Sommerfeld-Watson Transform

Theorem
Let $\map f z$ be a mapping with isolated poles.

Let $f$ go to zero faster than $\dfrac 1 {\size z}$ as $\size z \to \infty$.

Let $C$ be a contour that is deformed such that all poles of $\map f z$ are contained in $C$.

Then:
 * $\ds \sum \limits_{n \mathop = -\infty}^\infty \paren {-1}^n \map f n = \frac 1 {2 i} \oint_C \frac {\map f z} {\sin \pi z} \rd z$

Proof
From Cauchy's Residue Theorem:

This is for poles $z_k$ at order $N = 1$ because we say that simple poles exist for $\dfrac {\map f z} {\sin \pi z}$.

Using l'Hôpital's rule:

But $\sin \pi z$ has poles at $z_k = n$ for some $n \in \Z$ which implies:

Finally:
 * $\dfrac 1 {\cos \pi n} = \cos \pi n = \paren {-1}^n$

Therefore:


 * $\ds \frac 1 {2 i} \oint_C \map f z \rd z = \sum_{n \mathop = -\infty}^\infty \paren {-1}^n \map f n$