# Behavior of $I(n) = \int_{-\pi}^{\pi}\int_{-\pi}^{\pi}e^{ i n (x+y)}\,\frac{2\sin^2(x)\sin^2(y)}{2k -\cos(x)-\cos( y)}\,\mathrm{d}x\,\mathrm{d}y$

Consider the following integral: $$I(n) = \int_{-\pi}^{\pi}\int_{-\pi}^{\pi}e^{ i n (x+y)}\,\frac{2\sin^2(x)\sin^2(y)}{2k -\cos(x)-\cos( y)}\,\mathrm{d}x\,\mathrm{d}y,$$ where $$k>1$$ is a real number and $$n>0$$ is an integer number.

My question. Is it possible to characterize the asymptotic behavior of $$I(n)$$ as a function of $$n$$?

Numerics suggests that $$I(n)\sim \frac{1}{\sqrt n}\left(k-\sqrt{k^2-1}\right)^{2n}$$, however I have not yet been able to formally prove this fact.

This is not an homework question. Any help is welcome! Thanks in advance!

Progress. I briefly sketch my attempt to solve the problem, hoping that someone may find this useful. Note that $$g(x,y)$$ is analytic in a neighborhood of $$\mathbb{R}^2/\mathbb{Z}$$. Thus, we can shift the integral in the complex direction $$it$$ as long as $$g$$ remains analytic. This is indeed the case for all $$t < t_0 := \cosh^{-1} k = \ln \left(k + \sqrt{k^2-1}\right)= -\frac1{2\pi} \ln \left(k - \sqrt{k^2-1}\right).$$ Thus, by letting $$t=t_0$$, we get \begin{align*} I(n) &= e^{-2 n t_0}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}e^{ in(x+y)}\,g(x+it_0,y+it_0)\,\mathrm{d}x\,\mathrm{d}y\\ &= \left(k-\sqrt{k^2-1}\right)^{2n} I'(n), \end{align*} where \begin{align*} I'(n) &= \int_{-\pi}^{\pi}\int_{-\pi}^{\pi}e^{ in(x+y)}\,g(x+it_0,y+it_0)\,\mathrm{d}x\,\mathrm{d}y\\ &=\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\frac{e^{i n (x+y)}}{K}\frac{(K^2(2-K^2e^{-2i x})-e^{-2ix})(K^2(2-K^2e^{-2i y})-e^{-2 iy}) }{K(4k-K(e^{- i x}+e^{- i y}))-e^{ i x}-e^{ i y}}\,\mathrm{d}x\,\mathrm{d}y\\ \end{align*} where $$K:=k+\sqrt{k^2-1}$$. Although the integrand of $$I'(n)$$ blows up at $$x=0$$, $$y=0$$, $$I'(n)$$ still converges absolutely, and I think it should be possible to prove that $$I'(n)$$ exhibits subexponential decay. As a further observation, note that by using the change of variables $$z=e^{i x}$$, $$t=e^{iy}$$, $$I'(n)$$ can be rewritten as $$I'(n):=\oint_{|z|=1} \oint_{|t|=1} \frac{z^{n-2}t^{n-2}\left(2K^2z^2-K^4-z^4\right)\left(2K^2t^2-K^4-t^4\right)}{K\left(4kzt-K(z+t)\right)-(z+t)zt}\,\mathrm{d}z\,\mathrm{d}t.$$ Perhaps, this integral could be handled more easily using contour integration techniques; however it looks like pretty messy. I'm currently stuck here.

• I don't have a full answer for you, but I do have a couple things to point out. 1) $g(x,y)$ is an even function in both arguments, and given the region of integration $I(z)$ is real because the contribution from the imaginary component vanishes (Meaning we can replace the complex exponential with a cosine). 2) The integral wildly diverges for $0<k\leq1$, but will always converge smoothly (as in without hiccups) for $k>1$. Feb 18 '20 at 19:03
• As an edit, I think I made an error. The integral should converge for $k=1$. Feb 18 '20 at 19:09
• Thanks! I forgot to say that I’m interested in the case $k>1$ Feb 18 '20 at 20:08
• It seems related to the harmonic potential kernel of a killed simple random walk on the plane. Perhaps the techniques for expanding such potentials (such as in Fukia and Uchiyama, Kozma and Schreiber, etc) might be of a help? Feb 22 '20 at 19:16

I've proven the asymptotics in Mathoverflow 353430, for large integer $$n$$
$$J(n,\kappa) := \Big(\frac{2}{\pi}\Big)^2 \int_{-\pi}^\pi \int_{-\pi}^\pi \exp{(i\,n(x+y))}\frac{\sin^2x\,\sin^2y} {2\kappa - (\cos{x}+\cos{y}) }\, dx \,dy \sim$$ $$\sim \frac{8}{\sqrt{\pi \kappa n}}(\kappa^2-1)^{7/4} (\kappa - \sqrt{\kappa^2-1})^{2n}\quad, \quad (\kappa>1)$$