Is it true that there exists a holomorphic function whose derivative is $1/(z^2-1)$? How to tackle problems like these?

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    $\begingroup$ Write it as a taylor series with help of the geometric series. Then integrate termwise. $\endgroup$ – Paul K Dec 14 '16 at 9:03
  • $\begingroup$ In what domain? $\endgroup$ – lhf Dec 14 '16 at 9:04
  • $\begingroup$ In $\Bbb C$ i guess, since it is complex analysis? But it is not specified in this problem. $\endgroup$ – J.doe Dec 14 '16 at 9:05
  • $\begingroup$ @menag Then what? $\endgroup$ – J.doe Dec 14 '16 at 9:06
  • $\begingroup$ Alright, I got my answer, which is false. Can someone confirm that? $\endgroup$ – J.doe Dec 14 '16 at 9:11

$\dfrac1{z^2-1}$ is holomorphic in the open unit disk and so has a primitive there.

The primitive is given explicitly by integrating the Taylor series: $$ \int \frac1{z^2-1} =-\int \sum_{n=0}^{\infty} z^{2n} =-\sum_{n=0}^{\infty} \frac{z^{2n+1}}{2n+1} $$

Both series converge in the open unit disk.

There is no primitive on the whole complex plane because the function is not even defined on the whole complex plane.

  • $\begingroup$ Wait.. so you are saying that this statement is true? $\endgroup$ – J.doe Dec 14 '16 at 9:22

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