TITLE: Integral $\iint \limits_{{x,y \ \in \ [0,1]}} \frac{\log(1-x)\log(1-y)}{1-xy}dx\,dy=\frac{17\pi^4}{360}$
QUESTION [6 upvotes]: 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.

REPLY [4 votes]: Wow, RV and L are pretty fast. Well, even though their ways are more efficient, I worked out yet another way so I may as well post it. 
$$\int_{0}^{1}\log(1-y)\int_{0}^{1}\frac{\log(1-x)}{1-xy}dxdy$$
Let $u=xy, \;\ \frac{du}{x}=dy$
$$\int_{0}^{x}\frac{\log(1-u/x)}{1-u}du\int_{0}^{1}\frac{\log(1-x)}{x}dx$$
The left integral w.r.t u is a classic dilog: $\int_{0}^{x}\frac{\log(1-u/x)}{1-u}du=-Li_{2}(\frac{x}{1-x})$
Thus,  $$\int_{0}^{1}\frac{Li_{2}(\frac{x}{1-x})\log(1-x)}{x}dx$$
But,  $$-Li_{2}(\frac{x}{1-x})=\sum_{n=1}^{\infty}\frac{H_{n}}{n}x^{n}$$
giving:
$$\int_{0}^{1}\log(1-x)\sum_{n=1}^{\infty}\frac{H_{n}}{n}x^{n-1}dx$$
Also, $\int_{0}^{1}x^{n-1}\log(1-x)dx=\frac{H_{n}}{n}$
So, we finally obtain:
$$\sum_{n=1}^{\infty}\frac{H_{n}^{2}}{n^{2}}=\frac{17\pi^{4}}{360}$$