We were discussing in our tutoring, whether it the following statement is true: If $f$ is holomorph AND has an antiderivative, than the domain itself must be a star domain. I opposed this with the above mentioned function.The function is basically defined on the whole complex plane except the line between i and -i. Now, if we chose ANY kind of closed path $\gamma$, the result of integral over that path is 0. If that path goes around the above mentioned line, the residuums of the singularities cancel each other out, so in summary, for every chosen closed path, the integral is 0, and hence the function must have an antiderivative; and domain is NOT a star domain. BUT, if we write the function as:

$\frac{1}{z^2+1}=\frac{1}{(z-i)(z+1)}=\frac{1}{2i}\left(\frac{1}{z-i}-\frac{1}{z+i}\right)$, and write the integral as:

$\int\limits_\gamma \frac{1}{z^2+1}\text{d}z=\int\limits_\gamma \frac{1}{2i}\left(\frac{1}{z-i}-\frac{1}{z+i}\right)\text{d}z$, we must define two seperate logarithims, and to do so, chose two seperate star domains. And hence, the antiderivative of the original function must be a star domain after all. And because of this, i am confused. Where do i make a mistake? Is the condition "the integral over all the closed paths in a domain must be zero" not enough to show the existence of an antiderivative?

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    $\begingroup$ the domain would be the intersection of the domains of those two logarithms. The intersection of two star domains (about different "center" points) need not be a star domain. $\endgroup$ Commented Jul 29, 2021 at 14:49
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    $\begingroup$ "ANY" is the keyword here.You considered the case where the contour contained both the singularity point (which is great as the sum of residues summed up to 0) but you did forget to include the case where the contour contains any one of the two singularity at a time (which would give a non zero value of the integral ). Therefore it is a star domain at the end . Hope you see it $\endgroup$ Commented Jul 29, 2021 at 16:48
  • $\begingroup$ @masterOFnone exactly! but if we "delete" the connecting line between -i and i (basically [-i,i]), then it wont be possible to chose a contour which contains only one of the singularities, and hence the integral becomes zero for any chosen contour. But, if we delete the line [-i,i] from the complex plane, it wont be a star domain $\endgroup$
    – Ozzy
    Commented Jul 29, 2021 at 17:00
  • $\begingroup$ @Ozan for the second part use Cauchy Integral Formula (ofcos after distributing integral over the partial fractions). Why complicate things. (π-π=0). What's the argument.?(mind you i maybe wrong but i don't see how) $\endgroup$ Commented Aug 2, 2021 at 14:41

1 Answer 1


The statement in the beginning of your message is of course wrong. And the simplest counterexample if a constant function: it has an anti-derivative in ANY domain whatsoever.

Concerning the function in your title, it indeed has an anti-derivative in $C\backslash[-i,i]$, namely a branch of arctangent. To see this, since the domain is not simply connected, one has to check that the residue at $\infty$ is zero.


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