TITLE: Evaluation of $\int_{0}^{1} (1-x)^{-1/2} \ \ln^{2}(x) \ \ln^{2}(1-x) \ \mathrm{d}x$
QUESTION [6 upvotes]: Can the integral 
\begin{align}
\int_{0}^{1} \frac {\ln^{2}(x)~\ln^{2}(1-x)}{\sqrt{1-x}}~\mathrm{d}x
\end{align}
be evaluated in terms of $\zeta(3)$ and $\ln(2/\mathrm{e})$ ?

REPLY [9 votes]: A possible way to evaluate this integral is to start from
$$
B(s,t)=\int_0^1x^{s-1}(1-x)^{t-1}dx=\frac{\Gamma(s)\Gamma(t)}{\Gamma(t+s)}
$$
Thus
$$
\int_0^1\frac{\ln^2 x\ln^2(1-x)}{\sqrt{1-x}}dx=\left.\frac{\partial^4}{\partial^2s\partial^2t}B(s,t)\right\vert_{(s,t)=(1,\frac{1}{2})}
$$
This yields after simplification
$$
\int_0^1\frac{\ln^2 x\ln^2(1-x)}{\sqrt{1-x}}dx=
\frac{1}{3} \left(168 \zeta (3) (\ln 4-4)-3 \pi ^4+48
   (48+(\ln  4 -12) \ln  4 )+8 \pi ^2 (\ln
    64 -14)\right).
$$
In fact this simplification can be done using formulas about the Polygamma function,
like $(15)$ and $(16)$ here. I thank O.L. for this information.