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.
 A: Inspired by this answer, we write \begin{align}S&=\sum_{k\ge1}\frac{\operatorname B(k,1/2)}{(2k+1)^2}=\sum_{k\ge1}\frac{4^k}{k(2k+1)^2{2k\choose k}}=\sum_{k\ge1}\frac{\int_0^{\pi/2}\sin^{2k+1}x\,dx}{k(2k+1)}.\end{align} Notice that \begin{align}\sum_{k\ge1}\frac{x^{2k+1}}{k(2k+1)}&=\int_0^x\sum_{k\ge1}\frac{t^{2k}}k\,dt=-\int_0^x\log(1-t^2)\,dt\\&=2x-x\log(1-x^2)-2\operatorname{arctanh}x\end{align} through $\log(1-t^2)=\log(1-t)+\log(1+t)$, from which \begin{align}S&=\int_0^{\pi/2}\left(2\sin x-2\sin x\log\cos x-2\operatorname{arctanh}\sin x\right)\,dx\\&=2-2\int_0^1\log u\,du-2\int_0^{\pi/2}\log(\tan x+\sec x)\,dx\\&=4-4G.\end{align}
A: 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.
A: Consider
$$F(x)=\sum_{k=1}^{\infty}\frac{B(k, \frac{1}{2})}{(2k+1)^2}x^{2k+1}=\sum_{k=1}^{\infty}\frac{\sqrt{\pi }}{(2k+1)^2}\frac{ \Gamma (k)}{\Gamma \left(k+\frac{1}{2}\right)} x^{2 k+1}$$
$$F'(x)=\sum_{k=1}^{\infty}\frac{\sqrt{\pi }}{2k+1}\frac{ \Gamma (k)}{\Gamma \left(k+\frac{1}{2}\right)} x^{2 k}=2+\frac{2 x \sin ^{-1}(x)}{\sqrt{1-x^2}}-\frac{2 \sin ^{-1}(x)}{x \sqrt{1-x^2}}$$
$$F(x)=\int F'(x)\,dx=4x-2 i \text{Li}_2\left(-e^{i \sin ^{-1}(x)}\right)+2 i \text{Li}_2\left(e^{i \sin
   ^{-1}(x)}\right)-2 \sqrt{1-x^2} \sin ^{-1}(x)+$$ $$4 \sin ^{-1}(x) \tanh
   ^{-1}\left(e^{i \sin ^{-1}(x)}\right)$$
$$\int_0^1 F'(x)\,dx=4(1- C)$$
