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
$$ \displaystyle \int_0^\infty \frac{x(x^2-1)}{(1+x^2)^3\log x}dx $$
[解説]
$ \begin{eqnarray*} &&\int_0^\infty \frac{x(x^2-1)}{(1+x^2)^3\log x}dx\\ &=&\int_0^\infty \frac x{(1+x^2)^3}\int_0^2 x^t dtdx\\ &=&\int_0^2\int_0^\infty\frac{x^{t+1}}{(1+x^2)^3}dxdt\\ &=&\int_0^2\int_0^{\frac\pi2}\cos^4\theta\tan^{t+1}\theta d\theta dt~~~~~~~~~~(x=\tan\theta)\\ &=&\int_0^2\int_0^\infty \sin^{1+t}\theta\cos^{3-t}\theta d\theta dt\\ &=&\frac12\int_0^2 B\l\frac12t+1,2-\frac12t \r dt\\ &=&-\frac18\int_0^2 t(2-t)\Gamma\l\frac12t+1 \r\Gamma\l-\frac12t \r dt\\ &=&\frac\pi8\int_0^2 \frac{t(2-t)}{\sin\frac{\pi}2t}dt\\ &=&\frac1{\pi^2}\int_0^\pi \frac{t(\pi-t)}{\sin t}dt\\ &=&\frac2{\pi^2}\int_0^\pi t(\pi-t)\sum_{n=1}^\infty \sin^2\frac{\pi n}2\sin ntdt\\ &=&\frac2{\pi^2}\sum_{n=1}^\infty \sin^2\frac{\pi n}2\frac{2-2\cos\pi n}{n^3}\\ &=&\frac8{\pi^2}\sum_{n=1}^\infty \frac{\sin^4\frac{\pi n}2}{n^3}\\ &=&\frac8{\pi^2}\sum_{n=0}^\infty \frac1{(2n+1)^3}\\ &=&\frac{7\z(3)}{\pi^2} \end{eqnarray*} $
よって、この問題の解答は$\d\frac{7\z(3)}{\pi^2}$となります。