Proving $\frac{e^{z^2}}{\sqrt{\pi}}\int_{-\infty}^{z}e^{-t^2}dt$ is bounded for $\Re(z) \leq 0$ I've been given the following equations;
$$\psi_1(z)=\frac{e^{z^2}}{\sqrt{\pi}}\int_{-\infty}^{z}e^{-t^2}dt$$
(i.e. integrate over the straight line contour {$t + z: -\infty\lt t \lt 0 $})
$$\psi_2(z)=\frac{e^{z^2}}{\sqrt{\pi}}\int_{\infty}^{z}e^{-t^2}dt$$
(i.e. integrate over the straight line contour {$t + z: \infty\gt t \gt 0 $})
I am asked to show $\psi_1(z)$ is bounded for Re(z)$\leq$0 and similarly show $\psi_2(z)$ is bounded for Re(z)$\geq$0. Also I want to show the limit of $\psi_1(x)$ as $x\rightarrow -\infty$ and $\psi_2(x)$ as $x\rightarrow \infty$
What I have done so far is to show that these equations can be represented in terms of the error function, that is;
\begin{align}
\psi_1(z)&=\frac{e^{z^2}}{2}(\operatorname{erf}(z)+1) \\
\psi_2(z)&=\frac{e^{z^2}}{2}(\operatorname{erf}(z)-1)
\end{align}
Yet am having trouble finding an upper bound for the above functions.
 A: The answer is relative to the function $\psi_1(z)$, I prove something slightly stronger than the question, i.e. that $\psi_1(z)$ is bounded for $\Re(z)$ bounded. For $\psi_2(z)$ I strongly suspect that the proof is more or less the same.
I start with some manipulations on the function:
$$\psi_1(z)\cdot \sqrt{\pi}= e^{z^2}\int_{-\infty}^{z} e^{-t^2}dt= e^{\Re(z)^2 +2i \Re(z)\Im(z) -\Im(z)^2}\int_{-\infty}^{\Re(z)}e^{-t^2 -2ti \Im(z) +\Im(z)^2}dt=$$
$$= e^{\Re(z)^2} \int_{-\infty}^{\Re(z)} e^{-t^2 -2i\Im(z)(\Re(z)-t)}dt\leq e^{\Re(z)^2}\int_{-\infty}^{\Re(z)}e^{-t^2}dt.$$
Now we are reduced to consider $z=x\in\mathbb{R}$. Let us compute the limit:
$$\lim_{x \rightarrow -\infty} e^{x^2}\int_{-\infty}^{x}e^{-t^2}dt=\lim_{x \rightarrow -\infty} \frac{\int_{-\infty}^{x}e^{-t^2}dt}{e^{-x^2}}=...$$
By L'Hôpital's rule,
$$...=\lim_{x \rightarrow -\infty} \frac{e^{-x^2}}{-2xe^{-x^2}}=0.$$
In specific, by positivity of $\psi_1(z)$, we deduce that $\lim_{z \rightarrow -\infty} \psi_1(z)=0$.
Observing that $\psi_1(0)=\frac{1}{2}<\infty$ and by continuity of $\psi_1(z)$ we deduce it is bounded if $\Re(z)$ is bounded. We need this last condition because $\lim_{z \rightarrow +\infty} \psi_1(z)=+\infty$.
