# When dealing with not simply connected sets how can I prove that a function has/doesn't have a primitive?

Contex: Recently I asked a question about if $$\frac{1}{z} - \frac{1}{z-1}$$ is holomorphic on $$0<|z|<1$$ and it got me thinking say I wanted to prove that its primitive exists/doesn't exist. I know techniques through which to prove the function has a primitive, for example, if it was on an open disc and is holomorphic then the primitive exists or for example if the set is simply connected I could use Caychy's Integral Theorem and then Morera's theorem. However, I do not know about any techniques where we are dealing with sets that are not necessarily simply connected.

My question: What are techniques to check if a function has a primitive on a set that is not simply connected? (for example what can I do to prove $$\frac{1}{z} - \frac{1}{z-1}$$ does/nt have a primitive on $$0<|z|<1$$)

For holomorphic functions, the existence of a primitive is equivalent to path-independence of all contour integrals. If a primitive $$F$$ of $$f$$ exists, then $$\int_\gamma f(z)\mathrm dz=F(b)-F(a)$$, where $$a$$ and $$b$$ are the start and end points of $$\gamma$$ (you can show this using the definition of the contour integral and the fundamental theorem of calculus). And for the reverse, if the contour integral is independent of the path, only depending on the start and end points, and the domain is connected, then $$F(z):=\int_{z_0}^z f(w)\mathrm dw$$ is a primitive, where the integral goes along any arbitrary path connecting a fixed $$z_0$$ to $$z$$. You you can show this directly via the definition of differentiability, which will give you $$F'(z)=f(z)$$. For disconnected domains, you can do the same construction for each connected component of the domain separately and define the primitive piecewise.
Also, path independence of the integral is equivalent to the property that integrals along closed contours all vanish. So a holomorphic function $$f:D\to\mathbb C$$ has an antiderivative if and only if all integrals along closed contours vanish. To show that an antiderivative doesn't exist, you can simply try to find such a contour (and such a contour is guaranteed to exist if there is no antiderivative).
• Hi, thank you very much for the answer! I have two questions, 1) would you just keep trying different closed paths until you find one for which $\oint_\gamma f \neq 0$? For example what would you choose for $\frac{1}{z} - \frac{1}{z-1}$ on $0<|z|<1$? I tried using circles but it did not work. 2) For the homotopy version of Caychy's theorem isn't it assumed that the set is simply connected (so that all closed paths are homotopic to each other)? Nov 14, 2020 at 19:06
• 1) I'd test for a circle around each hole in the domain. The only hole in your domain is $0$, so I'd try a circle centered at $0$ (it will be $2\pi\mathrm i$). 2) That's just the regular version. The homotopy version is stronger than the regular version, precisely because it also holds for arbitrary open sets, not just simply connected ones. Nov 14, 2020 at 19:10
• Hi, thank you for the answer. for 1) I still get 0 ($2\pi i - 2\pi i)$so I will try something else. 2) I think this could be a mistake on my part. In my book I have Caychy's Theorem first given as if $f$ is holomorphic on a set $U$ and there are paths $\gamma$ and $\beta$ such that there is a homotopy between them then $f$ has a primitive in the region between the two paths. The homotopy version is listed as I have stated above and then there is another version that I have not yet come across to that is called The Global Cauchy Thoerem which deals with open sets (no matter if simply connected) Nov 14, 2020 at 19:32
• The integral over $\frac{1}{1-z}$ should be $0$. You should take a closer look at that part. And about the homotopy version: the fact you mention holds on open sets. And from that fact you can deduce that the integrals along homotopic paths are equal, which is the fact I mention. Nov 14, 2020 at 22:14
• Thank you very much for the notes. I double-checked and you are right, $\oint \frac{1}{z-1}=0$ since $log(z-1)$ is its primitive. Nov 14, 2020 at 23:53