How does one approach to solving integrals like: $ \oint_C f(z)e^{1/z}dz$ where $C: |z| = 1$
Let's say f(z) is a complex function and it is defined inside $|z| = 1$, so the only problem is $z = 0$. Is there some way like the Cauchy integral formula? The function in my case is quite a pain and I cannot figure out how to solve it.
Let's assume that $f$ is analytic inside $C$ such that
$$f(z) = \sum_{n=0}^{\infty} f_n z^n$$
$\forall |z| \lt 1$. Then expanding the exponential in its Laurent series about the essential singularity at $z=0$, we have the integral being
$$\oint_C dz \, \sum_{n=0}^{\infty} f_n z^n \, \sum_{m=0}^{\infty} \frac1{m! z^m} = \sum_{n=0}^{\infty} f_n \sum_{m=0}^{\infty} \frac1{m!} \oint_C dz \, z^{n-m}$$
We can exchange the order of summation and integration because the series converge. Now we see that, by Cauchy's theorem, the integral is equal to zero except when $m=n+1$, when it equals $i 2 \pi$. Thus, the integral is equal to
$$\oint_C dz \, f(z) e^{1/z} = i 2 \pi \sum_{n=0}^{\infty} \frac{f_n}{(n+1)!} $$
and without further information about $f$, that's as far as one can take this.