級数解説13

2020/08/26に るめなる 君が出題した問題です。

https://twitter.com/eturty12/status/1298590979885170690?s=21

[解説]

$\begin{array}{rcl}& & {\int }_0^1{\text{Li}}_2(x(1-x))dx\\ & =& {\int }_0^1\sum _{n=1}^{\mathrm{\infty }}\frac{x^n(1-x{)}^n}{n^2}dx\\ & =& \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}{\int }_0^1{x}^n(1-x{)}^{n}dx\\ & =& \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}B(n+1,n+1)\\ & =& \sum _{n=1}^{\mathrm{\infty }}\frac{{\mathrm{\Gamma }}^2(n+1)}{n^2\mathrm{\Gamma }(2n+2)}\\ & =& \sum _{n=1}^{\mathrm{\infty }}\frac{n{!}^2}{n^2(2n+1)(2n)!}\\ & =& \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2(\genfrac{}{}{0ex}{}{2n}{n})}{\int }_0^1{x}^{2n}dx\\ & =& {\int }_0^1\sum _{n=1}^{\mathrm{\infty }}\frac{x^{2n}}{n^2(\genfrac{}{}{0ex}{}{2n}{n})}dx\\ & =& 2{\int }_0^1{\arcsin }^2\frac{x}{2}dx\\ & =& 4{\int }_0^{\frac{1}{2}}{\arcsin }^{2}xdx\\ & =& 4{\int }_0^{\frac{\pi }{6}}{x}^2\cos xdx\\ & =& 4{[(x^2-2)\sin x+2x\cos x]}_0^{\frac{\pi }{6}}\\ & =& \frac{{\pi }^2}{18}+\frac{2\pi }{\sqrt{3}}-4\end{array}$

よって、この問題の解答は${\displaystyle \frac{{\pi }^2}{18}+\frac{2\pi }{\sqrt{3}}-4}$となります。