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I want to calculate the integral:$$\int_0^1\int_0^1\mathrm{ln}(|x-y|)\;\mathrm{d}x\;\mathrm{d}y,$$ but as you can see when $x=y$ the integrand goes to $-\infty$.

Does this integral has a solution? Could you give me any hint about how to solve it?

Thanks!

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    $\begingroup$ Try breaking it into when $x>y$ and $x<y$. $\endgroup$ – user223391 Jun 20 '15 at 22:58
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Hint:

After splitting up the interval of integration, changing the order of integration and exploiting symmetry we find

$$\begin{align} I &=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\ln{\left(\left|x-y\right|\right)}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\ln{\left(\left|x-y\right|\right)}+\int_{0}^{1}\mathrm{d}x\int_{x}^{1}\mathrm{d}y\,\ln{\left(\left|x-y\right|\right)}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\ln{\left(x-y\right)}+\int_{0}^{1}\mathrm{d}x\int_{x}^{1}\mathrm{d}y\,\ln{\left(y-x\right)}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\ln{\left(x-y\right)}+\int_{0}^{1}\mathrm{d}y\int_{0}^{y}\mathrm{d}x\,\ln{\left(y-x\right)}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\ln{\left(x-y\right)}+\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\ln{\left(x-y\right)}\\ &=2\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\ln{\left(x-y\right)}.\\ \end{align}$$

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$$\begin{eqnarray*}\iint_{(0,1)^2}\log(\left|x-y\right|)\,dx\,dy &=& 2 \int_{0}^{1}\int_{0}^{x}\log(x-y)\,dy\, dx\\&=&2\int_{0}^{1}x\int_{0}^{1}\log(x-xz)\,dz\,dx\\&=&2\iint_{(0,1)^2}x\log x+x\log(1-z)\,dz\,dx\\&=& 2\left(-\frac{1}{4}-\frac{1}{2}\right)=\color{red}{-\frac{3}{2}}.\end{eqnarray*}$$

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