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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 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 Whatever you do or do not do, answering here seems pointless. – Did Apr 10 '12 at 6:30
up vote 5 down vote accepted

\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.

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I'm realizing that I made mistake in my calculations... Thank you for your answer! – Chon Apr 9 '12 at 18:32

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