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?