Evaluation of $\int_{0}^{1} (1-x)^{-1/2} \ \ln^{2}(x) \ \ln^{2}(1-x) \ \mathrm{d}x$

Question

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 $\ln(2/\mathrm{e})$ ?

Answer 1

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.