does $\ f(z) = e^{z^2} $ have an antiderivative in the whole complex plane?
I think it does not but I am having trouble explaining why.
Since e is differentiable everywhere in the complex plane but zero, wouldn't it work the same for anti derivatives ?
Note that for all $z\in \mathbb{C}$, we have $$e^{z^2} = \sum_{n=0}^{+\infty} \frac{z^{2n}}{n!}.$$ Hence, the antiderivative exists and is given by $$z\mapsto \sum_{n=0}^{+\infty} \frac{z^{2n+1}}{(2n+1)n!}.$$ Here we have used holomorphicity to compute the antiderivative term by term.