積分解説18

2020/11/20に 白茶 さんが出題した問題です。

https://mathlog.info/articles/797

[解説]

$\begin{array}{rcl}& & {\int }_0^{\mathrm{\infty }}\frac{x(x^2-1)}{(1+x^2{)}^3\log x}dx\\ & =& {\int }_0^{\mathrm{\infty }}\frac{x}{(1+x^2{)}^3}{\int }_0^2{x}^{t}dtdx\\ & =& {\int }_0^2{\int }_0^{\mathrm{\infty }}\frac{x^{t+1}}{(1+x^2{)}^3}dxdt\\ & =& {\int }_0^2{\int }_0^{\frac{\pi }{2}}{\cos }^4\theta {\tan }^{t+1}\theta d\theta dt(x=\tan \theta )\\ & =& {\int }_0^2{\int }_0^{\mathrm{\infty }}{\sin }^{1+t}\theta {\cos }^{3-t}\theta d\theta dt\\ & =& \frac{1}{2}{\int }_0^{2}B(\frac{1}{2}t+1,2-\frac{1}{2}t)dt\\ & =& -\frac{1}{8}{\int }_0^{2}t(2-t)\mathrm{\Gamma }(\frac{1}{2}t+1)\mathrm{\Gamma }(-\frac{1}{2}t)dt\\ & =& \frac{\pi }{8}{\int }_0^2\frac{t(2-t)}{\sin \frac{\pi }{2}t}dt\\ & =& \frac{1}{{\pi }^2}{\int }_0^{\pi }\frac{t(\pi -t)}{\sin t}dt\\ & =& \frac{2}{{\pi }^2}{\int }_0^{\pi }t(\pi -t)\sum _{n=1}^{\mathrm{\infty }}{\sin }^2\frac{\pi n}{2}\sin ntdt\\ & =& \frac{2}{{\pi }^2}\sum _{n=1}^{\mathrm{\infty }}{\sin }^2\frac{\pi n}{2}\frac{2-2\cos \pi n}{n^3}\\ & =& \frac{8}{{\pi }^2}\sum _{n=1}^{\mathrm{\infty }}\frac{{\sin }^4\frac{\pi n}{2}}{n^3}\\ & =& \frac{8}{{\pi }^2}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1{)}^3}\\ & =& \frac{7\zeta (3)}{{\pi }^2}\end{array}$

よって、この問題の解答は${\displaystyle \frac{7\zeta (3)}{{\pi }^2}}$となります。