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Homework Help: Tough Integral

  1. Apr 21, 2012 #1
    1. The problem statement, all variables and given/known data

    Evaluate the integral

    2. Relevant equations

    3. The attempt at a solution

    So I'm essentially integrating log|z| around a circle of radius 1 centered at -1. Evaluating at the endpoints gives a singularity, but I feel like that shouldn't matter since they are only the endpoints. I guess my main issue is I don't know how to integrate log|z|. I was inclined to say that:

    [tex]\int_0^{2\pi}log|e^{i\theta}-1|d\theta = \int_0^{2\pi}log(1 - cos(\theta))d\theta.[/tex]

    Is this equality correct?

    I also tried u-du substitution with [itex]u = e^{i\theta} - 1[/itex] and [itex]d\theta = \frac{du}{i(u - 1)}[/itex], but then when I update the limits of integration I get that I'm integrating from zero to zero, I know there's a conceptual issue which explains why that happens and how rectify it, but I still don't understand it.

    Can anyone help me with this? Thanks.
  2. jcsd
  3. May 11, 2012 #2
    In the complex plane, log is a multivalued function. You can try defining a branch cut. I checked the solution on Mathematica, and it returned 0 as well.
  4. May 12, 2012 #3
    I don't think so. I get:

    [tex]\int_0^{2\pi} \log\left|e^{it}-1\right|dt=\int_0^{2\pi}\left[\log \sqrt{2}+1/2 \log \left(1-\cos(t)\right)\right] dt[/tex]
  5. May 12, 2012 #4
    \vert e^{i \theta} - 1 \vert = \sqrt{(\cos \theta - 1)^2 + \sin^2 \theta} = \sqrt{2 \, (1 - \cos \theta)} = \sqrt{2 \, 2 \, \sin^{2} \frac{\theta}{2}} = 2 \, \left\vert \sin \frac{\theta}{2} \right\vert

    But, if [itex]0 \le \theta \le 2 \pi[/itex], then [itex]\sin \frac{\theta}{2} \ge 0[/itex], and the absolute sign is unimportant.
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