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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.

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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

3 Answers 3

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...

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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 $$J_r(x) = \dfrac1{2\pi} \int_{-\pi}^{\pi} e^{-i (r \tau - x \sin(\tau))} d \tau$$

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


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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 at 21:03

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