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

${\displaystyle I(a):={\int }_0^{\frac{\pi }{2}}{\tan }^{-1}\frac{a}{\sqrt{\tan x}}dx}$
$={\displaystyle \mathrm{Im}{\int }_0^{\frac{\pi }{2}}\log (\sqrt{\tan x}+ia)dx}$
ここで、$I(0)=0$
(これは余談ですが、実部も0になりますね。)
${\displaystyle \frac{\partial I(a)}{\partial a}=\mathrm{Im}{\int }_0^{\frac{\pi }{2}}\frac{i}{\sqrt{\tan x}+ia}dx}$
$={\displaystyle {\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\tan x}}{\tan x+a^2}dx}$
$={\displaystyle {\int }_{-\infty }^{\infty }\frac{1}{x^2+a^2}\frac{x^2}{x^4+1}dx}$
$={\displaystyle \mathrm{Re}{\int }_{-\infty }^{\infty }\frac{1}{x^2+a^2}\frac{1}{x^2+i}dx}$
$={\displaystyle \mathrm{Re}2\pi i({\frac{1}{(z+ai)(z^2+i)}|}_{z=ai}+{\frac{1}{(z^2+a^2)(z-e^{-\frac{\pi }{4}i})}|}_{z=e^{\frac{3}{4}\pi i}})}$
$={\displaystyle \pi \mathrm{Re}(\frac{1}{a}\frac{1}{i-a^2}+\frac{1}{a^2-i}{e}^{-\frac{\pi }{4}i})}$
$I(a)={\displaystyle \pi \mathrm{Re}\int (\frac{1}{a}\frac{1}{i-a^2}+\frac{1}{a^2-i}{e}^{-\frac{\pi }{4}i})da}$
$={\displaystyle \pi \mathrm{Re}-i(\log a-\log (a+e^{\frac{\pi }{4}i}))+C}$
$={\displaystyle \pi \mathrm{Re}i(\log (a+e^{\frac{\pi }{4}i}))+C}$
${\displaystyle I(0)=-\frac{{\pi }^2}{4}+C}$
${\displaystyle \iff C=\frac{{\pi }^2}{4}}$
${\displaystyle I(\sqrt{2})=\pi \mathrm{Re}i(\log (\sqrt{2}+e^{\frac{\pi }{4}i}))+\frac{{\pi }^2}{4}}$
$={\displaystyle -\pi {\tan }^{-1}(\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}+\frac{1}{\sqrt{2}}})+\frac{{\pi }^2}{4}}$
$={\displaystyle \pi {\tan }^{-1}\frac{1}{2}(\because {\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{1}{2}=\frac{\pi }{4})}$
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