Show that $\int_0^{2\pi}\int_0^t \frac r {\sqrt{t^2-r^2}} e^{-i k r \cos\theta} \, d\theta \, dr=2\pi \frac{\sin(k t)} k$

Question

I'm trying to compute this integral:$$\int_0^{2\pi} \int_0^t \frac r {\sqrt{t^2-r^2}} e^{-i k r \cos\theta} \, d\theta \, dr$$ I can perform an integration by $\theta$ in terms of Bessel function, but maybe there is a simpler way?

Answer 1

Here's an alternative way (whether it is simpler or not is probably a matter of taste):

First note that by symmetry your integral reduces to $$ I \equiv \int \limits_0^{2 \pi} \int \limits_0^t \frac{r}{\sqrt{t^2-r^2}} \mathrm{e}^{-\mathrm{i} k r \cos(\theta)} \, \mathrm{d} r \, \mathrm{d} \theta = 4 \int \limits_0^{\pi/2} \int \limits_0^t \frac{r}{\sqrt{t^2-r^2}} \cos( k r \cos(\theta)) \, \mathrm{d} r \, \mathrm{d} \theta \, . $$ Now let $r = t \sqrt{u}$ and $\cos(\theta) = \sqrt{v}$ to find $$ I = t \int \limits_0^1 \int \limits_0^1 \frac{\cos(k t \sqrt{u v})}{\sqrt{(1-u)(1-v)v}} \, \mathrm{d} u \, \mathrm{d} v \, . $$ We can expand the cosine and interchange summation and integration by the dominated convergence theorem. Then using the beta function we get \begin{align} I &= t \sum \limits_{n=0}^\infty \frac{(-1)^n}{(2n)!} (kt)^{2n} \int \limits_0^1 u^n (1-u)^{-\frac{1}{2}} \, \mathrm{d} u \, \int \limits_0^1 v^{n-\frac{1}{2}} (1-v)^{-\frac{1}{2}} \, \mathrm{d} v \, \\ &= t \sum \limits_{n=0}^\infty \frac{(-1)^n}{(2n)!} (kt)^{2n} \operatorname{B} \left(n+1, \frac{1}{2}\right) \operatorname{B}\left(n+\frac{1}{2},\frac{1}{2}\right) \\ &= t \sum \limits_{n=0}^\infty \frac{(-1)^n}{(2n)!} (kt)^{2n} \frac{n! \sqrt{\pi}}{\Gamma \left(n+\frac{3}{2}\right)} \frac{\Gamma \left(n+\frac{1}{2}\right) \sqrt{\pi}}{n!} \\ &= 2 \pi t \sum \limits_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} (kt)^{2n}\\ &= 2 \pi t \operatorname{sinc} (kt) \, . \end{align}

Answer 2

I don’t expect an easier method because to use the simplest type of Bessel function is easy. Others but more complicate methods of course exist.

I will only repeat here, what perhaps you've calculated. Maybe someone likes to compare it with other methods.

$\displaystyle \int\limits_0^{2\pi}\int\limits_0^t \frac{r}{\sqrt{t^2-r^2}} e^{-ikr\cos \theta } d \theta dr = 2\pi t \int\limits_0^1 \frac{r}{\sqrt{1-r^2}}\int\limits_0^1 e^{-iktr\cos(2\pi \theta)} d\theta \, dr $

It follows with $\,x:=kt\,$ and $\,\int\limits_0^1 e^{-ixr\cos(2\pi \theta)} d\theta = J_0(xr)\,$ :

$\displaystyle \int\limits_0^1 \frac{r}{\sqrt{1-r^2}} J_0(xr) dr = \sum\limits_{k=0}^\infty \frac{(-1)^k (x/2)^{2k}}{k!^2} \int\limits_0^1\frac{r^{2k+1}}{\sqrt{1-r^2}}dr = \sum\limits_{k=0}^\infty \frac{(-1)^k x^{2k}}{(2k+1)!} = \frac{\sin x}{x}$