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

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?

• The answer is $\frac{2\pi}{n\color{red}{!}}$, not $\frac{2\pi}{n}$. – kobe Mar 11 '15 at 3:32

## 3 Answers

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 $\frac{2\pi}{n!}$.

• Perfect -- solved it easily after that! – user138798 Mar 12 '15 at 2:47

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

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

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

• Thanks for the hint Travis. I did try to use the subtitution, but did not know how to subsistute for cos (nx - sinx). – user138798 Mar 11 '15 at 2:58