Compute the integral $ \int_{-\pi}^{\pi}e^{e^{ix}}e^{-ikx}\text{d}x $

Question

I was told to compute the following integral $$ \int_{-\pi}^{\pi}e^{e^{ix}}e^{-ikx}\text{d}x $$ where $k\in\mathbb{Z}$ but I don't know how to start. Should I expand the exponentials in terms of trigonometric functions or what is the trick here? Can I use theory about Fourier series to simplify this? Any help is welcome.

Answer 1

Hint: Let $z=e^{ix}$. The integral becomes a contour integral along the unit circle, so Cauchy integral formula applies.

Answer 2

Write $$e^{e^{ix}}=\sum_{n=0}^\infty \frac{e^{nix}}{n!}$$ and use the fact that $$\int_{-\pi}^\pi e^{nix} \, dx= \begin{cases}2\pi & n = 0\\ 0 & \text{otherwise} \end{cases}$$

Answer 3

Put

$$z:=e^{ix}\;,\;\;-\pi\le x\le \pi\implies dz=iz\,dx\,,\,\,|z|=1\implies dx=-\frac izdz\implies$$

$$\int_{-\pi}^\pi e^{e^{ix}}e^{ikx}\,dx=\oint_C e^zz^k\left(-\frac iz\,dz\right)=\frac1i\oint_Ce^zz^{k-1}dz= \begin{cases}0,&k\ge1\\{}\\ \cfrac{2\pi}{(-k)!},&k<1\end{cases}$$

With $\;C=\;$ the unit circle. Use Cauchy's Integral Formula