I have to simplify an expression which looks vaguely like:
$$\int_0^{\pi} e^{ir\sin{\theta}} F(\theta)\,d\theta$$
where $r$ is very very big. If $F(\theta)$ is asymptomatic, for very large $r$ the rapid oscillations in $e^{ir\sin{\theta}}$ will cause the integral to go to zero everywhere except for where $\frac{d}{d\theta}\sin{\theta}=0$. I seem to recall that the solution to such integrals involves replacing the oscillating part of the integrand with something involving delta functions. For this case in particular, it seems to be the case that:
$$\lim_{r\rightarrow\infty}e^{ir\sin{\theta}}\propto\frac{e^{ir}}{\sqrt{r}} \delta\left(\theta-\frac{\pi}{2}\right),\;\theta\in\left[0,\pi\right],$$
where this is all understood to be inside an integral. I was wondering if any of you would know how to procedurally evaluate such limits in this and similar cases.
Write $\sin(\theta)\approx 1-\frac12 (\theta-\pi/2)^2$ for $\theta$ near $\pi/2$. Using the Method of Steepest Decent we find that
$$\int_0^{\pi}e^{ir\sin(\theta)}F(\theta)\,d\theta\approx \sqrt{\frac{2\pi}{ir}}e^{ir}F(\pi/2)$$
as $r\to \infty$.
Note that it makes no sense to write
$$\lim_{r\to \infty}e^{ir\sin(\theta)}=\sqrt{\frac{e^{ir}}{r}}\delta(\theta-\pi/2)$$
since $r$ appears on the right-hand side. However, we can write in distribution
$$\lim_{r\to \infty}\sqrt{\frac{ir}{2\pi}}e^{ir(\sin(\theta)-1)}\sim \delta(\theta-\pi/2)$$
for $\theta\in [0,\pi]$. If we extend the domain of $\theta$, then the limit gives rise to a Dirac Comb at $\theta =\pi/2+n\pi$.