# 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?

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W|A's answer is correct, as shown by the change of variables $(x,y)\to(x,z)$ with $z=xy$. – Did Apr 9 '12 at 15:01
@Didier Could you show me how to do that? – Pedro Tamaroff Apr 9 '12 at 23:48
@PeterT.off The OP does not seem interested... – Did Apr 9 '12 at 23:51
@Didier That's why I phrased it as show me. Should I open a new question? – Pedro Tamaroff Apr 9 '12 at 23:54
@PeterT.off: Whatever you do or do not do, answering here seems pointless. – Did Apr 10 '12 at 6:30