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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.

doraemonpaul
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Henrique Tyrrell
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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 08 '12 at 00:32
  • you had the cos multiplying the exponential, it has been fixed – Jean-Sébastien Nov 08 '12 at 00:32

4 Answers4

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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...

Robert Israel
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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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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!

Siddhant Das
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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 '15 at 21:03
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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

Michel
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