How can I prove the following equality:
$$\sum_{n\geq 0}^\infty\frac{1}{4^n(2n+1)^3}\binom{2n}{n} = \frac{\pi^3}{48} + \frac{\pi\ln^2(2)}{4}$$
Reference: American Mathematical Monthly (June 30, 2018) -Problem 12051 - 06, proposed by P. Ribeiro (Portugal)
I tried to related with an integral but coudl not find
thank's for ideas
We have that $$\int_0^1x^{2n}\ln(x)\,dx=\left[x^{2n+1} \left(\frac{\ln(x)}{(2n+1)}-\frac{1}{(2n+1)^2}\right) \right]_0^1=-\frac{1}{(2n+1)^2}$$ and, for $x\in[0,1]$, $$\sum_{n=0}^{\infty}\binom{2n}{n}\frac{x^{2n}}{4^n(2n+1)}=\frac{\arcsin(x)}{x}.$$ Hence \begin{align*} \sum_{n=0}^{\infty}\binom{2n}{n}\frac{1}{4^n(2n+1)^3}&= -\int_0^1\frac{\arcsin(x)\ln(x)}{x}\,dx\\&= -\left[\arcsin(x)\,\frac{\ln^2(x)}{2}\right]_0^1 +\frac{1}{2}\int_0^1\frac{\ln^2(x)}{\sqrt{1-x^2}}\,dx\\ &=\frac{1}{2}\int_0^1\frac{\ln^2(x)}{\sqrt{1-x^2}}\,dx =\frac{1}{2}\int_0^{\pi/2}\ln^2(\sin(t))\,dt=\frac{\pi^3}{48}+\frac{\pi\ln^2(2)}{4} \end{align*} where the last integral is known: Integrate square of the log-sine integral: $\int_0^{\frac{\pi}{2}}\ln^{2}(\sin(x))dx$ .
