No more hockey stick

I'm sharing my solution for proposed problem

Solution by proposer

Recall the ordinary generating function for binomial theorem, ie $\displaystyle B(j)= \sum_{k=0}^{n}{n\choose k}x^k =(1+x)^n$ . Now we perform jth times derivatives w.r.t x and multiply throughly by $\frac{1}{j!}$ we yields at $x=1$ $\frac{1}{j!}\frac{d^j}{dx^j}B(j) = \sum_{k=0}^{\infty}{n\choose k}(k)_jx^{k-j}= (n)_j (1+x)^{n-j} \Rightarrow \sum_{k=j}^n{n\choose k}{k\choose j} = 2^{n-j} {n\choose j}$ where $(n)_j$ is pochhammer polynomial and hence we have $\sum_{n=0}^{\infty}\sum_{j=0}^{n}{ n\choose j} \frac{d^j}{dx^j}\frac{2^j B(J)}{j!8^ n(2n+1)^3}=\sum_{n=0}^{\infty}\sum_{j=0}^{n}\frac{1}{4^n}{ n\choose j}^2\frac{1}{(2n+1)^3} =\sum_{n=0}^{\infty}\frac{1}{4^n(2n+1)^3}{2n\choose n}$ since by Vandermonde Convolution $\displaystyle \sum_{k=0}^{r}{m\choose k}{ n\choose n-k}={m+n\choose r}$ set $m= r=n$ we have $\displaystyle \sum_{k=0}^{r}{r\choose k}^2={2r\choose r}$ (case appears in our main problem) and hence are left to evaluate the latter sum. We recall the generating function for central binomial coefficients as $\displaystyle \sum_{k=0}^{\infty}{2n\choose n}x^{2k}=\frac{1}{\sqrt{1-4x^2}}$ so $\displaystyle \sum_{k=0}^{\infty}{2n\choose n}\frac{x^{2k}}{4^n}=\frac{1}{\sqrt{1-x^2}}$ and hence on integrating we have $\sum_{n=0}^{\infty}\frac{x^{2n}}{4^n(2n+1)}{2n\choose n}=\frac{\sin^{-1} x}{x}\Rightarrow \sum_{n=0}^{\infty}\frac{1}{4^n(2n+1)}\int_0^1 x^{2n} \ln x dx=\int_0^1\frac{\sin^{-1} x \ln x}{x}dx$ $=-0+\frac{1}{2}\int_{0}^1\frac{\ln^2 x}{\sqrt{1-x^2}}=\frac{1}{2}\int_0^{\frac{\pi}{2}}\ln^2(\sin x ) dx =\frac{\pi}{16}\frac{d^2}{dm^2}\left(\frac{1}{4^m}{2m\choose m}\right)_{m=0}$ $=\frac{\pi}{8}\frac{d}{dm}\left( \frac{1}{4^m}{2m\choose m}\left(\psi^0(2m+1)-\psi^0(m+1)-\ln 2\right)\right)_{m=0}=\frac{\pi}{16}\frac{1}{4^m}{2m\choose m}$ $\displaystyle \left(4\psi^0(m+1)+4\psi^0(2m+1)-2\psi^1(m+1)+4\psi^1(2m+1)\right)+4\psi^0(m+1)(\ln 4 -2\psi^0(m+1))-8\ln 2\psi^0(2m+1)+\ln^2 4)_{m=0}$ and hence $\frac{\pi}{16}(8\psi(1)+2\psi^1(1)+4\psi^1(1)(\ln 4-2\psi(1))-4\ln 4\psi (1)+\ln^2 4)=\frac{\pi}{16}\left(\frac{\pi^2}{3}+\ln^2 4\right)$ $\displaystyle=\frac{\pi^3}{48}+\frac{\ln^2(2)}{4}= \displaystyle \frac{\pi}{4}\left(\frac{\zeta(2)}{2}+\ln^2 2\right)$ which completes the proof.