Summation Formula (Complex Analysis)

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Theorem

Let $C_N$ be the square with vertices $\paren {N + \frac 1 2} \paren {\pm 1 \pm i}$ 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:

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


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

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


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:

\(\displaystyle \oint_{C_N} \pi \cot \paren {\pi z} \map f z \rd z\) \(=\) \(\displaystyle 2 \pi i \sum_{z_0 \mathop \in X_N \cap A_N} \Res {\pi \cot \paren {\pi z} \map f z} {z_0}\) Residue Theorem
\(\displaystyle \) \(=\) \(\displaystyle 2 \pi i \paren {\sum_{n \mathop \in A_N \setminus X_N} \Res {\pi \cot \paren {\pi z} \map f z} n + \sum_{z_0 \in X_N} \Res {\pi \cot \paren {\pi z} \map f z} {z_0} }\)

We then have:

\(\displaystyle \Res {\pi \cot \paren {\pi z} \map f z} n\) \(=\) \(\displaystyle \lim_{z \mathop \to n} \paren {\paren {z - n} \pi \cot \paren {\pi z} \map f z}\) Residue at Simple Pole
\(\displaystyle \) \(=\) \(\displaystyle \map f n \lim_{z_0 \mathop \to n} \paren {\frac {z - n} z + 2 \sum_{k \mathop = 1}^\infty \frac {z \paren {z - n} } {z^2 - k^2} }\) Mittag-Leffler Expansion for Cotangent Function
\(\displaystyle \) \(=\) \(\displaystyle \map f n \cdot 2 \lim_{z \mathop \to n} \paren {\frac z {z + n} }\)
\(\displaystyle \) \(=\) \(\displaystyle \map f n \frac {2 n} {2 n}\)
\(\displaystyle \) \(=\) \(\displaystyle \map f n\)

We also have:

\(\displaystyle \cmod {\oint_{C_N} \pi \cot \paren {\pi z} \map f z \rd z}\) \(\le\) \(\displaystyle \frac{\pi A M \paren {8N + 4} } {N^k}\) Estimation Lemma
\(\displaystyle \) \(=\) \(\displaystyle \frac {8 \pi A M} {N^{k - 1} } + \frac {4 \pi A M} {N^k}\)
\(\displaystyle \) \(\to\) \(\displaystyle 0\) as $N \to \infty$, as $k - 1 > 0$

So, taking $N \to \infty$:

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

Giving:

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

$\blacksquare$


Sources