Fourier transform and inverse calculations without residues and cauchy theorem

Question

Do you have any ideas of a function other than the Gaussian distribution for which we can do the numerical calculations (by hand) of its Fourier transform and retrieve the original function from its Fourier transform without using Fourier transform formulas/properties, Residues theorem and Cauchy theorem ?

Answer 1

Here, we reply to the comment "I was thinking more about functions like the sign function, $\exp(-aItI)$, or the rect function."

Let $f(t)=\text{sgn}(t)$. Note that the integral $\int_{-\infty}^\infty f(t) e^{i\omega t}\,dt$ fails to exist as either a Lebesgue integral or an improper Riemann integral.

Inasmuch as $f$ is locally integrable, its Fourier transform exists as a tempered distribution. To show this, we let $\phi\in \mathbb{S}$. Then, we have for any $\delta>0$

$$\begin{align} \langle \mathscr{F}\{\text{sgn}\}, \phi \rangle&=\langle \text{sgn}, \mathscr{F}\{ \phi \}\rangle\\\\ &=\int_{-\infty}^\infty \text{sgn}(t)\int_{-\infty}^\infty \phi(\omega)e^{i\omega t}\,d\omega\,dt\\\\ &=\lim_{L\to\infty}\int_{-L}^L \text{sgn}(t)\int_{-\infty}^\infty \phi(\omega)e^{i\omega t}\,d\omega\,dt\\\\ &=\lim_{L\to\infty}\int_{-\infty}^\infty \phi(\omega)\int_{-L}^L \text{sgn}(t)e^{i\omega t}\,dt\,d\omega\\\\ &=\lim_{L\to\infty}\int_{-\infty}^\infty \phi(\omega)\left(\frac{e^{i\omega L}+e^{-i\omega L}-2}{i\omega}\right)\,d\omega \\\\ &=\lim_{L\to\infty}\left(\int_{|\omega|\le \delta} \phi(\omega)\left(\frac{e^{i\omega L}+e^{-i\omega L}-2}{i\omega}\right)\,d\omega+\int_{|\omega|\ge \delta} \phi(\omega)\left(\frac{e^{i\omega L}+e^{-i\omega L}-2}{i\omega}\right)\,d\omega \right)\\\\ &=\int_{|\omega|\ge\delta}\frac{i2}{\omega}\phi(\omega)\,d\omega+O(\delta) \end{align}$$

Letting $\delta\to 0$, yields the result

$$\langle \mathscr{F}\{\text{sgn}\}, \phi \rangle=\lim_{\delta\to 0^+}\int_{|\omega|\ge\delta} \frac{i2}{\omega}\phi(\omega)\,d\omega$$

Therefore, in distribution we have $\mathscr{F}\{\text{sgn}\}(\omega)=\text{PV}\left(\frac{i2}{\omega}\right)$.

Now, if we wish to find the inverse Fourier transform of $F(\omega)=\text{PV}\left(\frac{i2}{\omega}\right)$, we proceed as follows.

$$\begin{align} \mathscr{F}\{F\}(t)&=\text{PV}\left(\frac1{2\pi}\int_{-\infty}^\infty \frac{i2}{\omega}e^{-i\omega t}\,d\omega\right)\\\\ &=\frac1{\pi}\int_{-\infty}^\infty \frac{\sin(\omega t)}{\omega}\,d\omega\\\\ &=\text{sgn}(t) \end{align}$$

as was to be shown!