# Evaluating $I=\oint \frac{\cos(z)}{z(e^{z}-1)}dz$ along the unit circle

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?

• On what curve are you integrating?
– Mark
Jul 22 '20 at 0:30
• Over which curve are you integrating? Jul 22 '20 at 0:30
• the curve is a circle of radius 1 Jul 22 '20 at 0:31
• centered at origin Jul 22 '20 at 0:31
• find residue at pole $z=0$
– user805287
Jul 22 '20 at 0:47

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.

• It is not a removable singularity at 0, is it?
– user581023
Jul 22 '20 at 7:21
• @rain1 After multiplying by (z-z0)^2 (here z0 =0) f(z) becomes holomorphic at z0, i.e. without singularity, this is why we expand it in Taylor series. Is it clear? Jul 22 '20 at 8:08
• I understand now, thank you
– user581023
Jul 22 '20 at 8:30
• @Sebasiano Thanks for the edit, however there should be $z^2 \frac{\cos(z)}{z \left(e^z -1\right)}= \frac{z \cos(z)}{ e^z -1}=1-\frac{z}{2}+O(z^2)$, this edit made a confusion. Jul 22 '20 at 19:42