4
$\begingroup$

I would like to find if possible a solution (closed form) for the following integral:

$$\frac{1}{2 \pi}\cdot\int\limits_0^{2\pi}\exp\bigg[-ia(\cos x+\sin x)\bigg]\,j_{0}(b\cos x)\,j_{0}(b\sin x)\,\mathrm dx$$

where $a,b$ are positive real constants and $j_{0}$ is the spherical bessel function of order zero ($j_0(z)=sinc(z)$). Anyone has a clever substituion or a known integral that might help me in finding the solution?

$\endgroup$
4
  • $\begingroup$ Really strange. One might guess that $J_{0.5}(z)$ is the Bessel function of the first kind of order one half, also known as: $$J_{\frac{1}{2}}(z)=\sqrt{\frac{2}{\pi z}}\sin(z).$$ Do our notations agree? $\endgroup$ May 22, 2015 at 13:55
  • $\begingroup$ @JackD'Aurizio Yes it is. It is order one half, correct! $\endgroup$
    – JFNJr
    May 22, 2015 at 14:00
  • $\begingroup$ How do you define $J_{\frac{1}{2}}(z)$ when $z<0$? $\endgroup$ May 22, 2015 at 15:43
  • $\begingroup$ @JackD'Aurizio yes you are right again...I really confused the notations for bessel function and spherical bessel function. The text is now edited in the right way. $\endgroup$
    – JFNJr
    May 22, 2015 at 15:49

1 Answer 1

0
$\begingroup$

I would note that the j_0's are just sin y/y and so you can rewrite the numerators as proportional to exp[ibcos x] - exp[-ibcos x]. Then the whole integral will involve only complex exponentials and 1/(b cos x* b sin x). You can then pull out a and b such that you have factors of the form exp[i cos x]^b. A change of variable u = cos x will then convert this to simpler exponentials, and simplify the denominator; what results is straightforward to integrate.

$\endgroup$
1
  • $\begingroup$ I agree about everything but the very last words: the result is not straightforward at all to integrate. $\endgroup$ Sep 1, 2015 at 20:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .