Evaluating the integral of a $\cos(\theta)$ within the exponential wrt $\theta$ I want to evaluate the following integral
$\int^{2\pi}_0 \, d\theta e^{- i k (x - x')\cos{\theta}}$,
where all of the variables are real and $i$ is the imaginary unit. The difficulty is the cosine term within the exponent. Are there any techniques that one could suggest to reduce this to an integral which is easier to evaluate. I have seen similar posts but have not been able to relate the problem to them.
Thanks for your help!
 A: Well, if you let $u=k(x-x')$, we're looking at $\int_0^{2\pi}e^{iu\cos\theta}d\theta$. Using the fact that our integrand is $2\pi$-periodic, we know that this is also equal to $\int_0^{2\pi}e^{iu\cos(\theta+\pi)}d\theta=\int_0^{2\pi}e^{-iu\cos\theta}d\theta$. Taking the mean of these two, we're now calculating:
$$\int_0^{2\pi}\cos(u\cos\theta)d\theta=2\pi J_0(u)$$
where $J_0$ is the $0^{th}$ Bessel function.
One way we can justify this is by expanding our integrand as $\sum_n\frac{(-1)^nu^{2n}}{(2n)!}\cos ^{2n}\theta$ and using the result that $\int_0^{2\pi}\cos^{2n}\theta d\theta=2\pi\frac{2n\choose n}{2^{2n}}$. 
The interchange of summation and integration is well-justified as the series converges extremely quickly. For an explicit comparison, the triangle inequality lets us dominate the series by the series $\cosh\vert u\vert$, which converges uniformly on compact subsets of $\mathbb{R}$ (indeed, of $\mathbb{C}$). 
Putting these together, we get:
$$\sum_n[\frac{(-1)^nu^{2n}}{(2n)!}][2\pi\frac{2n\choose n}{2^{2n}}]=2\pi\sum_n\frac{(-1)^n(u/2)^{2n}}{(n)!^2}=2\pi J_0(u)$$
by comparing to the series for the Bessel function.
