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I want to calculate the Fourier transform of the function $ f\left(t\right)=e^{-t^{2}} $

The answer should be $$ \mathscr{F}\left(f\right)[\omega]=\frac{1}{2\sqrt{\pi}}e^{-\frac{\omega^{2}}{4}} $$

I solved it in 2 different ways (and got the same result), but I doubt my second way because I cannot explain in 100%.

So the first way is to note that both $$ \intop_{-\infty}^{\infty}|f\left(t\right)|dt<\infty,\thinspace\thinspace\thinspace\intop_{-\infty}^{\infty}t|f\left(t\right)|dt<\infty $$

And thus by a well known theorem $$ \frac{d}{d\omega}\mathscr{F}\left(f\right)[\omega]=\mathscr{F}\left(g\right)[\omega] $$

Where $ g\left(t\right)=-itf\left(t\right) $.

Using integration by parts, we find that $$ \frac{d}{d\omega}\mathscr{F}\left(f\right)[\omega]=-\frac{\omega}{2}\mathscr{F}\left(f\right)[\omega] $$

Solving this differential equation we find: $$ \mathscr{F}\left(f\right)[\omega]=Ce^{-\frac{\omega^{2}}{4}} $$

And for $\omega=0 $ we have $$ \mathscr{F}\left(f\right)[0]=\frac{1}{2\pi}\intop_{-\infty}^{\infty}f\left(t\right)dt=\frac{1}{2\pi}\intop_{-\infty}^{\infty}e^{-t^{2}}dt=\frac{1}{2\sqrt{\pi}} $$

And hence we found $ C $.

This way require a lot of details, and I have much easier one.

The second way:

Note that :

$$ \mathscr{F}\left(f\right)[\omega]=\frac{1}{2\pi}\intop_{-\infty}^{\infty}e^{-t^{2}}e^{-i\omega t}dt=\frac{1}{2\pi}\intop_{-\infty}^{\infty}e^{-\left(t^{2}+i\omega t\right)}dt=\frac{1}{2\sqrt{\pi}}e^{-\frac{\omega^{2}}{4}}\intop_{-\infty}^{\infty}\frac{\sqrt{2}}{\sqrt{2\pi}}e^{-\left(t+\frac{i\omega}{2}\right)^{2}}dt $$

And note that the density of a normal distributed random variable is given by $ N\left(\mu,\sigma\right)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{\left(x-\mu\right)^{2}}{2\sigma^{2}}} $

So we got $ \frac{1}{2\sqrt{\pi}}e^{-\frac{\omega^{2}}{4}} $ multiplied by an integral (over the whole space) over a density of random distributed random variable with $ N\left(-\frac{i\omega}{2},\frac{1}{\sqrt{2}}\right) $.

We got the same result.

Finally, my question is:

When I learned probability theory, we never mentioned complex numbers, So my argument that this is an integral over a density of a normal distribution, may be not acceptable.

Is it legit?

Thanks in advance.

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  • $\begingroup$ You can make your second method more legit by just using a complex substitution to turn the integral into an actual canonical Gaussian integral. In fact I would say that that's relatively easier to make completely rigorous than the argument based on properties of Fourier transforms. $\endgroup$ – Izaak van Dongen Mar 11 at 18:51
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If you know some complex analysis, you can fill in the details of the second argument. For simplicity, I'm going to omit all constant factors. Here's the claim:

For any $\beta\in\Bbb{R}$, we have $\int_{\Bbb{R}}e^{-t^2}\,dt = \int_{\Bbb{R}}e^{-(t+i\beta)^2}\,dt$.

To prove this, consider for each $R>0$, the closed rectangular loop $\gamma_R$ oriented counter-clockwise passing through the four points $-R,R, R+i\beta, -R+i\beta$ (draw a picture, and maybe you would like to assume $\beta>0$ as well for ease of visualization). Then, the function $z\mapsto e^{-z^2}$ is holomorphic on $\Bbb{C}$ so by Cauchy's theorem, we have \begin{align} \int_{\gamma_R}e^{-z^2}\,dz &= 0. \end{align} Writing this out explicitly, we have \begin{align} \int_{-R}^{R}e^{-x^2}\,dx + \int_{0}^{\beta}e^{-(R+iy)^2}\,dy - \int_{-R}^Re^{-(x+i\beta)^2}\,dx -\int_{0}^{\beta}e^{-(-R+iy)^2}\,dy &= 0. \end{align} Here, the $+,-$ signs take the orientation the the curve $\gamma_R$ into account (again draw a picture). I leave it to you to justify why as $R\to \infty$, the two integrals $\int_0^{\beta}$ vanish, thereby leaving us with the equality \begin{align} \int_{\Bbb{R}}e^{-t^2}\,dt =\int_{\Bbb{R}}e^{-(t+i\beta)^2}\,dt. \end{align}

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