# Computing an Eisenstein series using complex analysis.

I'm havig difficulty with solving an exercise involving the computation of an Eisenstein series.

In this exercise we consider the cotangent function $$\text{cot(z)}=\frac{\cos(z)}{\sin(z)}$$.

a) Show that $$\text{cot}$$ has simple poles at all integer multiples of $$\pi$$ and is defined on all other points $$\mathbb{C} \setminus \pi \mathbb{Z}$$.

b) For fixed $$z_0 \in \mathbb{C} \setminus \mathbb{Z}$$ consider the function

$$f(z)=\frac{\pi \text{cot}( \pi z)}{z_0-z}.$$ Calcute the residues of $$f$$ at all of its singularities. (Hint: Use $$\underset{z=z_0}{ \text{res}} \ f(z)=\lim_{z \to z_0} (z-z_0) f(z)$$).

c) Show that $$\lim_{N \to \infty} \oint_{|z|=N+\frac{1}{2}} f(z)dz=0$$. (Hint: Consider the integral $$\oint_{|z|=N+ \frac{1}{2}} z^{-1} \pi \text{cot}(\pi z) dz$$.)

d) Use your results to calculate, for $$z \in \mathbb{C} \setminus \mathbb{Z}$$,

$$\lim_{N \to \infty} \sum_{n=-N}^{N} \frac{1}{z-n}.$$

e) (Bonus) Use the methods from the problems above to give a proof that

$$\zeta(2)=\frac{\pi^2}{6}.$$

I have problems with c) and d). I considered the integrals

$$\oint_{|z|=N+\frac{1}{2}} \frac{\pi \text{cot}( \pi z)}{z_0-z} dz \tag{1} \\$$

$$\oint_{|z|=N+\frac{1}{2}} \frac{\pi \text{cot}( \pi z)}{z} dz \tag{2}.$$

I know from b) that the residues of $$f$$ are

$$\underset{z=z_0}{\text{res}}f(z)=- \pi \text{cot}(\pi z_0).$$

$$\underset{z=k \pi}{\text{res}}f(z)=\frac{1}{z_0-k}, \ \ k \in \mathbb{Z}.$$

Moreover, it is clear that the integrand of (2) is almost the integrand of (1) (to see this set $$z_0=0$$ in the integrand of (1) and multiply by (-1)) and threfore the two integrals are almost equal. Given that I know the residues of $$f$$ I think I am supposed to use the residue theorem to compute (2) and then use the result to compute (1). But I do not see how to exactly proceed. I also do not see how to solve d) with help of the previous results. The only thing I see is that the denominator of $$\frac{1}{z-n}$$ resembles the denominator of $$f$$ so I expect the series will translate into an expression involving the integral computed in c). But so far I have not found an explicit strategy.

\Edit: Corrected the result for the residues on the real line.

## 1 Answer

First of all, the series $$\sum_{k\in} \dfrac{1}{z-k}$$ does not converge: you have to rearrange the terms ($$k$$ with $$-k$$) to have a convergent form.

Then, show that:

• your two functions are meromorphic
• they have the same pole
• the pole are of same degree
• the difference is a constant
• compute that constant with a suitable limit (I think you should look at the limit while $$t\to \infty$$ of the difference evaluated in $$it$$)

Thus, you have two different writings of the same function. One would give you $$\zeta(2)$$, the other $$\frac{\pi^2}{6}$$.

Edit I forgot to say what are the two functions. The one is the series and the other is $$z \mapsto \pi\mathrm{cotan} (\pi z)$$.