Evaluating $\int ^{2\pi} _{0} e^{-it} e^{( e^{it})} dt $

Question

I came across this question. Could you help me?

$$\int ^{2\pi} _{0} e^{-it} e^{( e^{it})} dt$$

The answer is $2\pi i$.

I was trying to use the Cauchy's integral formula but it was harder than the previous ones.

Answer 1

By expanding $e^{e^{it}}$ as a Taylor series in $e^{it}$, $$\int_{0}^{2\pi}e^{-it}e^{e^{it}}\,dt = \sum_{n\geq 0}\frac{1}{n!}\int_{0}^{2\pi} e^{(n-1)it}\,dt = \color{red}{2\pi} $$ (there is no $i$ in the final outcome) since the integral $\int_{0}^{2\pi}e^{mi\theta}\,d\theta$ equals zero for any $m\in\mathbb{Z}\setminus\{0\}$.