# Double integral over square converging to $0$

I am struggling to solve the following problem as part of a bigger project that I am working on.

Let $$\mathcal{S} \subset \mathbb{R}^2$$ be a square of length $$\sqrt{n}$$ centered at the origin, $$f:\mathcal{S} \times \mathcal{S} \rightarrow \mathbb{R}$$ given by $$f(x,y) = e^{-|x-y|}|x-y|^2*\Big( \frac{1}{\max(\min(|x|,|y|),1)} \Big)^2$$.

I want to show that $$\frac{1}{(\log n)^2} \int_{\mathcal{S}x \mathcal{S}} f(x,y) dx dy$$ converges to $$0$$ as $$n$$ goes to infinity. Notice that here $$x,y \in \mathcal{S}$$, so they are basically 2-d vectors.

Using Campbell-Hardy theorem, this is equivalent to showing that the expected value of the sum of $$f(x,y)$$ converges to $$0$$, where $$x,y$$ run over all the points in a Poisson Point Process of rate $$1$$ over this square. Using Python, I proved empirically that the claim above is true. Now, I am trying to find a rigorous math proof.

I tried breaking the integral in easier to compute integrals depending on the positions of $$x,y$$, but all of my approaches were futile so far.

• Are you sure $f$ does not depend on $n$? As it stands, the integral over that square is nondecreasing with $n$. – punctured dusk Mar 14 at 20:25
• I forgot a term for the limit part. Thanks for bringing this up, please check the updated version. – user548645 Mar 14 at 20:27