How to find the analytical form of Fourier transformation of this function?

Question

could anyone help me on this Fourier transformation? The function is $$f(t) = \frac{e^{-a t^2}}{b+t^2},$$ in which $a, b>0 $.

I have tried with Mathematica, but it fails to give an answer.

Answer 1

You want to use the convolution theorem of the Fourier Transform that lets you say

$$\mathscr{F}\left\{g(t)h(t)\right\} = \mathscr{F}\left\{g(t)\right\} * \mathscr{F}\left\{h(t)\right\}$$

and apply that to two factors of $f(t)$ of your choosing.

From a table lookup of the Fourier transform of

$$\mathscr{F}\left\{e^{-\pi t^2}\right\} = e^{-\pi s^2}$$

and the similarity theorem

$$\mathscr{F}\left\{g(ct)\right\} = \dfrac{1}{|c|}G\left(\dfrac{s}{c}\right)$$

we get

$$\begin{align*}\mathscr{F}\left\{e^{-a t^2}\right\} &= \mathscr{F}\left\{e^{-\pi\left({\sqrt{\dfrac{a}{\pi}}t}\right)^2}\right\}\\ \\ &=\sqrt{\dfrac{\pi}{a}}e^{-\pi\left({\sqrt{\dfrac{\pi}{a}}s}\right)^2}\\ \\ &=\sqrt{\dfrac{\pi}{a}}e^{-\dfrac{\pi^{2}s^2}{a}}\\ \end{align*}$$

We can then also perform a table lookup and find that

$$\mathscr{F}\left\{\dfrac{2d}{d^2+(2\pi t)^2}\right\} = e^{-d|s|}$$

So using the similarity theorem again

$$\begin{align*}\mathscr{F}\left\{\dfrac{1}{b+t^2}\right\} &=\dfrac{1}{2\sqrt{b}}\mathscr{F}\left\{\dfrac{2\sqrt{b}}{b+\left(2\pi\dfrac{1}{2\pi} t\right)^2}\right\}\\ \\ &= \dfrac{1}{2\sqrt{b}} 2\pi e^{-\sqrt{b}|2\pi s|}\\ \\ &= \dfrac{\pi}{\sqrt{b}} e^{-2\pi\sqrt{b}|s|}\\ \end{align*}$$

Putting this all together, your final answer becomes

$$\begin{align*}\mathscr{F}\left\{f(t)\right\} &= \mathscr{F}\left\{\dfrac{e^{-at^2}}{b+t^2}\right\}\\ \\ &= \mathscr{F}\left\{e^{-at^2} \dfrac{1}{b+t^2}\right\}\\ \\ &= \mathscr{F}\left\{e^{-at^2}\right\} * \mathscr{F}\left\{\dfrac{1}{b+t^2}\right\}\\ \\ &= \sqrt{\dfrac{\pi}{a}}e^{-\dfrac{\pi^{2}s^2}{a}} * \dfrac{\pi}{\sqrt{b}} e^{-2\pi\sqrt{b}|s|}\\ \end{align*}$$