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I have an integral of the following formbehavior $$ \int_0^{\infty}\frac{dk}{2\pi^2}\frac{\sin kr}{kr}k^2\exp\left[ -\frac{1}{2}(k R)^n \right] $$ I need to determine the asymptotic behavior for the integral at large distance $s\equiv r/R\rightarrow \infty$. For $n=1$ and $n=2$, it is very easy to solve these integrals to determine the asymptotics. For example, $n=1$ gives a Lorentzian and $n=2$ a Gaussian. Now the question is whether there is a way to generalize this for some arbitrary $n$? I used Mathematica to compute this for $n=$ and the result was in the form of generalized hypergeometric function and the asymptotic behavior was difficult to deduce.

Judas503
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  • Seems like the first thing to do is substitute $u = kR$ to get $1/(2\pi^2R^3 s)\int_0^\infty \sin(s u)u\exp(-u^n/2)du$. – eyeballfrog Jul 06 '22 at 21:16

1 Answers1

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Here's a partial answer valid for odd $n$. Even $n$ looks like the more interesting case.

Following the comment by @eyeballfrog let's look at the problem $$ F(s) = \int_0^\infty \sin(su) f(u) \, du $$ where $f(u) = u e^{-u^n/2}$. A `standard' approach to produce an asymptotic series for $F$ would be to apply integration by parts repeatedly to find $$ F(s) = -\frac{1}{s}\left[\cos(su)f(u)\right]_0^\infty + \frac{1}{s^2}\left[\sin(su)f'(u)\right]_0^\infty+ \frac{1}{s^3}\left[\cos(su)f''(u)\right]_0^\infty - \ldots $$ with the last term an integral that is asymptotically lower order than the last minus one due to Riemann--Lebesgue lemma.

Due to the behaviour of sines/cosines, even terms (corr. to even derivatives of $f$, and odd powers of $s$) in the above series are possibly nonzero. If we expand $f$ near $u=0$: $$ f(u) = u - \frac{u^{n+1}}{2} + \frac{u^{2n+1}}{2^2\cdot 2!} - \frac{u^{3n+1}}{2^3\cdot 3!} + \ldots $$ we see that if $n$ is even, all even derivatives of $f$ are zero, and so every term in the above expansion is zero. The integral will be asymptotically smaller than any power of $s$.

If $n$ is odd, the first nonzero term in the expansion of $F$ will be the one corresponding to the $(n+1)$th derivative of $f$, the second will correspond to the $(3n+1)$th derivative, and so on: $$ F(s) \sim \frac{(n+1)!(-1)^{(n-1)/2}}{2} \frac{1}{s^{n+2}} - \frac{(3n+1)!(-1)^{(3n-1)/2}}{2^3 3!} \frac{1}{s^{3n+2}} + \ldots $$

messenger
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    For even $n$, see this answer. For an even $n \geq 4$, we take a contour going from $-\infty$ to $\infty$ and passing through two stationary points of $\phi(u) = -u^n + 2 i u$ in such a way that $\operatorname {Re} \phi(u)$ is maximized at those points. For $n = 4$, we get $$F(s) = \sqrt {\frac \pi 3} \exp \left( -\frac {3 s^{4/3}} {2^{10/3}} \right) \left( \sin \left( \frac {3^{3/2} s^{4/3}} {2^{10/3}} \right)
    • o(1) \right).$$

    (For odd $n$, the main contribution comes from $[0, i \epsilon]$.)

     – Maxim Jul 07 '22 at 12:47
  • @Maxim I should've known there'd be a steepest descents answer! – messenger Jul 08 '22 at 00:19