An Integral Question $ \int_{A\times A} \frac{dxdy}{|x-y|^{2k}} < \infty $ Let $A\subset \mathbb R^2$ be bounded and $0<k<2$. 
I want to find the condition of $k$ such that
$$
\int_{A\times A} \frac{dxdy}{|x-y|^{2k}} < \infty
$$
I solved the case that $A\subset \mathbb R$ by direct integral calculation. But in this case I have no idea now. Please help me. I posted this yesterday, but it was deleted "not sufficient condition". But I could not understand why that post has been deleted... 
 A: Here is a plane of attack.  First, convert to polar coordinates with $(r_1,\theta_1)$, $(r_2,\theta_2)$ and study the convergence of the integral $I$ given by
$$\begin{align}
I&=\int_0^{2\pi}\int_0^L \int_0^{2\pi}\int_0^L \frac{r_1r_2}{\left(r_1^2+r_2^2-2r_1r_2\cos(\theta_2-\theta_1)\right)^{k}}\,dr_1\,\,d\theta_1\,dr_2\,d\theta_2\\\\
&=2\pi\int_0^L \int_0^Lr_1r_2\left(\int_0^{2\pi} \frac{1}{\left((r_1-r_2)^2+4r_1r_2\sin^2(\phi/2)\right)^{k}}\,d\phi\right)\,dr_1\,dr_2\tag1
\end{align}$$
Next, we note that the inner integral on the right-hand side of $(1)$ exhibits a singularity for $r_1=r_2$ and $\phi=0$.  We can examine equivalently, therefore, the behavior of the integral $J$ as expressed as
$$\begin{align}
J&=\int_0^{2\pi} \frac{1}{\left((r_1-r_2)^2+r_1r_2\phi^2\right)^{k}}\,d\phi\\\\
&=\frac{1}{\sqrt{r_1r_2}\,|r_1-r_2|^{2k-1}}\int_0^{\arctan\left(2\pi \sqrt{r_1r_2}/|r_1-r_2|\right)}\cos^{2k-1}(u)\,du \tag 2
\end{align}$$
where we used $\sin(\phi/2)\sim \phi/2$ as $\phi\to 0$.
The integral on the right-hand side of $(2)$ is bounded by $\pi/2$ when $k\ge 1/2$.  This boils the problem down to analyzing the convergence of the integral 
$$\int_0^L\int_0^L \frac{\sqrt{r_1r_2}}{|r_1-r_2|^{2k-1}}\,dr_1\,dr_2$$
for $1/2\le k<2$.

Note that we also have
$$I\le 4\pi^2 \int_0^L\int_0^L \frac{r_1r_2}{|r_1-r_2|^{2k}}\,dr_1\,dr_2$$
which reduces to the problem when $A\subset \mathbb{R} $.  This can be used to examine convergence for $0<k<1/2$.
