Why does $z\mapsto \exp(-z^2)$ have an antiderivative on $\mathbb C$?

Question

Why does $z\mapsto \exp(-z^2)$ have an antiderivative on $\mathbb C$?

So far I have seen the following results:

1. If $f\colon U\to\mathbb C$ has an antiderivative $F$ on $U$ then $\displaystyle\int_\gamma f(z)~\mathrm dz=F(\gamma(b))-F(\gamma(a))$ along a smooth curve $\gamma\colon[a,b]\to\mathbb C$.
2. If $f$ has an antiderivative then $\displaystyle\int_\gamma f(z)~\mathrm dz$ depends only on the start and end point of the curve $\gamma$.
3. If $f$ has an antiderivative, then $\displaystyle\oint_\gamma f(z)~\mathrm dz=0$ for all closed curves $\gamma$.

My problem is that I could use those results to show a function has no antiderivative (like $1/z$ yielding $2\pi\mathrm i\neq 0$ on $\partial B_r(0)$) but I am unsure what might work the other way round. What might be useful here in terms of complex analysis?

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EDIT

There is a follow-up question based on the problem above.

Use the result from above to calculate

$$\int_{-\infty}^\infty\exp(-x^2-\mathrm ikx)\,\mathrm dx, k\in\mathbb R.$$

I guess something constructive like GEdgar's suggestion might be useful here, isn't it?

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EDIT 2

Here is my solution for the follow-up question

$$ \begin{align*} \int_{-\infty}^\infty \exp(-x^2-\mathrm ikx)\,\mathrm dx &= \int_{-\infty}^\infty \exp(-(x^2+\mathrm ikx))\,\mathrm dx \\ &= \int_{-\infty}^\infty \exp\left(-\left(x^2+\mathrm ikx-\frac{k^2}{4}+\frac{k^2}{4}\right)\right)\,\mathrm dx \\ &= \int_{-\infty}^\infty \exp\left(-\left(\left(x+\frac{1}{2} \mathrm ik\right)^2+\frac{k^2}{4}\right)\right)\,\mathrm dx \\ &= \exp\left(-\frac{k^2}{4}\right)\int_{-\infty}^\infty \exp\left(-\left(x+\frac{1}{2} \mathrm ik\right)^2\right)\,\mathrm dx \\ &= \exp\left(-\frac{k^2}{4}\right)\int_{-\infty}^\infty \exp(-t^2)\,\mathrm dt \\ &= \exp\left(-\frac{k^2}{4}\right)\sqrt{\pi} \end{align*} $$

Answer 1

Another way.

$\exp(-z^2)$ is an entire function. Its power series at the origin converges for all $z$.
Take the anti-derivative of that series term-by-term. It still converges for all $z$. (Use the formula for radius of convergence.) So the series of antiderivatives is an antiderivative for $\exp(-z^2)$.

That argument works for all entire functions. In this case: $$ f(z) = \sum_{n=0}^\infty \frac{(-1)^n}{n!} z^{2n} \\ F(z) :=\sum_{n=0}^\infty \frac{(-1)^n}{n!}\;\frac{z^{2n+1}}{2n+1} \\ F'(z) = f(z) $$

Answer 2

The composition of functions with an antiderivative has an antiderivative.

$\exp$ has an antiderivative because it is its own derivative (and hence its own antiderivative) everywhere.

$z \mapsto -z$ has an antiderivative $z \mapsto \frac{-z^2}{2}$.

$z \mapsto z^2$ has antiderivative $z \mapsto \frac{z^3}{3}$.

Answer 3

Define, for some $z_0\in \mathbb{C}$ $$g(z)=\int_{z_0}^z e^{-t^2}dt$$ Now for any $2$ homotopic (within the domain where $e^{-z^2}$ is analytic) curves the value of this integral will be the same. Since $e^{-z^2}$ is analytic and $\mathbb{C}$ is simply connected, we find that the value of $g(z)$ is independent of the chosen contours. Hence this is a well defined function of $z$, and clearly $g'=f$.

more details: We know that holomorphic differentials are closed, i.e. $d(fdz)=0$. We also know that on a simply connected domain closed forms are exact. This means we can write, using stokes theorem, $$\int_{\gamma}f dz = \int_{\gamma}dg=\int_{\partial\gamma}g=g(\gamma(1))-g(\gamma(0))=g(z)-g(z_0)$$ which shows that the integral does not depend on the path $\gamma$ we chose.