I am trying to find the set of real numbers $s\in \mathbb{R}$ such that $$\iint_{(a,b)^2}\frac{1}{|x-y|^s}\mathrm{d}x\mathrm{d}y < \infty$$ where $(a,b)$ is an arbitrary finite open interval of $\mathbb{R}$. For now, except for the trivial case $s\leq 0$, I am not able to solve the problem.
My intuition is that as for the one dimensional case, I should have $s<1$ for the integral to converge, but i do not know how to prove it.
On
The command of Mathematica 13.2 on Windows 10
Integrate[1/RealAbs[x - y]^s, {x, a, b}, {y, a, b},
Assumptions -> s \[Element] Reals && a \[Element] Reals && b \[Element] Reals && a < b && s < 2]
results in $$\frac{2 (a-b)^2 (b-a)^{-s}}{(s-2) (s-1)} $$
and the command
Integrate[1/RealAbs[x - y]^s, {x, a, b}, {y, a, b}, Assumptions -> s \[Element] Reals && a \[Element] Reals && b \[Element] Reals && a < b && s >= 2]
says the integral diverges.
For $b>a$, with the change of variables $x=\frac{b-a}{2}u+\frac{b+a}{2}$ and $y=\frac{b-a}{2}v+\frac{b+a}{2}$ we can write $$ \iint_{(a,b)^2}\frac{1}{|x-y|^s}\mathrm{d}x\mathrm{d}y { = \iint_{(-1,1)^2}\frac{1}{(\frac{b-a}{2})^s|u-v|^s}\frac{(b-a)^2}{4}\mathrm{d}u\mathrm{d}v \\= (\frac{b-a}{2})^{2-s}\iint_{(-1,1)^2}\frac{\mathrm{d}u\mathrm{d}v}{|u-v|^s}. } $$ By defining $A=\{(u,v):u^2+v^2<1\}$ and $B=\{(u,v):u^2+v^2<2\}$, we have $$ { A\subseteq(-1,1)^2 \quad,\quad (-1,1)^2\subseteq B. } $$ Therefore $$ \iint_{A}\frac{\mathrm{d}u\mathrm{d}v}{|u-v|^s} \le \iint_{(-1,1)^2}\frac{\mathrm{d}u\mathrm{d}v}{|u-v|^s} \le \iint_{B}\frac{\mathrm{d}u\mathrm{d}v}{|u-v|^s}. $$ Also $$ \iint_{u^2+v^2<1}\frac{\mathrm{d}u\mathrm{d}v}{|u-v|^s} { = \int_0^{2\pi}\int_0^1\frac{r\mathrm{d}r\mathrm{d}\theta}{r^s|\cos\theta-\sin\theta|^s} \\= 2^{-\frac{s}{2}}\int_0^1r^{1-s}\mathrm{d}r \int_0^{2\pi}\frac{\mathrm{d}\theta}{|\sin\theta|^s} \\= 2^{2-\frac{s}{2}}\int_0^1r^{1-s}\mathrm{d}r \int_0^{\frac{\pi}{2}}\frac{\mathrm{d}\theta}{(\sin\theta)^s}. } $$ The latter integral converges when $s<1$ and diverges when $s\ge 1$. Therefore