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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.

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  • $\begingroup$ 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 $k>1$. $\endgroup$ Feb 18 '20 at 19:03
  • $\begingroup$ As an edit, I think I made an error. The integral should converge for $k=1$. $\endgroup$ Feb 18 '20 at 19:09
  • $\begingroup$ Thanks! I forgot to say that I’m interested in the case $k>1$ $\endgroup$
    – Ludwig
    Feb 18 '20 at 20:08
  • $\begingroup$ 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? $\endgroup$ Feb 22 '20 at 19:16
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I've proven the asymptotics in Mathoverflow 353430, for large integer $n$

$$ J(n,\kappa) := \Big(\frac{2}{\pi}\Big)^2 \int_{-\pi}^\pi \int_{-\pi}^\pi \exp{(i\,n(x+y))}\frac{\sin^2x\,\sin^2y} {2\kappa - (\cos{x}+\cos{y}) }\, dx \,dy \sim$$ $$ \sim \frac{8}{\sqrt{\pi \kappa n}}(\kappa^2-1)^{7/4} (\kappa - \sqrt{\kappa^2-1})^{2n}\quad, \quad (\kappa>1)$$

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