Analytic function need not have a primitive

Question

I am told that an analytic function need not possess a primitive in its domain of analyticity. However, if $f$ is analytic in the disk $\text{B}(a,r)$, then it has a primitive in that disk. Suppose $f$ is analytic in an open set $G$. Then for each $a\in G$, $\exists r_a>0$ such that the disk $\text{B}(a,r_a)\subset G$. In each of these disks $f$ has a primitive. Why can't we always "stitch" these "local primitives" to obtain a primitive for $f$ in the set $G$? When can we do such a thing?

Answer 1

If $\Omega$ is simply connected, then every holomorphic function on $\Omega$ has an anti-derivative. This is a standard theorem proven in almost every textbook.

Conversely, if $\Omega$ is not simply connected, you can always find an example of a holomorphic function on $\Omega$ that doesn't admit an anti-derivative. The canonical example is $f(z) = 1/z$ on the punctured disk (or punctured plane), and this example is easy to adapt to any non-simply connected domain.

Note that if $f$ has an anti-derivative $F$, then $$ \int_\gamma f(z)\,dz = F(b)-F(a) $$ if $\gamma$ is a curve from $a$ to $b$. (This is the complex analogue of the fundamental theorem of calculus.) In particular, the integral of $f$ along any closed curve in $\Omega$ is $0$, whereas e.g. $$ \int_{|z|=1} \frac{dz}{z} = 2\pi i. $$