Integral with residue theorem complex variable: ${\large \int_0^{2\pi} \! e^{\alpha e^{it}} \, \mathrm{d}t }$

Question

I need help computing the following integral: $${\large \int_0^{2\pi} \! e^{\alpha e^{it}} \, \mathrm{d}t }$$ I know I can use the complex variable residue theorem, and I know it should give $2\pi$, but I have no idea what contour to use or how to use the residue theorem for this problem, since it has no singularities. Any help would be greatly appreciated.

Answer 1

Setting $z=e^{it}$, we have $$ dz=iz\,dt, $$ and since $z$ runs in the unite circle $\{z\in \mathbb{C}:\, |z|=1\}$ when $t$ runs in $[0,2\pi]$, we get thanks to the Residue Theorem: $$ \int_0^{2\pi}e^{\alpha e^{it}}\,dt=\int_{|z|=1}\frac{e^{\alpha z}}{iz}\,dz=2\pi i\mathrm{Res}\left(\frac{e^{\alpha z}}{iz};0\right)=2\pi. $$