How to integrate $e^{r\cos x} \cos(r\sin x)$ 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.
 A: 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...
A: HINT
Look up Bessel functions. We have $$J_r(x) = \dfrac1{2\pi} \int_{-\pi}^{\pi} e^{-i (r \tau - x \sin(\tau))} d \tau$$
A: This integral can be obtained in closed form. I have written the complete answer on Quora. The link is posted below.
http://www.quora.com/How-do-I-solve-the-integral-int-limits-pi-_-0-e-cos-x-cos-sin-x-mathrm-d-x
I am posting the method below for $r=1$. The exact same steps can be take for any real $r$.
This integral certainly exists. Let's begin by rewriting $$\cos(\sin x) = \frac{e^{i \sin x}+e^{-i\sin x}}{2},$$
obtaining
$$\int_{-\pi}^{\pi} e^{\cos x}\cos(\sin x)~\mathrm{d}x = 2\int_0^{\pi} e^{\cos x}\cos(\sin x)~\mathrm{d}x = 2\,\Re \left[\int_0^{\pi} e^{e^{i x}}~\mathrm{d}x\right].$$ 
Consider the integral 
$$\int_0^{\pi} e^{e^{i x}}~\mathrm{d}x=\int_0^{\pi} \sum_{n=0}^{\infty} \frac{e^{inx}}{n!}~\mathrm{d}x = \sum_{n=0}^{\infty} \int_0^{\pi}\frac{e^{inx}}{n!}~\mathrm{d}x$$
$$=\int_0^{\pi}~\mathrm{d}x + \sum_{n=1}^{\infty} \int_0^{\pi} \frac{e^{inx}}{n!}~\mathrm{d}x$$
$$= \pi - i \sum_{n=1}^{\infty} \frac{(-1)^n-1}{n^2 (n-1)!}$$
The summation can be obtained in closed form in terms of the hyperbolic sine integral or $\mathrm{Shi}(z)$ at $z=1$. However it is not required, as we are interested in the real part alone.
Thus, 
$$\int_{-\pi}^{\pi} e^{\cos x}\cos(\sin x)~\mathrm{d}x = 2\pi$$
Cheers!
A: you should first solve the integral of
exp(rexp(ix)) by using Laplace transform of 
this exponential function then you take 
real part of this complex integral and the
problem is solved :D
