(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Find [tex]I = \int_0^{2\pi} \frac{1}{cos\phi+b} d\phi[/tex]

2. Relevant equations

Given above..

3. The attempt at a solution

This problem is an introductory problem to trigonometric functions and here is how the answer is obtained - but I have a question about it. First, here is the solution.

Use complex analysis. Make the substitution [tex]z=e^{i\phi}[/tex] and [tex]cos\phi=\frac{e^{i\phi}+e^{-i\phi}}{2}[/tex] to get [tex]\frac{1}{i}\oint_C {\frac{1}{z^2+2bz+1}dz}[/tex] integrated over the unit circle contour.

We now find the poles: [tex]z_\pm=-b\pm \sqrt{b^2-1}[/tex].

However, as b goes to infinity, you see that the [tex]z_-[/tex] goes to -infinity and [tex]z_+[/tex] goes to 0. Obviously, only [tex]z_+[/tex] stays inside the unit circle for |b|>1 so the solution is:

[tex]\frac{2\pi}{\sqrt{b^2-1}}[/tex]

This is fine and I have no problem with this. This only applies for |b|>1 though. What about for |b|<=1? Well, obviously the integral diverges if you look back at the original integral. Here's my problem - if you try to use contour integration for |b|<=1, you will have to include the residues of both poles because both poles will be on the unit circle as b decreases. Wouldn't the residues cancel giving an answer of zero? What if b is actually a function and you want to do further integrations afterward with respect to a different variable? Thanks for any help on this.

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# Homework Help: Complex Analysis of a trigonometric function integral

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