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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

2 Answers 2

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

HINT

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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