<s>[INST] Evaluate $\int^1_0 \log^2(1-x) \log^2(x) \, dx$ I have no idea where to even start. WolframAlpha cant compute it either.
$$\int^1_0 \log^2(1-x) \log^2(x) \, dx$$
I think it can be done with series, but I am not sure, can someone help a little so I can get a start??
Thanks!
 [/INST] Starting with the Beta function
\begin{align}
B(x, y) = \int_{0}^{1} t^{x-1} (1-t)^{y-1} \, dt
\end{align}
differentiate with respect to $x$ and $y$ twice. This leads to
\begin{align}
I &= \int_{0}^{1} t^{x-1} (1-t)^{y-1} \, \ln^{2}(t) \, \ln^{2}(1-t) \, dt \\
&= \partial_{x}^{2} \partial_{y}^{2}  B(x,y).
\end{align} 
Now the integral in question is the case for $x=y=1$. For this, it is seen that
\begin{align}
\partial_{x}^{2} \left. B(x,y) \right|_{x=1} = \frac{1}{y} \left[ \psi'(1) - \psi'(y+1) + \left( \psi(1) - \psi(y+1) \right)^{2} \right].
\end{align}
Differential with respect to $y$ yields
\begin{align}
\partial_{y}^{2} \partial_{x}^{2} \left. B(x,y) \right|_{x=1} &= \frac{1}{y} \left[ - \psi'''(y+1) - 2 \psi''(y+1) \left( \psi(1) - \psi(y+1) \right) + 2 \left( \psi'(y+1) \right)^{2} \right] \\
& \hspace{5mm} - \frac{2}{y^{2}} \left[ - \psi''(y+1) - 2 \psi'(y+1) \left( \psi(1) - \psi(y+1) \right) \right] \\
& \hspace{10mm} + \frac{2}{y^{3}} \left[ \psi'(1) - \psi'(y+1) + \left( \psi(1) - \psi(y+1) \right)^{2} \right].
\end{align}
Now
\begin{align}
\partial_{y}^{2} \partial_{x}^{2} \left. B(x,y) \right|_{x=y=1} &= 4 - \psi'''(2) + 4 \psi''(2) - 4 \psi'(2) + 2 \left( \psi'(2) \right)^{2} .
\end{align}
Since 
\begin{align}
\psi'(2) &= \frac{\pi^{2}}{6} -1  \\
\psi''(2) &= 2 - 2\zeta(3) \\
\psi'''(2) &= \frac{\pi^{4}}{45} - 6
\end{align}
then
\begin{align}
\partial_{y}^{2} \partial_{x}^{2} \left. B(x,y) \right|_{x=y=1} = 24 - \frac{4 \pi^{2}}{3} - \frac{\pi^{4}}{90} - 8 \zeta(3).
\end{align}
</s>