Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The title says everything. I'm studying fourier series and I've stumbled upon this question:

find the fourier series of $f(x) = e^{r\cos x} \cos(r\sin x)$. So that i need to integrate this function from $-\pi$ to $\pi$

I've tried integration by parts and a few u substitutions and got nowhere.

share|cite|improve this question
what do you mean it doesn't agree? To find the fourier series you have to find the Fourier coeficients and to do that you have to integrate the function and integrate that function times some cosine. – Henrique Tyrrell Nov 8 '12 at 0:32
you had the cos multiplying the exponential, it has been fixed – Jean-Sébastien Nov 8 '12 at 0:32

Hint: first note that $f(x)$ is the real part of $e^{r \cos x} e^{i r \sin x} = e^{r e^{ix}}$. Expand the "outer" exponential in a series...

share|cite|improve this answer
The "outer" is the first one? – Henrique Tyrrell Nov 8 '12 at 0:41
Yes, the first $\exp$ in $\exp(r \exp(ix))$. – Robert Israel Nov 8 '12 at 0:42


Look up Bessel functions. We have $$J_r(x) = \dfrac1{2\pi} \int_{-\pi}^{\pi} e^{-i (r \tau - x \sin(\tau))} d \tau$$

share|cite|improve this answer

This integral can be obtained in closed form. I have written the complete answer on Quora. The link is posted below.


share|cite|improve this answer
Can you please include the contents of the post you link to, or at least a sketch of it, in your answer? Link only answers are generally unwelcome here. If the link goes away, such an answer becomes entirely useless - although that is probably not likely to happen in this case. Still, we prefer to have the contents on site, and link-only answers run a risk of being deleted. – Daniel Fischer May 27 '15 at 21:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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