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Complex integral is zero but fn. is not analytic

  1. Aug 6, 2013 #1

    ppy

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    Hi
    I have been using a textbook which shows that ∫cos z/z^2 around the circle |z|=1 is zero by doing a Laurent expansion and finding the residue is zero.
    I was under the impression that only analytic functions have a integral of zero around a closed surface. ( the Cauchy-Goursat Theorem ). The above function is not analytic at z=0 so can non-analytic functions have a closed integral of zero ?

    Thanks
     
  2. jcsd
  3. Aug 6, 2013 #2
    Of course. Observe: $$\int\limits_{[0,2\pi)}\frac{\cos(e^{i\theta})}{e^{2i\theta}}ie^{i\theta}d\theta = i\int\limits_{[0,2\pi)}\frac{\cos(e^{i\theta})}{e^{i\theta}}d\theta=0.$$
     
  4. Aug 6, 2013 #3

    ppy

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    Thanks for that. So if the closed integral is zero it doesn't prove that the function is analytic ? Does it infer anything ?
    Thanks
     
  5. Aug 6, 2013 #4
    Not really.

    It implies that the contour integral is 0, though. :tongue:
     
  6. Aug 6, 2013 #5

    mathwonk

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    you are missing a hypothesis of morera's theorem that the integrand be continuous.
     
  7. Aug 6, 2013 #6
    This is true. :redface:

    If the integrand is continuous and the integral is 0, then, with a couple of other conditions, the integrand is holomorphic.
     
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