$$ \gdef\Res #1{\underset{#1}{\mathrm{Res}}} \gdef\hi{\hat {\pi }} \gdef\ti{\tilde {\pi }} $$
#35. 今日の積分35 (2022/01/20)
便利メモ/数学/今日の積分  Share on Twitter

$$\lim _{r\to 0+}\left (\frac {\ln r}2+\int _{0}^{1-r}\frac {dx}{(1-x)(1+x^{2})}\right ) =\frac {\pi }8+\frac {\ln 2}4$$

$$\begin {aligned} \lim _{r\to 0+}\left (\frac {\ln r}2+\int _{0}^{1-r}\frac {dx}{(1-x)(1+x^{2})}\right ) &=\lim _{r\to 0+}\int _{0}^{1-r}\left (-\frac {1}{2(1-x)}+\frac {1}{(1-x)(1+x^{2})}\right )dx\\ &=\frac {1}2\int _{0}^{1}\frac {1}{1-x}\left (-1+\frac {2}{1+x^{2}}\right )dx\\ &=\frac {1}2\int _{0}^{1}\frac {1-x^{2}}{(1-x)(1+x^2)}dx\\ &=\frac {1}2\int _{0}^{1}\frac {1+x}{1+x^{2}}dx\\ &=\frac {1}2\left [\arctan x+\frac {1}2\ln (1+x^{2})\right ]_0^1\\ &=\frac {1}2\left (\frac {\pi }4+\frac {\ln 2}2\right )\\ &=\textcolor {blue}{\frac {\pi }8+\frac {\ln 2}4}. \end {aligned}$$