How to prove that$$\lim_{a\to0+}\int_{E\times E}\frac{1}{(x-y)^2+a}dxdy=+\infty$$where $E$ is a Lebesgue-measurable set in $\mathbb{R}$ s.t. $m(E)>0$?
Because the set $E$ can be any mesurable set, I can't use Newton-Leibniz formula to simplify the integration. To be honest, I can't think of a place to start.
Furthermore, I guess that$$\int_{E\times E}\frac{1}{(x-y)^2}dxdy=+\infty$$ is also true under the same assupmtion. Is that true in fact?
1st Solution. By applying a suitable translation and restriction, it suffices to assume $E \subseteq [0, 1]$. Also, monotone convergence theorem shows that
$$ \lim_{a \to 0^+} \int_{E^2} \frac{1}{(x - y)^2 + a} \, \mathrm{d}x\mathrm{d}y = \int_{E^2} \frac{1}{(x - y)^2} \, \mathrm{d}x\mathrm{d}y. $$
Hence, it suffices to investigate the integral in the right-hand side. Let $n \geq 0$ be an integer. Then
\begin{align*} \int_{E^2} \frac{1}{(x - y)^2} \, \mathrm{d}x\mathrm{d}y &\geq \sum_{k=0}^{2^n - 1} \int_{([k/2^n, (k+1)/2^n] \cap E)^2} \frac{1}{(x - y)^2} \, \mathrm{d}x\mathrm{d}y \\ &\geq \sum_{k=0}^{2^n - 1} \int_{([k/2^n, (k+1)/2^n] \cap E)^2} \frac{1}{(1/2^n)^2} \, \mathrm{d}x\mathrm{d}y \\ &= 4^n \sum_{k=0}^{2^n - 1} m([k/2^n, (k+1)/2^n] \cap E)^2 \\ &\geq 2^n \left( \sum_{k=0}^{2^n - 1} m([k/2^n, (k+1)/2^n] \cap E) \right)^2\\ &= 2^n m(E)^2. \end{align*}
So by letting $n \to \infty$, the claim follows.
2nd Solution. Let $F \subseteq E$ be measurable and bounded. Also, write $a = \varepsilon^2$ for $\varepsilon > 0$. Then by Fubini's theorem,
\begin{align*} \int_{E^2} \frac{1}{(x - y)^2 + \varepsilon^2} \, \mathrm{d}x\mathrm{d}y &\geq \int_{F^2} \frac{1}{(x - y)^2 + \varepsilon^2} \, \mathrm{d}x\mathrm{d}y \\ &= \frac{1}{\varepsilon} \operatorname{Im} \left( \int_{F^2} \int_{0}^{\infty} e^{-(\varepsilon+i(x-y))t} \, \mathrm{d}t\mathrm{d}x\mathrm{d}y \right) \\ &= \frac{1}{\varepsilon} \int_{0}^{\infty} e^{-\varepsilon t} \operatorname{Im} \left( \int_{F^2} e^{iyt}e^{-ixt} \, \mathrm{d}x\mathrm{d}y \right) \, \mathrm{d}t \\ &= \frac{1}{\varepsilon} \int_{0}^{\infty} e^{-\varepsilon t} \left| \int_{F} e^{iyt} \, \mathrm{d}y \right|^2 \, \mathrm{d}t \\ &= \int_{0}^{\infty} e^{-\varepsilon^2 s} \left| \int_{F} e^{\varepsilon iys} \, \mathrm{d}y \right|^2 \, \mathrm{d}s. \tag{$t = \varepsilon s$} \end{align*}
So, by the Fatou's lemma,
\begin{align*} \liminf_{\varepsilon \to 0^+} \int_{E^2} \frac{1}{(x - y)^2 + \varepsilon^2} \, \mathrm{d}x\mathrm{d}y &\geq \int_{0}^{\infty} \liminf_{\varepsilon \to 0^+} e^{-\varepsilon^2 s} \left| \int_{F} e^{\varepsilon iys} \, \mathrm{d}y \right|^2 \, \mathrm{d}s \\ &= \int_{0}^{\infty} m(F)^2 \, \mathrm{d}s \\ &= \begin{cases} \infty, & \text{if $m(F) > 0$}, \\ 0, & \text{if $m(F) = 0$}. \end{cases} \end{align*}
Now by choosing $F$ large enough so that $m(F) > 0$, we are done.