Evaluate the following along the unit circle: $$I=\oint \frac{\cos(z)}{z(e^{z}-1)}dz$$
I tried doing it by $$f(z)=\frac{\cos(z)}{e^{z}-1}$$
Then the integral would be: $$I=2\pi if(0)$$
The problem is that $f(0)$ gives me $\frac{1}{0}$.
So how do I solve it?
This integral can be expressed with help of the residue theorem: $$I=\oint \frac{\cos(z)}{z(e^{z}-1)}dz=2\pi i \sum_{k}\; \mathrm{Res}(f,a_k)$$
There is only one removable singularity at $z=0$ and you should be able to find the redsidue of $f$ at $z=0$, just find the coeficient at $z^1$ of the following series expansion of $z^2 f(z)$: $$ z^2 \frac{\cos(z)}{z\left(e^{z}-1\right)}=\frac{z \cos(z)}{e^{z}-1}=1-\frac{z}{2}+O(z^2)$$ i.e. $$\mathrm{Res}\left(\frac{\cos(z)}{z\left(e^{z}-1\right)},0\right)=-\frac{1}{2}$$ and so $$I=\oint \frac{\cos(z)}{z(e^{z}-1)}dz=- i \pi$$
For a bit more involved calculation based on the Cauchy integral theorem see e.g. this answer or using simply the residue theorem see this one.