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The bilateral Laplace transform of a Gaussian function could be established as: $$e^{x^2/2}=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}e^{-xy}e^{-y^2/2} dy$$

Then what should be a similar relation for a Gaussian function with complex variable:$e^{z^2/2}$?

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The usual Laplace transform uses $\int_{0}^\infty$. Here you have a bilateral Laplace transform. –  Robert Israel Aug 17 '12 at 20:54
    
I wrote what I think is an elegant account of how to find this (bilateral) Laplace transform, but that's not necessarily on topic for your final question, so I deleted it. If you're interested, I'll un-delete it. –  Michael Hardy Aug 17 '12 at 21:29
    
@RobertIsrael : If I were to call it a "two-sided Laplace transform", would that be too Germanic for you? –  Michael Hardy Aug 18 '12 at 3:52
    
I have nothing against "two-sided". Too Germanic would be "zweiseitig". –  Robert Israel Aug 19 '12 at 5:00

2 Answers 2

Let $z$ be a complex number. The integral is convergent for all $z \in \mathbb{C}$. Completing the square we get: $$ \int_{-\infty}^\infty \mathrm{e}^{-z \cdot y} \mathrm{e}^{-y^2/2} \mathrm{d} y = \mathrm{e}^{z^2/2}\int_{-\infty}^\infty \mathrm{e}^{-(y+z)^2/2} \mathrm{d} y $$ The latter integral multiplying the $\mathrm{e}^{z^2/2}$ is a constant, independent of $z$. Evaluating it at $z=0$ establishes the result.

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+1. Nice and smooth. –  Did Aug 17 '12 at 22:25

$$e^{x^2/2}=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}e^{-xy}e^{-y^2/2} \, dy$$

I'm not sure of the meaning of the question, but maybe you mean what if $x$ is complex and not necessarily real (so you're calling it $z$), but of course the transform is still defined by an integral from $-\infty$ to $\infty$ along the real line.

Here's how I would think about that:

  • Use Morera's theorem to show that $z\mapsto \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}e^{-zy}e^{-y^2/2}\,dy$ is an analytic function; and then

  • Show that if its value is $e^{z^2/2}$ when $z$ is real, then it must continue to be $e^{z^2/2}$ when $z$ is not real.

The second step is done by recalling that if two functions of a complex variable agree on the real line and both are analytic, then the agree off the real line too.

For the first step, we need to show that if one integrates this function of $z$ along any simple closed curve, one gets $0$. Morera's theorem says that that implies the function is analytic.

So we have $$ \begin{align} & {} \quad \int_\text{closed curve} (\text{function of }z)\,dz = \int_\text{closed curve} \left(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-zy} e^{-y^2/2}\, dy\right) \, dz \\[10pt] & = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \left( \int_\text{closed curve} e^{-zy} e^{-y^2/2} \, dz \right)\, dy\tag{by Fubini's theorem} \\[10pt] & = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty 0 \, dy\tag{by Cauchy's theorem} \\[10pt] & = 0. \end{align} $$ Fubini's theorem say the two itereted integrals (i.e. integrating in two different orders: $y$ first or $z$ first) are equal if the double integral of the absolute value of the function is finite.

Cauchy's theorem says when you integrate an analtic function around a simple closed curve---(the function is analytic everywhere in the region bounded by the curve)---then you get $0$.

Later edit: Two question not address above are:

  • For what values of $z$ is this function of $z$ actually defined by the integral? I.e. the function that you integrate should be integrable in the sense that the integral of its absolute value is finite. That's assumed before you apply Fubini's theorem. And if that doesn't happen, we'd have to worry about what sort of convergence of improper integrals we should think about.

  • Can the function of $z$ be analytically continued into regions where the integral that defines it doesn't behave well?

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