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?


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?


Here is my solution for the follow-up question

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


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: $$ f(z) = \sum_{n=0}^\infty \frac{(-1)^n}{n!} z^{2n} \\ F(z) :=\sum_{n=0}^\infty \frac{(-1)^n}{n!}\;\frac{z^{2n+1}}{2n+1} \\ F'(z) = f(z) $$


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 $z \mapsto \frac{z^3}{3}$.

  • $\begingroup$ Really? $\exp(\exp(x))$ doesn't have an antiderivative though, does it? $\endgroup$ – Lundborg May 1 '16 at 12:17
  • 2
    $\begingroup$ @Neutronic Not in closed form, certainly, but there is a function which differentiates to $\exp \circ \exp$ everywhere. $\endgroup$ – Patrick Stevens May 1 '16 at 12:18
  • $\begingroup$ @PatrickStevens ... Still, we need a proof of that fact. $\endgroup$ – GEdgar May 1 '16 at 13:38

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.

  • $\begingroup$ So far we haven't been talking about homotopy so I find it difficult to fully understand your proof. I have a feeling you are trying to pick arbitrary curves which start and end point are fixed and showing that the integrals along those curves is always the same. Can you confirm this? I am struggling to understand how you deduce this exactly. $\endgroup$ – Christian Ivicevic May 1 '16 at 13:04
  • $\begingroup$ Yes that is what I am doing. I'll include a bit more detail in my answer $\endgroup$ – user2520938 May 1 '16 at 13:07

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