Computing $ \iint_{[0,1]^2} \frac{-x\ln(xy)}{1-xy} \mathrm dx \mathrm dy $ I would like to compute $$ \iint_{[0,1]^2} \frac{-x\ln(xy)}{1-xy} \mathrm dx \mathrm dy $$
Without going into detail, here is what I found:
$$ \int_{0}^{1}(\int_{0}^{1} \frac{-x\ln(xy)}{1-xy} \mathrm dx ) \mathrm dy=\int_{0}^{1}(-\sum_{n=0}^{\infty} \int_{0}^{1} x^{n+1}y^n\ln(xy) \mathrm dx)\mathrm dy $$
$$ \int_{0}^{1}\sum_{n=1}^{\infty}\frac{y^n}{(n+1)^2}\mathrm dy=\sum_{n=1}^{\infty} \frac{1}{(n+1)^3}\approx0.202 $$
However Wolfram gives: $$ \iint_{[0,1]^2} \frac{-x\ln(xy)}{1-xy} \mathrm dx \mathrm dy=1 $$
Where is the problem?

\begin{align*}\int_0^1x^{n+1}y^n\ln(xy)dx&=y^n\int_0^1x^{n+1}\ln xdx+y^n\ln y\int_0^1x^{n+1}dx\\
&=y^n\left[\frac{x^{n+2}}{n+2}\ln x\right]_0^1-y^n\int_0^1\frac{x^{n+2}}{n+2}\frac 1xdx
+y^n\ln y\frac 1{n+2}\\
&=-\frac{y^n}{(n+2)^2}+y^n\frac{\ln y}{n+2}\\
&=\frac{y^n}{n+2}\left(\ln y-\frac 1{n+2}\right)
\end{align*}
and 
\begin{align*}
\int_0^1\frac{-x\ln(xy)}{1-xy}&=-\sum_{n=0}^{+\infty}\int_0^1\int_0^1x^{n+1}y^n\ln(xy)dxdy\\
&=-\sum_{n=0}^{+\infty}\frac 1{n+2}\int_0^1y^n\left(-\frac{1}{n+2}+\ln y\right)dy\\
&=-\sum_{n=0}^{+\infty}\frac 1{(n+2)^2(n+1)}-\sum_{n=0}^{+\infty}\frac 1{n+2}\frac 1{n+1}\left(-\int_0^1y^{n+1}\frac 1ydy\right)\\
&=-\sum_{n=0}^{+\infty}\frac 1{(n+2)^2(n+1)}+\sum_{n=0}^{+\infty}\frac 1{(n+2)(n+1)^2}\\
&=-\sum_{n=0}^{+\infty}\frac 1{(n+2)(n+1)}\left(\frac 1{n+2}-\frac 1{n+1}\right)\\
&=\sum_{n=0}^{+\infty}\frac 1{(n+2)(n+1)}\\
&=\sum_{j=1}^{+\infty}\frac{j+1-j}{(j+1)j}\\
&=\sum_{j=1}^{+\infty}\frac 1j-\frac 1{j+1}=1
\end{align*}
so Wolfram is right.
