# Thread: Help me fix this residue integration

1. ## Help me fix this residue integration

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

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

This is the part which I'm not really sure. I'm guessing the $\displaystyle z_{1}$ 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?

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

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.

2. You need to compute this:

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

3. Why is z = 1/6 chosen?

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

4. 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?

5. 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.

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