$$ \gdef\Res #1{\underset{#1}{\mathrm{Res}}} \gdef\hi{\hat {\pi }} \gdef\ti{\tilde {\pi }} $$
#36. 今日の積分36 (2022/01/21)
便利メモ/数学/今日の積分  Share on Twitter

$$\begin {aligned} \int _{0}^{1}\frac {\ln ^{2}x}{(1-x)(1+x^{2})}dx &=\frac {35}{32}\zeta (3)+\frac {\pi ^{3}}{32} \end {aligned}$$
$$\begin {aligned} \int _{0}^{1}\frac {\ln ^{2}x}{(1-x)(1+x^{2})}dx &=\frac {1}2\int _{0}^{1}\left (\frac {1}{1-x}+\frac {1+x}{1+x^{2}}\right )\ln ^{2}xdx\\ &=\zeta (3)+\Re \int _{0}^{1}\frac {(1-i)\ln ^{2}x}{x-i}dx\\ &=\zeta (3)-\Re \left \{(1-i)\operatorname{Li} _{3}(-i) \right \}\\ &=\zeta (3)-\Re \left \{ (1-i)\left (-\frac {3}{32}\zeta (3)-\frac {\pi ^{3}}{32}i\right )\right \} \\ &=\zeta (3)+\frac {3}{32}\zeta (3)+\frac {\pi ^{3}}{32}\\ &=\textcolor {blue}{\frac {35}{32}\zeta (3)+\frac {\pi ^{3}}{32}}. \end {aligned}$$