Studying for qualifying exams. Complex analysis identity

Question

I am currently studying for my qualifying exam and I am having real difficulty with the following question. I need to find the value of the following integral for the path below. $$\displaystyle\int\limits_\gamma \frac{2z+1}{e^{\pi z}-1}dz$$ I have attempted to use the residue theorem but I don't know how to break up the regions of the integral, or how to sum them. I am also having difficulty applying said theorem to this case. I have solved such things with unit circle paths but this one is tough. Any ideas on how to solve this?

**The path:

Answer 1

You could use residue theorem to compute this.

Residue Theorem: $$ \int_\gamma \frac{2z+1}{e^{\pi z}-1} dz = 2\pi i \sum\limits_{k=1}^n Res(f,a_k)$$ where $a_k$ are all possible singular points inside $\gamma$, and $Res(f,a_k)$ is the residue at $a_k$.

Here for this integrand $f=\frac{2z+1}{e^{\pi z}-1}$, there are only 2 singular points inside your $\gamma$, i.e. $a_1 = 0$, $a_2 = 2i$. We need to find the residue of them respectively.

For $a_1 = 0$, since $e^{\pi z}-1 = \sum_{n=1} \frac{(\pi z)^n}{n!}$, thus $Res(f,a_1) = \frac{2*0+1}{\pi} = \frac{1}{\pi}$.

For $a_2 = 2i$, since $e^{\pi z}-1 = e^{\pi (z-2i)} - 1 = \frac{(\pi z-2\pi i)^n}{n!}$, thus $Res(f,a_2) = \frac{2* 2i +1}{\pi}$.

So $$ \int_\gamma \frac{2z+1}{e^{\pi z}-1} dz = 2\pi i (\frac{1}{\pi} + \frac{4i+1}{\pi}) = -8+4i$$