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


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 $\ln{\Gamma(x)}$.
