TITLE: Evaluating $\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}$.
QUESTION [9 upvotes]: My attempt.
$$\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}$$
$$=\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k}\binom{2k}{k}-\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k\left(2k+1\right)}\binom{2k}{k}$$
The first sum can be evaluated easily if one uses the central binomial coefficient generating function
, the closed form is $2\zeta \left(2\right)$.

For the remaining sum consider the $\arcsin$ series expansion.
$$\sum _{k=0}^{\infty }\frac{x^{2k+1}}{4^k\left(2k+1\right)}\binom{2k}{k}=\arcsin \left(x\right)$$
$$\sum _{k=1}^{\infty }\frac{x^k}{4^k\left(2k+1\right)}\binom{2k}{k}=\frac{\arcsin \left(\sqrt{x}\right)}{\sqrt{x}}-1$$
$$-\sum _{k=1}^{\infty }\frac{1}{4^k\left(2k+1\right)}\binom{2k}{k}\int _0^1x^{k-1}\ln \left(1-x\right)\:dx=-\int _0^1\frac{\arcsin \left(\sqrt{x}\right)\ln \left(1-x\right)}{x\sqrt{x}}\:dx$$
$$+\int _0^1\frac{\ln \left(1-x\right)}{x}\:dx$$
$$\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k\left(2k+1\right)}\binom{2k}{k}=-2\int _0^1\frac{\arcsin \left(x\right)\ln \left(1-x^2\right)}{x^2}\:dx-\zeta \left(2\right)$$
But I got stuck with:
$$\int _0^1\frac{\arcsin \left(x\right)\ln \left(1-x^2\right)}{x^2}\:dx$$
Anything I try yields more complicated stuff, is there a way to calculate the main sum or the second one (or the integral) elegantly\in a simple manner?

REPLY [4 votes]: It seemed something was missing, so with the right tools the proof isn't difficult.
$$\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}$$

Consider:
$$\sum _{k=1}^{\infty }\frac{x^k}{4^k}H_k\binom{2k}{k}=\frac{2}{\sqrt{1-x}}\ln \left(\frac{1+\sqrt{1-x}}{2\sqrt{1-x}}\right)$$
$$\sum _{k=1}^{\infty }\frac{H_k}{4^k}\binom{2k}{k}\int _0^1x^{2k}\:dx=2\int _0^1\frac{\ln \left(1+\sqrt{1-x^2}\right)}{\sqrt{1-x^2}}\:dx-2\int _0^1\frac{\ln \left(\sqrt{1-x^2}\right)}{\sqrt{1-x^2}}\:dx$$
$$-2\ln \left(2\right)\int _0^1\frac{1}{\sqrt{1-x^2}}\:dx$$
$$=2\int _0^1\frac{\ln \left(1+x\right)}{\sqrt{1-x^2}}\:dx-2\int _0^1\frac{\ln \left(x\right)}{\sqrt{1-x^2}}\:dx-\pi \ln \left(2\right)$$

$$\int _0^1\frac{\ln \left(1+x\right)}{\sqrt{1-x^2}}\:dx=\frac{\pi }{2}\ln \left(2\right)-\int _0^1\frac{\arcsin \left(x\right)}{1+x}\:dx$$
$$=\frac{\pi }{2}\ln \left(2\right)-\int _0^{\frac{\pi }{2}}\frac{x\cos \left(x\right)}{1+\sin \left(x\right)}\:dx=\int _0^{\frac{\pi }{2}}\ln \left(1+\sin \left(x\right)\right)\:dx$$
$$=4\int _0^1\frac{\ln \left(1+t\right)}{1+t^2}\:dt-2\int _0^1\frac{\ln \left(1+t^2\right)}{1+t^2}\:dt$$
This means that:
$$\int _0^1\frac{\ln \left(1+x\right)}{\sqrt{1-x^2}}\:dx=-\frac{\pi }{2}\ln \left(2\right)+2G$$

Thus:
$$\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}=-\pi \ln \left(2\right)+4G$$

Bonus.
$$\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}=\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k}\binom{2k}{k}-\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k\left(2k+1\right)}\binom{2k}{k}$$
And so we find that:
$$\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k\left(2k+1\right)}\binom{2k}{k}=2\zeta \left(2\right)+2\pi \ln \left(2\right)-8G$$
And in the body of the question we had:
$$\int _0^1\frac{\ln \left(1-x^2\right)\arcsin \left(x\right)}{x^2}\:dx=-\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k\left(2k+1\right)}\binom{2k}{k}-\frac{1}{2}\zeta \left(2\right)$$
Hence:
$$\int _0^1\frac{\ln \left(1-x^2\right)\arcsin \left(x\right)}{x^2}\:dx=-\frac{3}{2}\zeta \left(2\right)-\pi \ln \left(2\right)+4G$$