Integral $\iint \limits_{{x,y \ \in \ [0,1]}} \frac{\log(1-x)\log(1-y)}{1-xy}dx\,dy=\frac{17\pi^4}{360}$ Hi I am trying to integrate $$
\mathcal{I}:=\iint \limits_{{x,y \ \in \ [0,1]}} \frac{\log(1-x)\log(1-y)}{1-xy}dx\,dy=\int_0^1\int_0^1 \frac{\log(1-x)\log(1-y)}{1-xy}dx \,dy
$$
A closed form does exist. I tried to write
\begin{align}
\mathcal{I} &=\int_0^1 \log(1-y)dy\int_0^1 \log(1-x)\frac{dx}{1-xy} \\
&= \int_0^1\log(1-y)dy \ \int_0^1 \sum_{n\geq0}(xy)^n\, \ln(1-x) \ dx \\
&= \sum_{n\geq 0}\frac{1}{n+1}\int_0^1 \log(1-y) y^n\, dy \\
&= \sum_{n\geq 0}\frac{1}{n+1}\int_0^1 \log(1-y)y^n\, dy = ?
\end{align}
I was able to realize that 
$$
\mathcal{I}=\sum_{n\geq 1}\left(\frac{H_n}{n}\right)^2=\frac{17\zeta_4}{4}=\frac{17\pi^4}{360},\qquad \zeta_4=\sum_{n\geq 1} n^{-4} 
$$
however this does not help me solve the problem.  How can we calculate $\mathcal{I}$? Thanks.

Consider the double integral 
\begin{align}
I = \int_{0}^{1} \int_{0}^{1} \frac{\ln(1-x) \ \ln(1-y)}{1-xy} \ dx dy
\end{align}
which, upon expansion into series form, becomes
\begin{align}
I &= \sum_{n=0}^{\infty} \ \int_{0}^{1} x^{n} \ln(1-x) dx \ \int_{0}^{1} y^{n} \ln(1-y) dy \\
&= \sum_{n=0}^{\infty} \left( \int_{0}^{1} x^{n} \ \ln(1-x) \ dx \right)^{2}.
\end{align}
Without much difficulty it can be shown that
\begin{align}
\int_{0}^{1} x^{n} \ \ln(1-x) \ dx &= \left. \partial_{\mu} B(n+1, \mu+1) \right|_{\mu = 1} \\
&= B(n+1,1) \left( \psi(1) - \psi(n+2) \right) \\
&= \frac{H_{n+1}}{n+1}. 
\end{align}
Now, returning to the primary integral, it is seen that
\begin{align}
I = \sum_{n=0}^{\infty} \left( \frac{H_{n+1}}{n+1}\right)^{2} = \sum_{n=1}^{\infty} \left( \frac{H_{n}}{n} \right)^{2} = \frac{17 \zeta(4)}{4}.
\end{align}
Hence
\begin{align}
\int_{0}^{1} \int_{0}^{1} \frac{\ln(1-x) \ \ln(1-y)}{1-xy} \ dx dy = \frac{17 \zeta(4)}{4}.
\end{align}
