級数解説06

$\begin{array}{rcl}& & {\displaystyle {\int }_0^1\frac{x^{n-1}}{\sqrt{1-x}}dx}\\ & =& B(n,\frac{1}{2})\\ & =& \frac{\mathrm{\Gamma }(n)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(n+\frac{1}{2})}\\ & =& \frac{2^{n}n!}{n(2n-1)!!}\\ & =& \frac{2^{2n}n{!}^2}{n(2n)!}\\ & =& \frac{2^{2n}}{n(\genfrac{}{}{0ex}{}{2n}{n})}◻\end{array}$

$\begin{array}{rcl}& & \sum _{n=1}^{\mathrm{\infty }}\frac{2^{4n}}{n^3{(\genfrac{}{}{0ex}{}{2n}{n})}^2}\\ & =& \sum _{n=1}^{\mathrm{\infty }}\frac{2^{2n}}{n^2(\genfrac{}{}{0ex}{}{2n}{n})}{\int }_0^1\frac{x^{n-1}}{\sqrt{1-x}}dx\\ & =& {\int }_0^1\frac{1}{x\sqrt{1-x}}\sum _{n=1}^{\mathrm{\infty }}\frac{2^{2n}{x}^n}{n^2(\genfrac{}{}{0ex}{}{2n}{n})}dx\\ & =& 2{\int }_0^1\frac{{\arcsin }^2\sqrt{x}}{x\sqrt{1-x}}dx\\ & =& 4{\int }_0^{\frac{\pi }{2}}\frac{t^2}{{\sin }^{2}t\cos t}\sin t\cos tdt(x={\sin }^{2}t)\\ & =& 4{\int }_0^{\frac{\pi }{2}}\frac{t^2}{\sin t}dt\\ & =& 4{[t^2\log \tan \frac{t}{2}]}_0^{\frac{\pi }{2}}-8{\int }_0^{\frac{\pi }{2}}t\log \tan \frac{t}{2}dt\\ & =& 16{\int }_0^{\frac{\pi }{2}}t\sum _{n=0}^{\mathrm{\infty }}\frac{\cos (2n+1)t}{2n+1}dt\\ & =& 16\sum _{n=0}^{\mathrm{\infty }}\frac{1}{2n+1}{\int }_0^{\frac{\pi }{2}}t\cos (2n+1)tdt\\ & =& 8\sum _{n=0}^{\mathrm{\infty }}\frac{\pi (-1{)}^n(2n+1)-2}{(2n+1{)}^3}\\ & =& 8\pi \sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n+1{)}^2}-16\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1{)}^3}\\ & =& 8\pi G-14\zeta (3)◻\end{array}$