How to evaluate $\sum_{k=1}^{\infty}\frac{B\left(k, \frac{1}{2}\right)}{(2k+1)^2}$, where $B(x, y)$ is the Beta function? I am trying to evaluate this sum:
$$\sum_{k=1}^{\infty}\frac{B\left(k, \frac{1}{2}\right)}{(2k+1)^2}$$ where $B(x, y)$ is the Beta function.
Checking with WolframAlpha gives beautiful result: $4-4G$ where $G$ is the Catalan constant.
I tried to use the integral definition of Beta function:
$$B\left(k, \frac{1}{2}\right)=\int_{0}^{1}x^{k-1}(1-x)^{\frac{-1}{2}}dx,$$ so $$\sum_{k=1}^{\infty}\frac{B(k, \frac{1}{2})}{(2k+1)^2}=\sum_{k=1}^{\infty}\frac{1}{(2k+1)^2}\int_{0}^{1}x^{k-1}(1-x)^{\frac{-1}{2}}dx.$$ After changing the order of summation and integration: $$\int_{0}^{1}\frac{1}{\sqrt{1-x}}dx\sum_{k=1}^{\infty}\frac{x^{k-1}}{(2k+1)^2}.$$ And getting stuck, because the latter summation will lead to Lerch transcendent: $$\int_{0}^{1}\frac{\Phi\left(x,2,\frac{3}{2}\right)}{4\sqrt{1-x}}dx.$$ And I don't know how to progress further.

Let's denote
$$I=\int_{0}^{1}\frac{1}{\sqrt{1-x}}\sum_{k=1}^{\infty}\frac{x^{k-1}}{(2k+1)^2}\,dx$$
Using $\frac{1}{(2k+1)^2}=-\int_0^1t^{2k}\ln t dt$
$$I=-\int_{0}^{1}\frac{dx}{\sqrt{1-x}}\sum_{k=1}^{\infty}x^{k-1}\int_0^1t^{2k}\ln t dt$$
Changing the order of summation and integration and performing summation first hand
$$=-\int_{0}^{1}\frac{dx}{\sqrt{1-x}}\int_0^1\frac{t^2\ln t}{1-xt^2}dt$$
Making the substitutions $x\to 1-x$ and $\,s^2=x$
$$=-\int_{0}^{1}\frac{dx}{\sqrt x}\int_0^1\frac{t^2\ln t}{1-(1-x)t^2}dt=-2\int_0^1ds\int_0^1\frac{t^2\ln t}{1-(1-s^2)t^2}dt$$
Changing the order of integration and integrating over $s$
$$=-2\int_0^1\frac{t^2\ln t}{1-t^2}dt\int_0^1\frac{ds}{1+\frac{t^2}{1-t^2}s^2}=-2\int_0^1\frac{t\ln t}{\sqrt{1-t^2}}\arctan\frac{t}{\sqrt{1-t^2}}\,dt$$
Making the substitution $t=\sin x$ and integrating by part
$$=-2\int_0^\frac{\pi}{2}x\,\ln(\sin x)\,\sin x\,dx=2\cos x\ln(\sin x)\,\Big|_0^\frac{\pi}{2}-2\int_0^\frac{\pi}{2}\Big(\ln(\sin x)+\frac{x\,\cos x}{\sin x}\Big)\cos x\,dx$$
$$I=2-2\int_0^\frac{\pi}{2}\frac{x}{\sin x}(1-\sin^2x)\,dx=4-2\int_0^\frac{\pi}{2}\frac{x}{\sin x}dx$$
To evaluate the last integral we note that $\frac{dx}{\sin x}=d\big(\ln\tan\frac{x}{2}\big)$. Integrating by part
$$J=\int_0^\frac{\pi}{2}\frac{x}{\sin x}dx=x\,\ln\tan\frac{x}{2}\,\Big|_0^\frac{\pi}{2}-\int_0^\frac{\pi}{2}\ln\tan\frac{x}{2}\,dx=-2\int_0^\frac{\pi}{4}\ln\tan x\,dx$$
Making the substitution $x=\arctan t$
$$J=-2\int_0^1\frac{\ln t}{1+t^2}dt=2\,G$$
Taking all together,
$$\boxed{\,\,I=4-2\int_0^\frac{\pi}{2}\frac{x}{\sin x}dx=4-4\,G\,\,}$$
where $G$ denotes Catalan's constant.
