π^3

僕も$\pi^3$を含む形になる級数を求めたい！！！

\begin{align} \frac{\pi^3}{8}&=\arcsin^31\\&=6\int_{0< t_1< t_2< t_3<1}\frac{dt_1}{\sqrt{1-t_1^2}}\frac{dt_2}{\sqrt{1-t_2^2}}\frac{dt_3}{\sqrt{1-t_3^2}}\\&=6\int_0^1\frac{dt_3}{\sqrt{1-t_3^2}}\int_0^{t_3}\frac{dt_2}{\sqrt{1-t_2^2}}\int_0^{t_2}\frac{dt_1}{\sqrt{1-t_1^2}}\\&=6\sum_{n=0}^\infty\frac{\binom{2n}{n}}{2^{2n}(2n+1)}\int_0^1\frac{dt_3}{\sqrt{1-t_3^2}}\int_0^{t_3}\frac{t_2^{2n+1}}{\sqrt{1-t_2^2}}dt_2\\&=24\sum_{n=0}^\infty\frac{\binom{2n}{n}}{\binom{2n+2}{n+1}(2n+1)(2n+2)}\sum_{m=n+1}^\infty\frac{\binom{2m}{m}}{2^{2m}}\int_0^1\frac{t_3^{2m}\sqrt{1-t_3^2}}{\sqrt{1-t_3^2}}dt_3\\&=6\sum_{0< n_1\le n_2}\frac{\binom{2n_2}{n_2}}{2^{2n_2}(2n_1-1)^2(2n_2+1)} \end{align}

$\displaystyle \frac{\pi^3}{192}=\sum_{0< n_1< n_2}\frac{\binom{2n_2-2}{n_2-1}}{2^{2n_2}(2n_1-1)^2(2n_2-1)}$

おまけ

$\displaystyle \frac{1}{2}\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}=\left(\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\right)^3$