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?

  • $\begingroup$ This answer might help. $\endgroup$ – John Jul 8 '14 at 18:53
  • $\begingroup$ You can use Maple or Mathematica to get an answer for this integral. $\endgroup$ – 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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.