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.

  • $\begingroup$ 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

I've proven the asymptotics in Mathoverflow 353430, for large integer $n$

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.