複素積分9

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/1447414790981230600?s=21

${\displaystyle {\int }_{-\infty }^{\infty }\frac{\sin r{\tan }^{-1}x}{(1+x^2{)}^{r/2}(e^{\pi x}+1)}dx}$
$={\displaystyle \mathrm{Im}{\int }_{-\infty }^{\infty }\frac{1}{(1-ix{)}^r(e^{\pi x}+1)}dx}$
$={\displaystyle \mathrm{Im}2\pi i\sum _{n\ge 0}\frac{1}{(1+(2n+1){)}^r(-\pi )}}$
$={\displaystyle -2^{1-r}\zeta (r)}$
$={\displaystyle \eta (r)-\zeta (r)}$