Show $ \int_0^1 \int_0^1 \log \left| x-y\right|\,dx\,dy>-\infty.$ I'm struggling to show that $ \int_0^1 \int_0^1 \log \left| x-y\right|\,dx\,dy >-\infty$ which means that $f(x,y):=\log| x-y|$ is in $L^1([0,1]^2,\lambda\otimes \lambda)$. The part "$<+\infty$" is not difficult.
The lower bound $1-\frac{1}{u} \leq \log u$ for all $u>0$, reduces the problem to $ \int_0^1 \int_0^1  \frac{1}{\left| x-y\right|}\,dx\,dy>-\infty$.
Any help/hint would be appreciated.
 A: By setting $f(x,y)=\log\left|x-y\right|$ we have $f(x,y)=f(y,x)$ and $f\leq 0$ for any $(x,y)\in(0,1)^2$, hence
$$ \iint_{(0,1)^2}f(x,y)\,dx\,dy = 2\int_{0}^{1}\int_{0}^{x}f(x,y)\,dy\,dx \stackrel{y\mapsto xz}{=} 2\iint_{(0,1)^2}x\,f(x,xz)\,dz\,dx$$
and the original integral boils down to:
$$ 2\int_{0}^{1}\int_{0}^{1}x \log(1-z)+x\log(x)\,dx\,dz = 2\int_{0}^{1}x(\log x-1)\,dx=\color{red}{-\frac{3}{2}}.$$
A: Proceeding we note from symmetry that we can write the integral of interest as
$$\begin{align}
\int_0^1\int_0^1 \log(|x-y|)\,dx\,dy&=2\int_0^1\int_y^1 \log(x-y)\,dx\,dy\\\\
&=2\int_0^1 \left.\left((x-y)\log(x-y)-(x-y)\right)\right|_{x=y}^{x=1}\,dy\\\\
&=2\int_0^1\left( (1-y)\log(1-y)-(1-y)\right)\,dy\\\\
&=-\frac32
\end{align}$$
A: Showing
$$  \int_0^1 \int_0^1 \log \left| x-y\right|dxdy>-\infty $$
 is equivalent to showing
$$  \int_0^1 \int_0^1 \log \frac1{\left| x-y\right|}dxdy<\infty. $$
Noting
$$ x^{\frac12}\ln\frac1x\le C $$
for some constant, so one has
$$  \int_0^1 \int_0^1 \log \frac1{\left| x-y\right|}dxdy\le C\int_0^1 \int_0^1 \frac1{\sqrt{\left| x-y\right|}}dxdy<\infty. $$
