Sommerfeld-Watson Transform

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Theorem

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

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

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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:

\(\ds \oint_C \map f z \rd z\)

\(=\)

\(\ds 2 \pi i \, \sum \limits_{z_k} R_k(z_k)\)

\(\ds \)

\(=\)

\(\ds 2 \pi i \, \sum_{z_k} \lim_{z \mathop \to z_k} \paren {\paren {z - z_k} \frac {\map f z} {\sin \pi z} }\)

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

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Using l'Hôpital's rule:

\(\ds \oint_C \map f z \rd z\)

\(=\)

\(\ds 2 \pi i \sum_{z_k} \lim_{z \mathop \to z_k} \paren {\frac {\partial_z \paren {z - z_k} \map f z} {\partial_z \sin \pi z} }\)

\(\ds \)

\(=\)

\(\ds 2 \pi i \sum_{z_k} \lim_{z \mathop \to z_k} \paren {\frac {\map f z + \paren {z - z_k} \map {f'} z} {\pi \cos \pi z} }\)

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

\(\ds \oint_C \map f z \rd z\)

\(=\)

\(\ds 2 \pi i \sum_{n \mathop = -\infty}^\infty \lim_{z \mathop \to n} \paren {\frac {\map f z + \paren {z - n} \map {f'} z} {\pi \cos \pi z} }\)

\(\ds \)

\(=\)

\(\ds 2 i \sum_{n \mathop = -\infty}^\infty \paren {\frac {\map f n} {\cos \pi n} }\)

Finally:

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

Therefore:

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

$\blacksquare$

Source of Name

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