Cauchy's integral formula reads
$$ \frac{1}{2\pi i} \oint \frac{dz}{(z-z_0)}f(z) = f(z_0)$$
So, my question is what is the results for the following integral
$$ \frac{1}{2\pi i} \oint \frac{dz}{\prod_i(z-z_i)}f(z) $$, if it exists of course? Is it just the sum of the contours over each pole separately?
Here is what you are looking for $$\frac{1}{2\pi i} \oint \frac{dz}{\prod_i(z-z_i)}=\frac{1}{2\pi i}\sum_i \oint\frac{\alpha_i dz}{(z-z_i)} $$ where $$\frac{1}{\prod_i(z-z_i)} = \sum_i \frac{\alpha_i dz}{(z-z_i)}$$
for example $$\frac{1}{(z-1)(z-2)}=\frac{1}{(z-2)}-\frac{1}{(z-1)} $$