分母に中央二項係数がある級数メモ

\begin{align} f(x)&=\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^3\binom{2n}n} \\ f’(x)&=\frac2x\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2\binom{2n}n} \\&=\frac{4}{x}\arcsin^2x \\f(1)&=\int_{0}^{1}f’(x)dx \\&=4\int_{0}^{1}\frac{\arcsin^2x}{x}dx \\&=4\int_{0}^{\frac{\pi}2}\frac{x^2}{\tan x}dx \\&=4\Big[x^2\log\sin x\Big]_0^{\frac{\pi}2}-8\int_{0}^{\frac{\pi}2}x\log\sin xdx \\&=-8\int_{0}^{\frac{\pi}2}x\log\sin xdx \\&=8\log2\int_{0}^{\frac{\pi}2}xdx+8\sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{\frac{\pi}2}x\cos 2nx dx \\&=\pi^2\log2-4\sum_{n=0}^{\infty}\frac1{(2n+1)^3} \\&=\pi^2\log2-\frac72\zeta(3) \end{align}

\begin{align} f(x)&=\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^3\binom{2n}n} \\ f’(x)&=\frac2x\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2\binom{2n}n} \\&=\frac{4}{x}\arcsin^2x \\f\left(\frac{i}2\right)&=4\int_{0}^{\frac{i}2}\frac{\arcsin^2x}{x}dx \\&=-4\int_{0}^{\frac12}\frac{(\sinh^{-1}x)^2}{x}dx \\&=-4\int_{0}^{\sinh^{-1}\frac12}\frac{x^2}{\tanh x}dx \\&=4\left[\frac12\mathrm{Li}_3(e^{2x})-x\mathrm{Li}_2(e^{2x})+\frac{x^3}3-x^2\log(1-e^{2x})\right]_0^{\sinh^{-1}\frac12} \\&=2\mathrm{Li}_3(\phi^2)-4\log\phi\mathrm{Li}_2(\phi^2)+\frac43\log^3\phi-4\log^3\phi-4i\pi\log\phi-2\mathrm{Li}_3(1) \\&=-\frac25\zeta(3) \end{align}