Complex integral is zero but fn. is not analytic

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  • #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
 

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  • #2
611
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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
Of course. Observe: $$\int\limits_{[0,2\pi)}\frac{\cos(e^{i\theta})}{e^{2i\theta}}ie^{i\theta}d\theta = i\int\limits_{[0,2\pi)}\frac{\cos(e^{i\theta})}{e^{i\theta}}d\theta=0.$$
 
  • #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
 
  • #4
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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
Not really.

It implies that the contour integral is 0, though. :tongue:
 
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  • #5
mathwonk
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you are missing a hypothesis of morera's theorem that the integrand be continuous.
 
  • #6
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you are missing a hypothesis of morera's theorem that the integrand be continuous.
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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