Evaluating $\int_0^{2 \pi} e^{\cos x} \cos (nx - \sin x) \,dx$ using complex analysis

Question

I'm taking a complex analysis course and doing some practice computing residues & evaluating integrals. I pulled out an old book called "The Cauchy Method of Residues: Theory and Applications, Volume I"

On page 196-197, there are some interesting integrals to evaluate. I'm at 5.4.3.10.: I was able to do question 1, but was stumped at how to even begin with question 2:

Evaluate the integral $$\int_0^{2 \pi} e^{\cos x} \cos (nx - \sin x) \,dx ,$$ where $n$ is an natural number.

The answer is simply $\frac{2 \pi}{n}$. Any hints?

Answer 1

Note

$$\int_0^{2\pi} e^{\cos x}\cos(nx - \sin x)\, dx = \operatorname{Re} \int_0^{2\pi} e^{\cos x + i(nx - \sin x)}\, dx = \operatorname{Re} \int_0^{2\pi} e^{e^{-ix}} e^{inx}\, dx.$$

Using the parametrization $z = e^{-ix}$, $0 \le x \le 2\pi$ for the unit circle $|z| = 1$, we have

$$\int_0^{2\pi} e^{e^{-ix}}e^{inx}\, dx = \frac{1}{i}\int_{|z| = 1} e^{z} \frac{dz}{z^{n+1}}.$$

Now show that this contour integral is $\frac{2\pi}{n!}$.

Answer 2

Hint: Substitute $z := e^{i x}$, which transforms the integral into a contour integral along the unit circle; we have, e.g., $$\cos x = \frac{1}{2}(e^{i x} + e^{-i x}) = \frac{1}{2} \left(z + \frac{1}{z}\right).$$

Answer 3

Since $ \cos (nx - \sin x) $ is the real part of the function $ e^{i (nx - \sin x) }$ then you should consider the integral

$$ I = \int_{0}^{2\pi}e^{ \cos x}e^{i (nx - \sin x) }dx = \int_{0}^{2\pi}e^{ i n x}e^{\cos x - i\sin x }dx =\int_{0}^{2\pi}e^{ i n x}e^{e^{-i x} }dx. $$