TITLE: How to show $\sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} = 4 \int_{0}^{\frac{1}{2}} \frac{\arcsin^{2}(x)}{x} \ dx$?
QUESTION [5 upvotes]: $$\sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} = 4 \int_{0}^{\frac{1}{2}} \frac{\arcsin^{2}(x)}{x} \ dx.$$
Someone please show that this equation is correct !?

REPLY [4 votes]: It is known that 
$$
\arcsin(x)^2=\frac{1}{2}\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^2\binom{2n}{n}}
$$
see Companion to Concrete Mathematics - Mathematical Techniques and Various Applications Z. A. Melzak p. 108
Then
$$
\begin{aligned}
\int_{0}^{1/2}\frac{4\arcsin(x)^2}{x}
&=\int_{0}^{1/2}\frac{2}{x}\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^2\binom{2n}{n}}dx\\
&=\sum_{n=1}^\infty\frac{2^{2n+1}}{n^2\binom{2n}{n}} \int_{0}^{1/2}x^{2n-1}dx\\
&=\sum_{n=1}^\infty\frac{2^{2n+1}}{n^2\binom{2n}{n}} \frac{\left(\frac{1}{2}\right)^{2n}}{2n}\\
&=\sum_{n=1}^\infty\frac{1}{n^3\binom{2n}{n}}
\end{aligned}
$$