# Asymptotic decay rate of an oscillatory integral

Consider the following oscillatory integral $$I(n):=\int_{-\pi}^\pi\int_{-\pi}^\pi e^{i n(x+y)}\frac {(1 - \cos(2x)) (1 - \cos(2y))} {2k - (\cos x + \cos y)}\ \mathrm{d}x\,\mathrm{d}y.$$ where $$n\in\mathbb{N}$$, $$n>2$$, $$k\in\mathbb{R}$$, $$k>1$$, and $$i$$ is the imaginary unit.

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

• The function $(1-\cos(2x))(1-\cos(2y)) / (2k - \cos x - \cos y)$, by your assumption, is a smooth function on the torus. So its Fourier transform decays faster than any polynomial, no? – Willie Wong Feb 25 at 13:50
• @WillieWong Yes, correct. However, I would like to explicitly characterize the decay rate of $I(n)$ as a function of parameter $n$. – Ludwig Feb 25 at 13:52
• @WillieWong not sure I understand your point. To clarify, I would like to find an asymptotic estimate of the form $I(n)\sim f(n)$, where $f(n)$ decays exponentially. Numerics suggests that $f(n)=c^{2n}/\sqrt{n}$, where $c=k-\sqrt{k^2-1}$ but I'm looking for a formal proof. – Ludwig Feb 25 at 13:57
• As often with parameter-dependent integrals, Feyman's trick (taking derivatives with respect to the parameter $k$) might be useful here. – B K Feb 25 at 14:23
• These are things that you could've included in your question to start with. Frequently when people talk about asymptotics of oscillatory integral they are talking about stationary phase issues. In your case you might as well just titled the question about asymptotic decay rates of the Fourier series. – Willie Wong Feb 25 at 14:25

The asymptotics for large $$n$$ is $$J(n,\kappa) := \Big(\frac{2}{\pi}\Big)^2 \int_{-\pi}^\pi \int_{-\pi}^\pi \exp{(i\,n(x+y))}\frac{\sin^2x\,\sin^2y} {2\kappa - (\cos{x}+\cos{y}) }\, dx \,dy \sim$$ $$\sim \frac{8}{\sqrt{\pi \kappa n}}(\kappa^2-1)^{7/4} (\kappa - \sqrt{\kappa^2-1})^{2n}\quad, \quad (\kappa>1)$$ The proof consists of 5 parts. The first part is to write the 2-dimensional fourier transform in a 1-dimensional form, albeit with special functions as part of the integrand. In Parts 2-5 asymptotic analysis will be undertaken. The proof is already long, and I'm not going to justify every switch of $$\sum$$ and $$\int$$, nor do some of the details that a rigorous saddle point analysis requires, like proving the tails decay sufficiently fast. I have checked the result numerically, and for $$\kappa$$ = 2.2, I get the following errors between the true value and the asymptotic approximation: n=50, 2.46%; n=100, 1.23%, n=200, 0.61%

Part 1: Show $$J(n,\kappa) = \int_0^\infty \exp{(-2\kappa t)}\Big(2I_n(t) - I_{n-2}(t) - I_{n+2}(t) \Big)^2 dt$$ where $$I_n(t)$$ is the modified Bessel function.

With $$r=1/(2\kappa)$$,expand the integrand in a power series in $$r$$ to get $$J(n,\kappa) =r \sum_{m=0}^\infty r^m \Big(\frac{2}{\pi}\Big)^2 \int_{-\pi}^\pi \int_{-\pi}^\pi \big(\cos{x}+\cos{y})^m \sin^2{x} \sin^2{y} \,e^{i\,n(x+y)} dx \, dy =$$ $$= r \sum_{m=0}^\infty r^m m! \sum_{k=0}^m b_{m-k} \, b_k \, \text{ where }b_k=\frac{2}{k!\,\pi}\int_{-\pi}^\pi \cos^k y \sin^2 y \, e^{i\,n y} dy$$ By Cauchy product of power series, where $$B(u)=\sum_k u^k\,b_k,$$ $$J(n,\kappa) =r \sum_{m=0}^\infty r^m m! [u^m] B(u)^2$$ where the 'coefficient of' operator has been used. By the Borel transform we get $$J(n,\kappa) = \int_0^\infty dt \exp{(-2\kappa t)} \sum_{m=0}^\infty t^m [u^m] B(u)^2 = \int_0^\infty \exp{(-2\kappa t)}B(t)^2 dt$$ We work on $$B(t)$$ now. $$B(t) = \frac{2}{\pi} \int_{-\pi}^\pi dy \, e^{i\,n\,y} \sin^2 y \sum_{k=0}^\infty\frac{(t\, \cos{y})^k}{k!}$$ The interior summation is $$\exp{(t \cos{y} )}$$ which also has an expansion involving modified Bessel functions $$\exp{(t \cos{y} )} = I_0(t)+2\sum_{j=1}^\infty I_j(t) \cos{j\,y}$$ We thus must consider the integral $$\frac{2}{\pi} \int_{-\pi}^\pi dy \, e^{i n \, y} \sin^2{y} \cos{j\,y}$$ Mathematica knows that $$\frac{2}{\pi}\int_{-\pi}^\pi e^{i n \, y} \sin^2{y} \, dy = \frac{-2}{\pi} \frac{4 \sin{(\pi z)}}{z(z+2)(z-2)}$$ which, for integer $$z,$$ is non-zero only if $$z=0$$ or $$\pm 2.$$ The penultimate formula says we have $$z=n \pm j,$$ and the formula under the 'Part 1' heading follows by keeping only the non-zero terms in the infinite sum over Bessel functions.

Part 2: a useful simplification

$$2I_n(u)-I_{n-2}(u)-I_{n+2}(u)=\frac{2}{u}\Big( (n+1)I_{n+1}(u)-(n-1)I_{n-1}(u) \Big)$$ This follows from applying the recursion $$I_{n-1}(u)-I_{n+1}(u)=\frac{2n}{u}I_n(u)$$ twice. The equation of Part 1 will thus consist of two BesselI's squared, and a cross term.

Part 3: Show $$\int_0^\infty e^{-2 \kappa\,u}\,I_n(u)^2 \frac{du}{u^2}= \frac{1}{n\,\pi(4n^2-1)} \int_0^1 \frac{x^{-n}}{\sqrt{x(1-x)}}\big(\kappa-\sqrt{\kappa^2-x}\big)^{2n} \big(\kappa + 2n\sqrt{\kappa^2 - x} \big)\,dx$$

Use Gradshteyn 6.592.6 and 6.611.4, respectively: $$\frac{1}{\pi} \int_0^1 \frac{dx}{\sqrt{x(1-x)}}I_{2n}(2u\sqrt{x}) = I_n(u)^2 \, ,$$ $$\int_0^\infty e^{-a\,x} I_{\nu}(b\,x)\,dx = b^{-\nu}\frac{\big(a-\sqrt{a^2-b^2}\big)^{\nu}}{\sqrt{a^2-b^2}} .$$

Then integrate twice with respect to the argument of the exponential ($$a,$$ to begin with).

Part 4: Asymptotics of 'cross term'
$$M_n:=\int_0^\infty e^{-2 \kappa\,u}\,I_{n-1}(u)\,I_{n+1}(u) \frac{du}{u^2}= \int_0^\infty e^{-2 \kappa\,u}\,I_{n}(u)^2 \frac{du}{u^2} \big(1+\cal{O}(1/n)\big)$$ Use the series form for product of Bessel functions, Gradsteyn 8.442.1. Then $$M_n=\int_0^\infty \frac{du}{u^2}e^{-2 \kappa\,u} \sum_{m=0}^\infty \frac{(u/2)^{2n+2m}}{m!} \frac{(2n+m+1)_m}{(n+m+1)!(n+m-1)!} =$$ $$=\int_0^\infty \frac{du}{u^2}e^{-2 \kappa\,u} \sum_{m=0}^\infty \frac{(u/2)^{2n+2m}}{m!} \frac{(2n+m+1)_m}{(n+m)!(n+m)!}\Big(1 - \frac{1}{n+m+1} \Big)$$ $$=\int_0^\infty \frac{du}{u^2}e^{-2 \kappa\,u} I_n(u)^2\big(1+\cal{O}(1/n)\big)$$ Part 5: Put pieces together $$J(n,\kappa) \sim 4\int_0^\infty \frac{du}{u^2}e^{-2 \kappa u}\Big( \big((n+1)I_{n+1}(u)\big)^2 + \big((n-1)I_{n-1}(u)\big)^2 -2(n+1)(n-1)I_{n}(u)^2\Big)$$ and using the result of Part 3, $$\frac{\pi}{4}J(n,\kappa)\sim \int_0^1 \frac{dx\,x^{-n}}{\sqrt{x(1-x)}}\big( \kappa - \sqrt{\kappa^2-x}\big)^{2n} \cdot$$ $$\cdot \Big\{ \frac{n+1}{4(n+1)^2-1} \frac{\big( \kappa - \sqrt{\kappa^2-x}\big)^2}{x}\big(\kappa+2(n+1)\sqrt{\kappa^2-x}\big) -$$ $$\frac{n-1}{4(n-1)^2-1} \frac{x}{\big( \kappa - \sqrt{\kappa^2-x}\big)^2}\big(\kappa+2(n-1)\sqrt{\kappa^2-x}\big) +$$ $$\frac{2(n+1)(n-1)}{n(4n^2-1)} \big(\kappa+2n\sqrt{\kappa^2-x}\big) \Big\}$$ Expand curly brackets as $$n \to \infty.$$ $$\Big\{...\Big\}=\frac{2}{x}\big(\kappa^2-x\big)^{3/2}+\kappa(\kappa^2/x-1)\frac{1}{n}+...$$ Since earlier we approximated the cross-term to order 1/n, we'll keep only the first term in this expansion. By letting $$x \to u^2$$ then $$\frac{\pi}{16} J(n,\kappa) \sim \int_0^1 \frac{du \, u}{\sqrt{(1-u)(1+u)}}\big(\kappa/{u}-\sqrt{ (\kappa/u)^2-1}\big)^{2n}\big((\kappa/u)^2-1)^{3/2}$$
Use the following transformation, good for positive functions $$g,$$ $$\int_0^1g(t)/\sqrt{1-t}=2\int_0^1 g(4u(1-u))\,du$$ This has the effect of putting the strong growth behavior at the endpoint of the domain as $$u \to 1$$ to the region near $$u=1/2.$$ In fact, the following integral is well-setup for a classic saddle point analysis: $$\frac{\pi}{128}J(n,\kappa) \sim \int_0^1 \frac{du \, u(1-u)}{\sqrt{1+4u(1-u)}} \Big(\big(\frac{\kappa}{4u(1-u)}\big)^2-1\Big)^{3/2}\cdot$$ $$\Big(\frac{\kappa}{4u(1-u)}-\sqrt{ \big(\frac{\kappa}{4u(1-u)}\big)^2-1}\Big)^{2n}$$ Define $$h(u)=-\log\Big(\frac{\kappa}{4u(1-u)}-\sqrt{ \big(\frac{\kappa}{4u(1-u)}\big)^2-1}\,\Big)$$ We find $$h'(u)=0 \implies u=1/2.$$ A Taylor expansion about u=1/2 leads to $$h(u)=-\log\Big(\kappa-\sqrt{\kappa^2-1}\Big) +\frac{4\kappa}{\sqrt{\kappa^2-1}}(u-1/2)^2+\cal{O}((u-1/2)^4)$$ Define $$g(u)$$ as below and keep its first term in a Taylor expansion about u=1/2 $$g(u) = \frac{ u(1-u)}{\sqrt{1+4u(1-u)}} \Big(\big(\frac{\kappa}{4u(1-u)}\big)^2-1\Big)^{3/2}=\frac{\big(\kappa^2-1)^{3/2}}{4\sqrt{2}}+\cal{O}((u-1/2)^2)$$ Then $$\frac{\pi}{128}J(n,\kappa) \sim \Big(\kappa-\sqrt{\kappa^2-1}\Big)^{2n} \frac{\big(\kappa^2-1)^{3/2}}{4\sqrt{2}} \int_{-\infty}^\infty \exp{\Big(\frac{-8\kappa \,n}{\sqrt{\kappa^2-1}}(u-1/2)^2 \Big)} du$$ Use of the Gaussian integral and algebra completes the proof.

The leading order exponential can be (almost) found by Paley-Wiener (which also applies to Fourier series).

Your integrand extends to a complex analytic function on the strip $$|\Im x|, |\Im y| < a = \cosh^{-1}(k) = \ln(k + \sqrt{k^2 - 1})$$ So the Payley-Wiener theorem says that (after performing the Fourier transform separately in $$x$$ and $$y$$) for such functions the Fourier series has the decay $$I(n) \lesssim_{\epsilon} e^{-2an + n\epsilon}$$ for any $$\epsilon > 0$$. (The factor of 2 coming from the fact that you are looking at the Fourier series in $$x$$ times the Fourier series in $$y$$.) Plugging in you get

$$I(n) \lesssim_\epsilon \left(\frac{1}{k + \sqrt{k^2 - 1}} \right)^{2n} \cdot e^{n\epsilon}$$

(Note that $$(k + \sqrt{k^2 - 1})^{-1} = k - \sqrt{k^2 - 1}$$.)

For the precise leading order you will probably need to do some contour integrals by hand.

• Thanks! One quick question: what does $\lesssim_\varepsilon$ mean? – Ludwig Feb 25 at 14:29
• bounded above by, up to a constant depending on $\epsilon$. (So I guess I could've written: for every $\epsilon > 0$ there exists a $C(\epsilon)$ (independent of $n$) such that $I(n) \leq C(\epsilon) e^{-2an + n\epsilon}$.) In general the constant $C(\epsilon)$ may blow-up as $\epsilon \searrow 0$; in your case since you have a very explicit integrand you may be able to show by doing the contour integral that there is a uniform bound on $C(\epsilon)$ in which case you can get rid of the $e^{n\epsilon}$ loss. I am not sure how to get the $\sqrt{n}$ factor though. – Willie Wong Feb 25 at 14:31
• I see, thanks. So, assuming that I could get rid of the $e^{n\epsilon}$ term, what I get is the upper bound $I(n)\le (k-\sqrt{k^2-1})^{2n}$. Is it possible to show that such a bound is tight? (btw, I'm not completely sure about the $\sqrt{n}$ factor: it's just a numerical guess) – Ludwig Feb 25 at 14:51
• In terms of rates: yes. Paley-Wiener is if-and-only-if: is if you have a strictly faster exponential decay rate, it will contradict the fact that your integrand has poles when \cos(x) = \cos(y) = k. – Willie Wong Feb 25 at 16:23
• math.lsa.umich.edu/~rauch/555/fouriercomplex.pdf gets to it very quickly (as an exercise with hints). – Willie Wong Feb 25 at 18:42