Sommerfeld-Watson Transform

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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 $C$.


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


We know from the Residue Theorem:

\(\ds \oint_C f \left({z}\right) \, \mathrm d z\) \(=\) \(\ds 2 \pi i \, \sum \limits_{z_k} R_k(z_k)\)
\(\ds \) \(=\) \(\ds 2 \pi i \,\sum_{z_k} \lim_{z \to z_k} \left({\left({z - z_k}\right) \frac {f \left({z}\right)} {\sin \pi z} }\right)\)

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:

\(\ds \oint_C f \left({z}\right) \, \mathrm d z\) \(=\) \(\ds 2 \pi i \sum_{z_k} \lim_{z \to z_k} \left({\frac {\partial_z \left({z - z_k}\right) f \left({z}\right)} {\partial_z \sin \pi z} }\right)\)
\(\ds \) \(=\) \(\ds 2 \pi i \sum_{z_k} \lim_{z \to z_k} \left({\frac {f \left({z}\right) + \left({z - z_k}\right) f' \left({z}\right)}{\pi \cos \pi z} }\right)\)

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

\(\ds \oint_C f \left({z}\right) \, \mathrm d z\) \(=\) \(\ds 2 \pi i \sum_{n \mathop = -\infty}^\infty \lim_{z \mathop \to n} \left({\frac {f \left({z}\right) + \left({z - n}\right) f' \left({z}\right)} {\pi \cos \pi z} }\right)\)
\(\ds \) \(=\) \(\ds 2 i \sum_{n \mathop = -\infty}^\infty \left({\frac {f \left({n}\right)} {\cos \pi n} }\right)\)


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


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


Source of Name

This entry was named for Arnold Johannes Wilhelm Sommerfeld and George Neville Watson.