Let $\forall z \in \mathbb{C}, g(z) = \int_0^\infty{e^{-z t} e^{-t^2} dt}$.
Prove that there is a $z \in \mathbb{C}$ such that $g(z) = 0$.
Note that $g(z) = \frac{e^{z^2/4} \sqrt{\pi}}{2} . erfc(z/2)$ so the zeros of $g$ are the zeros of the $ercf$ function (up to a trivial homothetic transformation).
Note also that $g$ is an entire function and that any zero satisfies $\Re(z)<0$ (a fact I was able to prove).
Thanks for any help!
We can estimate
\begin{align} \lvert g(z)\rvert &\leqslant \int_0^{\infty} e^{\lvert z\rvert t} e^{-t^2}\,dt \\ &= e^{\lvert z\rvert^2/4} \int_0^{\infty} e^{- (t - \lvert z\rvert/2)^2}\,dt \\ &= e^{\lvert z\rvert^2/4} \int_{\lvert z\rvert/2}^{\infty} e^{-u^2}\,du \\ &< \sqrt{\pi}\cdot e^{\lvert z\rvert^2/2}\,. \end{align}
Thus $g$ is of order $\leqslant 2$, and since the first inequality is an equality for negative real $z$, the order of $g$ is exactly $2$. (If the order were not an integer, we'd already be done because an entire function of fractional order attains every complex value infinitely often.)
If $h$ is an entire function of finite order without zeros, then $h(z) = e^{P(z)}$ for some polynomial $P$, and consequently $h'(z) = P'(z)\cdot h(z)$. For $g$, we compute - since differentiation under the integral is legitimate -
\begin{align} g'(z) &= \int_0^{\infty} \biggl(\frac{d}{dz} e^{-zt}\biggr)e^{-t^2}\,dt \\ &= \int_0^{\infty} (-t)e^{-zt}e^{-t^2}\,dt \\ &= \frac{1}{2} \int_0^{\infty} e^{-zt}\biggl(\frac{d}{dt}e^{-t^2}\biggr)\,dt \\ &= \frac{1}{2}e^{-zt}e^{-t^2}\biggr\rvert_0^{\infty} - \frac{1}{2}\int_0^{\infty} \biggl(\frac{d}{dt}e^{-zt}\biggr)e^{-t^2}\,dt \\ &=-\frac{1}{2} + \frac{z}{2} g(z)\,. \end{align}
If $g$ had no zeros, from this calculation and the above remark it would follow that
$$Q(z)\cdot g(z) \equiv 1$$
for some polynomial $Q$ (namely $Q(z) = z - 2P'(z)$ where $g(z) = e^{P(z)}$) and hence
$$\frac{1}{g(z)} = e^{-P(z)} = Q(z)$$
is a polynomial. But since the left hand side has no zeros it follows that $Q(z) \equiv 1$ and hence that $g$ is constant. From that we can obtain various contradictions. For example
$$0 = g'(0) = \frac{0\cdot g(0)-1}{2} = -\frac{1}{2}\,.$$
Looking a bit closer at this argument shows that it proves more than just the existence of one zero, it proves the existence of infinitely many zeros. For if $g$ had only finitely many zeros, then we would have the factorisation
$$g(z) = R(z)\cdot e^{P(z)}$$
with polynomials $R,P$, and hence $g'(z) = (R'(z) + R(z)P'(z))\cdot e^{P(z)} = Q(z)\cdot g(z)$. This contradicts $g'(z) = -\frac{1}{2} + \frac{z}{2}g(z)$ by exactly the same argument.