# On the Residue Theorem

On the Residue Theorem Swapnil Sunil Jain December 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 $z_{0},z_{1},...,z_{n}$ then

 $\displaystyle\int_{\gamma}f(z)dz$ $\displaystyle=$ $\displaystyle 2\pi i\sum_{k=0}^{n}Res_{z=z_{k}}f(z)$

where $Res_{z=z_{k}}f(z)$ 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 $p$, $Res_{z=p}f(z)=\lim_{z\to p}\Big{[}(z-p)f(z)\Big{]}$

For a double pole $p$, $Res_{z=p}f(z)=\lim_{z\to p}\Big{[}\frac{d}{dz}(z-p)^{2}f(z)\Big{]}$

For a n-tuple pole $p$, $Res_{z=p}f(z)=\lim_{z\to p}\Big{[}\frac{1}{(n-1)!}\frac{d^{(n-1)}}{dz^{(n-1)}}% (z-p)^{n}f(z)\Big{]}$

## 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 $\frac{dz}{d\theta}=ie^{i\theta}=iz$ 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

 $\displaystyle\int_{\gamma}g(z)\frac{dz}{iz}$ (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

 $\displaystyle 2\pi i\sum_{k=0}^{n}Res_{z=z_{k}}(\frac{g(z)}{iz})=\frac{2\pi i}% {i}\sum_{k=0}^{n}Res_{z=z_{k}}(\frac{g(z)}{z})=2\pi\sum_{k=0}^{n}Res_{z=z_{k}}% (\frac{g(z)}{z})$

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

Title On the Residue Theorem OnTheResidueTheorem1 2013-03-11 19:29:41 2013-03-11 19:29:41 swapnizzle (13346) (0) 1 swapnizzle (0) Definition