\pi}{4}i})\big)+C$
$=\displaystyle\pi\Re i\big(\log(a+e^{\frac{\pi}{4}i})\big)+C$
$\displaystyle I(0)=-\frac{\pi^2}{4}+C$
$\displaystyle\Leftrightarrow C=\frac{\pi^2}{4}$
$\displaystyle I(\sqrt2)=\pi\Re i\big(\log(\sqrt2+e^{\frac{\pi}{4}i})\big)+\frac{\pi^2}{4}$
$=\displaystyle-\pi\tan^{-1}\big(\frac{\frac{1}{\sqrt2}}{\sqrt2+\frac{1}{\sqrt2}}\big)+\frac{\pi^2}{4}$
$=\displaystyle\pi\tan^{-1}\frac{1}{2}\quad\big(\because\tan^{-1}\frac{1}{3}+\tan^{-1}\frac{1}{2}=\frac{\pi}{4}\big)$
今まで解けた積分の中で1番難しかったなという印象です。
手伝ってくれた数学人くんのmathlogはこちら↓
https://mathlog.info/users/1958/articles