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Cauchy's Integral Formula Problem

  1. Jun 22, 2010 #1
    1. The problem statement, all variables and given/known data

    Use Cauchy's Integral Formula to compute the following integral:

    [tex]\int_{|z|=2\pi}\frac{\pi \sin(z)}{z(z-\pi)}dz[/tex]

    2. Relevant equations

    [tex]\int_C \frac{f(z)dz}{z-z_0}=2\pi if(z_0)[/tex]

    3. The attempt at a solution

    If we let

    [tex]f(z)=\frac{\pi \sin(z)}{z}[/tex]




    [tex]\int_{|z|=2\pi}\frac{\pi \sin(z)/z}{(z-\pi)}dz=0[/tex]

    Is this correct? I'm concerned about the fact that the function isn't differentiable at z=pi and that pi is an interior point of C...Part of me thinks that this integral cannot be evaluated with Cauchy's formula, but I do not yet understand the subject matter well enough to know for sure.
  2. jcsd
  3. Jun 22, 2010 #2
    I think you need to look at where the integral has the socalled simple pole

    and [tex]z= \pi[/tex] is a simple pole.
  4. Jun 23, 2010 #3
    Cheers Susanne! Going to try that angle :smile:
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