Solve $\int_0^1\ln^2\Gamma(x)\,\mathrm{d}x$ I want to solve the following integral but after some work I didn't find a way to go. Could anyone give me a hint? 
\begin{equation}
I=\int_{0}^{1}\ln^2\Gamma(x)\,\mathrm{d}x
\end{equation}
The answer is 
\begin{equation}
I=\frac{\ln^2 (2\pi)}{3}+\frac{\pi^2}{48}+\frac{\gamma\ln(2\pi)}{6}+\frac{\gamma^2}{12}+\frac{\zeta''(2)}{2\pi^2}-\frac{\zeta'(2)\ln (2\pi)}{\pi^2}-\frac{\gamma\zeta'(2)}{\pi^2}
\end{equation}
They only give a hint (using the Fourier Series) which I looked up at https://de.wikipedia.org/wiki/Gammafunktion.
\begin{equation}
\ln\Gamma(x) = \left(\tfrac{1}{2}-x\right) \bigl(\gamma + \ln(2\pi)\bigr) + \frac{1}{2} \ln\frac{\pi}{\sin(\pi x)} + \frac{1}{\pi} \sum_{k=2}^\infty \frac{\ln k}{k} \sin(2\pi k x)
\end{equation}
Want I have tried so far:


*

*squared the series

*integration by parts and the the fourier series

 A: Theorem $6.1$ from the paper A generalized polygamma function by Olivier Espinosa and Victor H. Moll will bring the light over your question. (see the special case $k=k'=1$)
A: Use Parseval's Theorem as @James Arathoon metioned and use the Fourier Series given here:
Integral that arises from the derivation of Kummer's Fourier expansion of $\ln{\Gamma(x)}$.
A: An almost-answer, but for brevity some of the details have been skipped.
If you square your formula for $\ln\Gamma(x)$, you'll find many of the terms integrate to $0$ on $[0,\,1]$ due to being of the form $o(x-\tfrac12)$ for odd $o$. Let $f(x)\sim g(x)$ denote the equivalence relation $\int_0^1(f(x)-g(x))dx=0$, so $f(x)\sim\int_0^1f(t)dt$. Before we go any further, I'll mention the famous result $\ln\sin(\pi x)\sim-\ln2$, and something with a similar proof,$$\ln^2\sin(\pi x)\sim\pi\ln4+\ln^22-2\ln\sin\frac{\pi x}{2}\ln\cos\frac{\pi x}{2}.$$Oh, and one more thing I'll need in a moment, for integer $k\ne0$: $x\sin(2\pi kx)\sim\frac{-1}{2\pi k}$. So$$\begin{align}\ln^2\Gamma(x)&=\color{red}{(\gamma+\ln(2\pi))^2(\tfrac12-x)^2}+\color{orange}{\frac14\ln^2\frac{\pi}{\sin(\pi x)}}+\color{limegreen}{\frac{1}{2\pi^2}\sum_{k\ge2}\frac{\ln^2k}{k^2}}\\&+\color{blue}{\frac{\gamma+\ln(2\pi)}{\pi}(1-2x)\sum_{k\ge2}\frac{\ln k}{k}\sin(2\pi kx)}\\&\sim\color{red}{\frac{(\gamma+\ln(2\pi))^2}{12}}\\&+\color{orange}{\frac14(\ln^2\pi+\ln4\ln\pi+\pi\ln4+\ln^22-2\ln\sin\tfrac{\pi x}{2}\ln\cos\tfrac{\pi x}{2})}\\&+\color{limegreen}{\frac{\zeta^{\prime\prime}(2)}{2\pi^2}}-\color{blue}{\frac{\gamma+\ln(2\pi)}{\pi^2}\zeta^\prime(2).}\end{align}$$So now we just need to evaluate $\int_0^1\ln\sin\frac{\pi x}{2}\ln\cos\frac{\pi x}{2}dx$.
