Integral $\int_0^{2π} e^{e^{ix}} dx$

One trick I learnt a while ago on AoPS was to use Feynman's trick in this case.

Let $f(x)=f =e^{ix}$ for simplicity. And consider: $$I(t)=\int_0^{2π} e^{\large tf} dx\Rightarrow I'(t)=\int_0^{2\pi} f e^{\large tf}dx$$ But since $(tf)' = \left(te^{ix}\right)'=it e^{ix} =it f$

$$\Rightarrow I'(t)=\frac{1}{it} \int_0^{2\pi} \left(e^{tf}\right)'dx =\frac{e^{tf}}{it}\bigg|_0^{2\pi} $$ Now $e^{ix}$ is periodic with $T=2\pi $ and $e^{2\pi i} = e^0 =1$ so we have:$$I'(t)=\frac{e^t -e^t}{it}=0$$ And since if the derivative of a function is $0$ then the original function must be a constant, this implies that $I(t)$ is simply a constant, and we can set any value we want to obtain the answer. $$I(t)=I(0)=\int_0^{2\pi} dx=2\pi$$

For any $n\in\mathbb{Z}\setminus\{0\}$ we have $\int_{0}^{2\pi}e^{nix}\,dx = 0.$ It follows that

$$ \int_{0}^{2\pi}e^{e^{ix}}\,dx = \sum_{n\geq 0}\frac{1}{n!}\int_{0}^{2\pi}e^{nix}\,dx =\frac{1}{0!}\int_{0}^{2\pi}e^{0ix}\,dx=2\pi.$$

Use $z= e^{it}$ and integrate in the circumference of radius 1. Sou your integral becomes

$$ \int_{C} \frac{e^z}{iz}dz$$ which has a singularity in $z=0$ which is $\frac{1}{i}$ and using the resiude theorem then the integral is $2\pi i \; Res_0(f) = 2\pi i \frac{1}{i} = 2\pi$