Graphical proof of flat/spread-out Fourier transforms of Gaussian distribution

Question

The most general form of the Gaussian or normal density function is

$$f(x) = \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{(x-m)^2}{2 \sigma^2}},$$

where m is the mean and $\sigma$ is the standard deviation.

How can we justify the statement that sharply peaked Gaussian's have flat, spread-out Fourier transforms, and vice versa by graphically investigating the behavior of f versus F as $\sigma$ varies.

Answer 1

Since \begin{align} f(x) = \frac{1}{\sqrt{2\pi \sigma^2}} e^{-(x-m)^2/2\sigma^2} \end{align} then the fourier transform of $f$ is given by \begin{align} F(\xi) =&\ \frac{1}{\sqrt{2\pi \sigma^2}}\int^\infty_{-\infty} e^{-2\pi i x\xi}e^{-(x-m)^2/2\sigma^2}\ dx = \frac{e^{-2\pi i m \xi}}{\sqrt{2\pi \sigma^2}}\int^{\infty}_{-\infty}e^{-2\pi i u\xi} e^{-\pi u^2/(2\pi\sigma^2)}\ du\\ =&\ e^{2\pi i m\xi} \int^\infty_{-\infty}e^{-2\pi i w(\sqrt{2\pi}\sigma \xi)} e^{-\pi w^2}\ dw= e^{2\pi im\xi}e^{-2\pi^2\sigma^2\xi^2}. \end{align} Hence we see that as $\sigma\rightarrow 0$ then $f(x)$ will peak at the origin whereas $F(\xi) \rightarrow e^{2\pi im\xi}$ which is ''flat".

Answer 2

Hint: Compute the Fourier transform of the $f$ you've written. You'll notice that $\sigma$ appears in the FT, but in a slightly different place. How does the shape of $f$ vary with $\sigma$, and how does the shape of the FT vary with $\sigma$? (You can finesse the complex-valued nature of the FT by considering just the absolute value of the FT.) Maybe plot these for a few values of $\sigma$, observing which values give sharply peaked density/FT, and which ones cause these to spread out.