Summation Formula (Complex Analysis)

Theorem
Let $C_N$ be the square with vertices $\left({N + \frac 1 2}\right) \left({\pm 1 \pm i}\right)$ for some real $N > 0$.

Let $f$ be a function meromorphic on $C_N$.

Let $\left\vert{f\left({z}\right)}\right\vert < \dfrac M {\left\vert{z}\right\vert^k}$, for constants $k > 1$ and $M$ independent of $N$, for all $z$ on $C_N$.

Let $X$ be the set of poles of $f$.

Then:
 * $\displaystyle \sum_{n \in \Z \setminus X} f \left({n}\right) = - \sum_{z_0 \in X} \operatorname{Res} \left({\pi \cot \left({\pi z}\right) f \left({z}\right), z_0}\right)$

If $X \cap \Z = \O$, this becomes:


 * $\displaystyle \sum_{n \mathop = -\infty}^\infty f \left({n}\right) = - \sum_{z_0 \in X} \operatorname{Res} \left({\pi \cot \left({\pi z}\right) f \left({z}\right), z_0}\right)$

Proof
By Summation Formula: Lemma, there exists a constant $A$ such that:


 * $\displaystyle \left\vert{\cot \left({\pi z}\right)}\right\vert < A$

for all $z$ on $C_N$.

Let $X_N$ be the poles of $f$ contained in $C_N$.

From Poles of Cotangent Function, $\cot \left({\pi z}\right)$ has poles at $z \in \Z$.

Let $A_N = \left\{n \in \Z : -N \le n \le N\right\}$

We then have:

We then have:

We also have:

So, taking $N \to \infty$:


 * $\displaystyle 0 = 2 \pi i \left({\sum_{n \mathop \in \Z \setminus X} f \left({n}\right) + \sum_{z_0 \mathop \in X} \operatorname{Res} \left({\pi \cot \left({\pi z}\right) f \left({z}\right), z_0}\right)}\right)$

Giving:


 * $\displaystyle \sum_{n \mathop \in \Z \setminus X} f \left({n}\right) = - \sum_{z_0 \mathop \in X} \operatorname{Res} \left({\pi \cot \left({\pi z}\right) f \left({z}\right), z_0}\right)$