# Fourier series of $f(x)=e^{\cos(x^2)}$

I would like to write the $2\pi$-periodic function $f(x)=e^{\cos(x^2)}\;$ $(0 \leq x \leq 2\pi)\;$ as a Fourier series, but I am unable to carry out the integration. In order to write it as a Fourier series, I try to compute the Fourier coefficients

$$c_n = \frac{1}{2\pi} \int_0^{2 \pi} e^{\cos(x^2)} e^{-inx}dx.$$

This is not a homework problem. However, it is taken from a set of problems in a course I took a couple of years ago. I am trying to refresh my skills in Fourier analysis.

Even for $n=0$, this does not look like a function with an elementary function antiderivative. There might be some special trick given the limits of intergration. –  alex.jordan Jul 21 '12 at 17:59
@alex.jordan WolframAlpha says that $c_1 = 0$. At the moment I don't see why. –  Cocopuffs Jul 21 '12 at 18:24
Wait! This is not a $2\pi$-periodic function: $e^{\cos(x^2)}\neq e^{\cos((x+2\pi)^2)}$. Or do you mean to restrict it to the interval $[0,2\pi]$ and then induce a periodic function? –  alex.jordan Jul 21 '12 at 19:23
Maybe it would help if you wrote $e^{\cos(x^2)} = e^{\frac 12 e^{ix^2}}e^{\frac 12 e^{-ix^2}}$, then used the Taylor series of $\exp$ to get $e^{\cos(x^2)} = \sum a_k e^{ikx^2}$ for some $a_k$ and went on from there? At least, this would be an easy way to do it, if you had to find the Fourier series of $e^{\cos(x)}$... –  Sam Jul 21 '12 at 21:53