級数解説06

$ \begin{eqnarray*} &&\d\int_0^1\frac{x^{n-1}}{\sqrt{1-x}}dx\\ &=&B\l n,\frac12 \r\\ &=&\frac{\Gamma(n)\Gamma\l\frac12 \r}{\Gamma\l n+\frac12\r}\\ &=&\frac{2^nn!}{n(2n-1)!!}\\ &=&\frac{2^{2n}n!^2}{n(2n)!}\\ &=&\frac{2^{2n}}{n\binom{2n}n}\qed \end{eqnarray*} $

$ \begin{eqnarray*} &&\sum_{n=1}^\infty\frac{2^{4n}}{n^3\binom{2n}n^2}\\ &=&\sum_{n=1}^\infty\frac{2^{2n}}{n^2\binom{2n}n}\int_0^1\frac{x^{n-1}}{\sqrt{1-x}}dx\\ &=&\int_0^1\frac1{x\sqrt{1-x}}\sum_{n=1}^\infty\frac{2^{2n}x^n}{n^2\binom{2n}n}dx\\ &=&2\int_0^1\frac{\arcsin^2\sqrt x}{x\sqrt{1-x}}dx\\ &=&4\int_0^{\frac\pi2}\frac{t^2}{\sin^2 t\cos t}\sin t\cos tdt~~~~~~~~~~(x=\sin^2 t)\\ &=&4\int_0^{\frac\pi2}\frac{t^2}{\sin t}dt\\ &=&4\left[t^2\log\tan \frac t2 \right]_0^{\frac\pi2}-8\int_0^{\frac\pi2}t\log\tan\frac t2dt\\ &=&16\int_0^{\frac\pi2}t\sum_{n=0}^\infty\frac{\cos(2n+1)t}{2n+1} dt\\ &=&16\sum_{n=0}^\infty\frac1{2n+1}\i{\frac\pi2}t\cos(2n+1)tdt\\ &=&8\sum_{n=0}^\infty\frac{\pi(-1)^n(2n+1)-2}{(2n+1)^3}\\ &=&8\pi\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}-16\sum_{n=0}^\infty\frac1{(2n+1)^3}\\ &=&8\pi G-14\z(3)\qed \end{eqnarray*} $