Complex Analysis: Integrating rational functions

[Niles]

Homework Statement

Hi all.

My question has to do with integrating rational functions over the unit circle. My example is taken from here (page 2-3): http://www.maths.mq.edu.au/%7Ewchen/lnicafolder/ica11.pdf

We wish to integrate the following

[tex]
\int_0^{2\pi } {\frac{{d\theta }}{{a + \cos \theta }}}.
[/tex]

According to the .pdf, we integrate along a unit circle, and we define [itex] z = e^{i\theta}[/itex]. When rewriting the integrating using z, we find that the integrand has to poles: One inside the unit circle and one outside. When we use the residue theorem, we only use the pole inside the unit circle.

My question: How are we even allowed just to say: "We choose only to integrate along a unit circle, and thus we only look at poles inside this circle"? If I claim that we should integrate along a circle big enough to include all poles, then who can say argument against my claim?

I hope you understand me. Niles.

 

[Landau]

The reasoning is not "we must integrate along a unit circle, therefore we substitute z=e^(it)", but "let's substitute z=e^(it), then - since t ranges from 0 to 2*pi - we are integrating along a unit circle".
If you want to substitue z=Re^(it), so that you integrate along a non-unit circle, you could try that, but the question is whether that will work. The substitution z=e^(it) is easy to work with.

 

[Niles]

If you want to substitue z=Re^(it), so that you integrate along a non-unit circle, you could try that, but the question is whether that will work. The substitution z=e^(it) is easy to work with.

I have not yet seen a proof of that substituting z = exp(it) works as well, only that it apparently makes things easy.

 

[Landau]

It turns out to work as shown in the pdf you're referring to, right?

 

[Niles]

In the PDF the author uses z = exp(it), and I have seen this done in other notes on this topic as well. But none of the authors explain why they are allowed to do this other than it works out in the end.