On the Residue Theorem

On the Residue TheoremSwapnil Sunil JainDecember 26, 2006

On the Residue Theorem

The Residue Theorem

If $\gamma$ is a simply closed contour and f is analytic within the region bounded by $\gamma$ except for some finite number of poles $z_0,z_1,\mathrm{\dots },z_n$ then

${\displaystyle {\int }_{\gamma }}f(z)𝑑z$$=$$2\pi i{\displaystyle \sum _{k=0}^n}Re{s}_{z=z_k}f(z)$

where $Re{s}_{z=z_k}f(z)$ is the reside of $f(z)$ at $z_k$.

Calculating Residues

The Residue of $f(z)$ at a particular pole $p$ depends on the characteristic of the pole.

For a single pole $p$, $Re{s}_{z=p}f(z)={lim}_{z\to p}\left[(z-p)f(z)\right]$

For a double pole $p$, $Re{s}_{z=p}f(z)={lim}_{z\to p}\left[\frac{d}{dz}{(z-p)}^{2}f(z)\right]$

For a n-tuple pole $p$, $Re{s}_{z=p}f(z)={lim}_{z\to p}\left[\frac{1}{(n-1)!}\frac{d^{(n-1)}}{d{z}^{(n-1)}}{(z-p)}^{n}f(z)\right]$

Evaluation of Real-Valued Definite Integrals

We can use the Residue theorem to evaluate real-valued definite integral of the form

${\displaystyle {\int }_0^{2\pi }}f(\sin (n\theta ),\cos (n\theta ))𝑑\theta$(1)

If we let $z=e^{i\theta }$, then $\frac{dz}{d\theta }=i{e}^{i\theta }=iz$ which implies that $d\theta =\frac{dz}{iz}$. Then using the identity $\cos (n\theta )=\frac{1}{2}(z^n+z^{-n})$ and $\sin (n\theta )=\frac{1}{2i}(z^n-z^{-n})$, we can re-write (1) as

${\displaystyle {\int }_{\gamma }}g(z){\displaystyle \frac{dz}{iz}}$(2)

where $g(z)=f(\frac{1}{2i}(z^n-z^{-n}),\frac{1}{2}(z^n+z^{-n}))$ and $\gamma$ is a contour that traces the unit circle. Then, by the Residue theorem, (2) is equal to

$2\pi i{\displaystyle \sum _{k=0}^n}Re{s}_{z=z_k}({\displaystyle \frac{g(z)}{iz}})={\displaystyle \frac{2\pi i}{i}}{\displaystyle \sum _{k=0}^n}Re{s}_{z=z_k}({\displaystyle \frac{g(z)}{z}})=2\pi {\displaystyle \sum _{k=0}^n}Re{s}_{z=z_k}({\displaystyle \frac{g(z)}{z}})$

where $z_k$ are the poles of $\frac{g(z)}{z}$.