defenite integral involve bessel function I have an integral which involves Bessel function as follows:
$I=\int_{r=0}^a \int_{\theta=0}^{2\pi}(e^{-jkr\cos(\theta-\phi)}d\theta)rdr$
I have tried with
$e^{-jkr\cos(\theta-\phi)}=\sum \limits_{m=-\infty}^{+\infty}J_m(kr)e^{jm\theta}e^{-jm(\phi+\frac{\pi}{2})}$
Finally I have arranged the integral as
$\sum \limits_{m=-\infty}^{+\infty}[e^{-jm(\phi+\frac{\pi}{2})}(\frac{1}{jm})(e^{j2\pi m}-1)\int_{r=0}^a J_m(kr)rdr]$
Now how to calculate the last integral?
 A: By making use of the expansion
\begin{align}
e^{- i kr \cos(\theta - \phi)} = \sum_{- \infty}^{\infty} i^{m} J_{m}(k r) \ e^{i m (\theta - \phi)} 
\end{align}
it can be seen that this is also
\begin{align}
e^{- i kr \cos(\theta - \phi)} = J_{0}(kr) + 2 \sum_{m=1}^{\infty} \cos m(\theta - \phi + \pi/2) \ J_{m}(kr).
\end{align}
Now the integral over $\theta$ yields
\begin{align}
\int_{0}^{2\pi} e^{- i kr \cos(\theta - \phi)} \ d\theta = 2 \pi J_{0}(k r) + 2 \sum_{m=1}^{\infty} \frac{1}{m} \left[ \sin m(2 \pi - \phi + \pi/2) - \sin m(\pi/2 - \phi) \right] \ J_{m}(kr).  
\end{align}
Since $\sin m(2 \pi - \phi + \pi/2) - \sin m(\pi/2 - \phi) = 0$ when $m$ is an integer, then the expression for the integral becomes
\begin{align}
\int_{0}^{2\pi} e^{- i kr \cos(\theta - \phi)} \ d\theta = 2 \pi J_{0}(k r).
\end{align}
Using the relation
\begin{align}
\int J_{0}(kr) \ r dr = \frac{r}{k} \ J_{1}(kr)
\end{align}
then the second integral becomes
\begin{align}
\int_{0}^{a} \int_{0}^{2\pi} e^{- i kr \cos(\theta - \phi)} \ d\theta \ r dr = \frac{2 \pi a}{k} \ J_{1}(ka). 
\end{align}
