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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.

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Welcome to Math.SE! I have a few observations. Can you prove that: $\int_{-\infty}^{z} e^{-t^2}dt \leq e^{\Im(z)^2}\int_{-\infty}^{\Re(z)} e^{-t^2}dt$? Then one can observe that $\int_{-\infty}^{\Re(z)} e^{-t^2}dt\leq \int_{-\infty}^{\infty} e^{-t^2}dt = \sqrt{\pi}$. Observed all this I do suspect that your claim is uncorrect, are you sure of it? Finally IF this is homework it is not a problem, but you should add the tag for it! –  Giovanni De Gaetano Oct 2 '12 at 12:56
@GiovanniDeGaetano, My claim came about when entering the function into wolfram alpha and seeing the exact result. I've proved this to myself and am quite sure of it. In regards to your upper bound, I feel that as there is still the exponential term out the front which will explode at -$\infty$, that it isn't as simple as this. –  Jack Moon Oct 2 '12 at 14:27
How do you define the integral from $-\infty$ to $z$? –  Davide Giraudo Oct 2 '12 at 14:41
Integration over the straight line contour ${t+z:-\infty \le t \le 0}$ –  Jack Moon Oct 2 '12 at 14:47
@JackMoon, actually I did simply point out a clue from which I suspected the uncorrectedness of your claim. I'm aware I didn't prove anything. I'll try now to do something more formal. –  Giovanni De Gaetano Oct 3 '12 at 9:09

1 Answer 1

up vote 1 down vote accepted

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 $\lim_{z \rightarrow +\infty} \psi_1(z)=+\infty$.

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Hoping that everything is fine I observe that your claim was indeed correct! –  Giovanni De Gaetano Oct 3 '12 at 9:29
Thank you Giovanni, are you able to clarify what L'Hôpital's rule implies? –  Jack Moon Oct 3 '12 at 12:17
You're welcome! So, this is the Wikipedia page of L'Hôpital's rule: en.wikipedia.org/wiki/L'Hôpital's_rule. Essentially I simply switched from the limit of the quotient to the limit of the quotient of the derivatives. Is this your question? Or perhaps you are puzzled by the actual computation of the limits? –  Giovanni De Gaetano Oct 3 '12 at 12:21
I was just wanting clarification on the statement of the theorem. The limit calculation is fine. –  Jack Moon Oct 3 '12 at 12:26
You need a reputation of at least 15 and I'm afraid I'm new here and only have 8. I've tried already. –  Jack Moon Oct 3 '12 at 12:29

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