Sommerfeld-Watson Transform

Theorem
Suppose a function $f(z)$ has isolated poles and goes to zero faster than $\dfrac{1}{|z|}$ as $|z| \to \infty$.

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

Proof
We know from the Residue Theorem:


 * $\displaystyle \oint_C dz \,f(z) = 2 \pi i \,\sum\limits_{z_k} R_k(z_k) = 2 \pi i \,\sum\limits_{z_k} \lim\limits_{z \to z_k} \left( (z-z_k) \frac{f(z)}{\sin \pi z}\right)$

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

Using l'Hôpital's rule:
 * $\displaystyle \oint_C dz \,f(z) = 2 \pi i \,\sum\limits_{z_k} \lim\limits_{z \to z_k} \left(\frac{ \partial_z(z-z_k) f(z)}{\partial_z \sin \pi z}\right)$


 * $\displaystyle \,\,\,\,\,\,\, = 2 \pi i \,\sum\limits_{z_k} \lim\limits_{z \to z_k} \left(\frac{f(z)+ (z-z_k)f'(z)}{\pi \cos \pi z}\right)$

But $\sin \pi z$ has poles at $z_k = n$ for some $n \in \mathbb{Z}$ which implies:
 * $\displaystyle \oint_C dz \,f(z) = 2 \pi i \,\sum\limits_{n = -\infty}^{\infty} \lim\limits_{z \to n} \left(\frac{ f(z) + (z-n) f'(z)}{\pi \cos \pi z}\right)$
 * $\displaystyle = 2 i \,\sum\limits_{n = -\infty}^{\infty}\left(\frac{f(n)}{\cos \pi n}\right)$

Finally $\dfrac{1}{\cos \pi n} = \cos \pi n = (-1)^n$, therefore:
 * $\displaystyle \frac{1}{2 i} \oint_C dz \,f(z)= \,\sum\limits_{n = -\infty}^{\infty}(-1)^n f(n)$