Sommerfeld-Watson Transform

Theorem
Let $f \left({z}\right)$ be a mapping with isolated poles.

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

Let $C$ be a contour that is deformed such that all poles of $f \left({z}\right)$ are contained in $C$.

Then:
 * $\displaystyle \sum \limits_{n \mathop = -\infty}^\infty \left({-1}\right)^n f \left({n}\right) = \frac 1 {2 i} \oint_C \frac {f \left({z}\right)} {\sin \pi z} \, \mathrm d z$

Proof
We know from the Residue Theorem:

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

Using l'Hôpital's rule:

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

Finally:
 * $\dfrac 1 {\cos \pi n} = \cos \pi n = \left({-1}\right)^n$

Therefore:


 * $\displaystyle \frac 1 {2 i} \oint_C f \left({z}\right) \, \mathrm d z = \sum_{n \mathop = -\infty}^\infty \left({-1}\right)^n f \left({n}\right)$