# Complex integral is zero but fn. is not analytic

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

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: $$\int\limits_{[0,2\pi)}\frac{\cos(e^{i\theta})}{e^{2i\theta}}ie^{i\theta}d\theta = i\int\limits_{[0,2\pi)}\frac{\cos(e^{i\theta})}{e^{i\theta}}d\theta=0.$$

Thanks for that. So if the closed integral is zero it doesn't prove that the function is analytic ? Does it infer anything ?
Thanks

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:

• 1 person
mathwonk
This is true. 