I know how to do this using the residue theorem, but I do not know how to integrate it term by term.
I wrote $e^{1/z} = \sum_{n=0}^{\infty} \frac{e^{-in\theta}}{n!}$, and
$$ \int e^{1/z}dz = \sum_{n=0}^{\infty} \int_{0}^{2\pi}\frac{e^{-in\theta}}{n!}ie^{i\theta}d\theta = \sum_{n=0}^{\infty}\frac{i}{n!}\int_{0}^{2\pi}e^{(1-n)i\theta}d\theta. $$
Let $I = \int_{0}^{2\pi}e^{(1-n)i\theta}d\theta$, $I = 0$ so the integral is $0$.
However, when I do the integral using the residue theorem, I got $2\pi i$. What is the problem?
If we define $$I_n = \int_0^{2\pi} e^{(1-n)i\theta}\,d\theta,$$ then $I_n = 0$ for any integer $n\neq 1$; however, $I_1 = 2\pi.$