Help me fix this residue integration

**geft** · November 8th 2010, 09:28 AM

$\int_{0}^{2\pi} \frac{d\theta}{37 - 12cos\theta} = \oint_{C} \frac{dz/iz}{37 - 12(\frac{z}{2} + \frac{1}{2z})}$
$= \oint_{C} \frac{dz/iz}{37 - 6z - \frac{6}{z}}$
$= \oint_{C} \frac{dz}{i(37z - 6z^{2} - 6)}$
$= \oint_{C} \frac{dz}{-i(6z^{2} - 37z + 6)}$
$= \frac{-1}{i} \oint_{C} \frac{dz}{(6z - 1)(z - 6)}$

$z_{1} = \frac{1}{6}, z_{2} = 6$

This is the part which I'm not really sure. I'm guessing the $z_{1}$ is outside because it's less than unity, but I start having trouble when it comes to identifying more complicated poles. What is the right way?

$Res = \left|\frac{1}{6z - 1} \right|_{z = 6} = \frac{1}{35}$
$\Rightarrow 2\pii(\frac{-1}{i})(\frac{1}{35}) = \frac{-2\pi}{35}$

The answer is not a negative, so it may be a careless mistake, but I've pored through the equations without spotting an error.

PS: Is \\ the tag for line break? I can't seem to get it to work.

**Ackbeet** · November 8th 2010, 09:48 AM

You need to compute this:

$\displaystyle-\frac{1}{i}(2\pi i)\text{Res}[\frac{1}{(6z-1)(z-6)},z\to 1/6]=-\frac{2\pi}{6}\left(\frac{1}{z-6}\right)\Big|_{z\to 1/6}=-\frac{2\pi}{6}\left(\frac{1}{1/6-36/6}\right)=-\frac{\pi}{3}\left(-\frac{6}{35}\right)=\frac{2\pi}{35}.$

**geft** · November 9th 2010, 12:28 AM

Why is z = 1/6 chosen?

What about this one? Picking the other pole just changes the residue sign. The answer is 2pi/3.

$\int_{0}^{2\pi} \frac{d\theta}{5 - 4sin\theta}$
$= \oint_{C} \frac{dz/iz}{5 - \frac{2z}{i} - \frac{2}{iz}}$
$= - \oint_{C} \frac{dz}{2z^{2} - 5iz - 2}$
$= - \oint_{C} \frac{dz}{(4z - 8i)(49 - 2i}$
$z_{1} = 2i, z_{2} = \frac{1}{2}$
$Res = \left|\frac{1}{4z - 2i} \right|_{z = 2i} = \frac{1}{6i}$
$\Rightarrow 2\pi i(\frac{1}{6i})(-1) = \frac{-\pi}{3}$

**Ackbeet** · November 9th 2010, 01:32 AM

You pick z = 1/6, because that pole is inside the contour, and the other one is not. Only poles inside a contour have any effect whatsoever on the value of the integral. The implied contour over which you're integrating is the unit circle. So the correct poles at which to evaluate the residues are the ones inside that contour. Make sense?

**geft** · November 9th 2010, 03:45 AM

How do you check if the pole is inside the contour? In my previous question, you told me to plot the circle, but the function here seems to be quadratic.

**HallsofIvy** · November 9th 2010, 04:04 AM

Do you know what the contour is? Whether the integrand is quadratic or not has nothing to do with the contour.

The original problem was to integrate $\int_{0}^{2\pi} \frac{d\theta}{37 - 12cos\theta}$
which you changed into the contour integral $\oint_{C} \frac{dz/iz}{37 - 12(\frac{z}{2} + \frac{1}{2z})}$ . How did you do that?
What was "z" in terms of " $\theta$ ". As $\theta$ goes from 0 to $2\pi$ , how does z change?

Thread: Help me fix this residue integration

LinkBack

Thread Tools

Display

Help me fix this residue integration

Similar Math Help Forum Discussions

residue

residue

Choosing a contour of integration to use the residue theore

Inverse Fourier Transform by Residue Theorem and Contour Integration

Integration using residue

Search Tags