Summation Formula (Complex Analysis)

Theorem
Let $C_N$ be the square with vertices $\paren {N + \dfrac 1 2} \paren {\pm 1 \pm i}$ for some $N \in \N$.

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

Let $\cmod {\map f z} < \dfrac M {\cmod z^k}$, for constants $k > 1$ and $M$ independent of $N$, for all $z \in \partial C_N$.

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

Then:
 * $\displaystyle \sum_{n \in \Z \setminus X} \map f n = - \sum_{z_0 \mathop \in X} \Res {\pi \cot \paren {\pi z} \map f z} {z_0}$

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


 * $\displaystyle \sum_{n \mathop = -\infty}^\infty \map f n = -\sum_{z_0 \mathop \in X} \Res {\pi \cot \paren {\pi z} \map f z} {z_0}$

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


 * $\displaystyle \cmod {\cot \paren {\pi z} } < A$

for all $z$ on $C_N$.

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

From Poles of Cotangent Function, $\cot \paren {\pi z}$ has poles at $z \in \Z$.

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

We then have:

We then have:

We also have:

So, taking $N \to \infty$:


 * $\displaystyle 0 = 2 \pi i \paren {\sum_{n \mathop \in \Z \setminus X} \map f n + \sum_{z_0 \mathop \in X} \Res {\pi \cot \paren {\pi z} \map f z} {z_0} }$

which gives:


 * $\displaystyle \sum_{n \mathop = -\infty}^\infty \map f n = -\sum_{z_0 \mathop \in X} \Res {\pi \cot \paren {\pi z} \map f z} {z_0}$