# 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?

• This answer might help. – John Jul 8 '14 at 18:53
• You can use Maple or Mathematica to get an answer for this integral. – Mhenni Benghorbal Jul 8 '14 at 19:30

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}