Summation Formula (Complex Analysis)
Theorem
Let $N \in \N$ be an arbitrary natural number.
Let $C_N$ be the square embedded in the complex plane $\C$ with vertices $\paren {N + \dfrac 1 2} \paren {\pm 1 \pm i}$.
Let $f$ be a meromorphic function on $\C$ with finitely many poles.
Suppose that:
- $\ds \int_{C_N} \paren {\pi \cot \pi z} \map f z \rd z \to 0$
as $N \to \infty$.
Let $X$ be the set of poles of $f$.
Then:
- $\ds \sum_{n \mathop \in \Z \mathop \setminus X} \map f n = - \sum_{z_0 \mathop \in X} \Res {\pi \map \cot {\pi z} \map f z} {z_0}$
If $X \cap \Z = \O$, this becomes:
- $\ds \sum_{n \mathop = -\infty}^\infty \map f n = -\sum_{z_0 \mathop \in X} \Res {\pi \map \cot {\pi z} \map f z} {z_0}$
Proof
By Summation Formula: Lemma, there exists a constant $A$ such that:
- $\cmod {\map \cot {\pi z} } < A$
for all $z$ on $C_N$.
Since $f$ has only finitely many poles, we can take $N$ large enough so that no poles of $f$ lie on $C_N$.
Let $X_N$ be the set of poles of $f$ contained in the region bounded by $C_N$.
From Poles of Cotangent Function, $\map \cot {\pi z}$ has poles at $z \in \Z$.
Let $A_N = \set {n \in \Z : -N \le n \le N}$
We then have:
| \(\ds \oint_{C_N} \pi \map \cot {\pi z} \map f z \rd z\) | \(=\) | \(\ds 2 \pi i \sum_{z_0 \mathop \in X_N \mathop \cap A_N} \Res {\pi \map \cot {\pi z} \map f z} {z_0}\) | Cauchy's Residue Theorem | |||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \pi i \paren {\sum_{n \mathop \in A_N \mathop \setminus X_N} \Res {\pi \map \cot {\pi z} \map f z} n + \sum_{z_0 \mathop \in X_N} \Res {\pi \map \cot {\pi z} \map f z} {z_0} }\) |
We then have, for each integer $n$:
| \(\ds \Res {\pi \map \cot {\pi z} \map f z} n\) | \(=\) | \(\ds \lim_{z \mathop \to n} \paren {\paren {z - n} \pi \map \cot {\pi z} \map f z}\) | Residue at Simple Pole | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map f n \lim_{z \mathop \to n} \paren {\frac {z - n} z + 2 \sum_{k \mathop = 1}^\infty \frac {z \paren {z - n} } {z^2 - k^2} }\) | Mittag-Leffler Expansion for Cotangent Function | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map f n \cdot 2 \lim_{z \mathop \to n} \paren {\frac z {z + n} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map f n \frac {2 n} {2 n}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map f n\) |
Note that by hypothesis:
- $\ds \int_{C_N} \paren {\pi \cot \pi z} \map f z \rd z \to 0$
So, taking $N \to \infty$:
- $\ds 0 = 2 \pi i \paren {\sum_{n \mathop \in \Z \mathop \setminus X} \map f n + \sum_{z_0 \mathop \in X} \Res {\pi \map \cot {\pi z} \map f z} {z_0} }$
which gives:
- $\ds \sum_{n \mathop \in \Z \mathop \setminus X} \map f n = -\sum_{z_0 \mathop \in X} \Res {\pi \map \cot {\pi z} \map f z} {z_0}$
$\blacksquare$
Sources
- 1999: Jerrold E. Marsden and Michael J. Hoffman: Basic Complex Analysis (3rd ed.): $4.4.1$: Summation Theorem