Fourier Transform of $sgn(t)$

Question

I'm trying to find the Fourier transform of $sgn(t)$ where

$$ sgn(t) = \begin{cases} 1, & t > 0 \\ 0, & t = 0 \\ -1, & t < 0. \end{cases} $$

By definition,

$$ X(\omega) = \int_{-\infty}^{\infty} sgn(t) e^{-i\omega t} \, dt = \int_0^{\infty} e^{-i\omega t} \, dt + \int_{-\infty}^0 -e^{-i\omega t} \, dt. $$

However, that second integral diverges. Where is the flaw in my methodology?

Answer 1

You have correctly deduced that taking the necessary integral in any usual fashion does not give an answer that is a function.

However, there is a way to get the answer as long as you've defined the fourier transform of the functions $f(x) = 1$ and $f(x) = \delta(x)$. In particular, we note that

$$ \operatorname{sgn}(t) = -1 + 2\int_{-\infty}^t \delta(\tau)\,d\tau $$

Answer 2

There can be several ways to get the FT of $\operatorname{sgn}(t)$. Note that $\operatorname{sgn}(t)$ is not integrable and cannot directly do the FT. But it can be achieved by limiting process, for instance, we can construct $\operatorname{sgn}(t)=\int_R\frac{\sin(\omega t)}{\pi\omega}\mathrm d\omega$ or by Heaviside functions with a decaying factor $e^{-\alpha t}$ with $\alpha>0$. $$\operatorname{sgn}(t)=u(t)-u(-t)=\lim_{\alpha\to0}\left[e^{-\alpha t}u(t)-e^{\alpha t}u(-t)\right].$$ Then we can evaluate Fourier integrals. $$\mathscr{F}[\operatorname{sgn}(t)]=\lim_{\alpha\to0}\left\{\mathscr{F}\left[e^{-\alpha t}u(t)\right]-\mathscr{F}\left[e^{\alpha t}u(-t)\right]\right\}=\lim_{\alpha\to0}\left\{\frac{1}{\alpha+i\omega}-\frac{1}{\alpha-i\omega}\right\}=\frac{2}{i\omega}$$

A second way is to use $$\mathscr{F}[u(t)]=\pi\delta(\omega)+\frac{1}{i\omega}=\mathscr{F}\left[\frac12+\frac12\operatorname{sgn}(t)\right]=\pi\delta(\omega)+\frac12\mathscr{F}[\operatorname{sgn}(t)].$$ Therefore $\mathscr{F}[\operatorname{sgn}(t)]=\frac{2}{i\omega}$.

EDIT: On the side way to complete $\mathrm{sinc}$ function approach, I found that it might be even better to recover Fourier transform from Laplace transform as below.

Let $f(t)$ be constructed as \begin{align} f(t)=e^{-at}u(t)-e^{at}u(-t) \end{align} which is an old function since $f(-t)=-f(t)$ and if a limit as $a\to0$ is to take, $f(t)$ becomes $\operatorname{sgn}(t)$ and still satisfies $\operatorname{sgn}(-t)=-\operatorname{sgn}(t)$. Therefore, its Fourier transform can be readily obtained by the known Laplace transform $\mathscr{L}[\mathbf{1}(t)]=\frac{1}{s}$. \begin{align} F(\omega)&=\mathscr{F}[x(t)]=\int_{-\infty}^{\infty}\bigl(e^{-at}u(t)-e^{at}u(-t)\bigr)e^{-i\omega t}\mathrm dt\\ &=2\int_0^{\infty}e^{-at}e^{-i\omega t}\mathrm dt=2\int_0^{\infty}e^{-(a+i\omega)t}\mathrm dt. \end{align} If we recall the Laplace domain notation $s=a+i\omega$, the right integral is exactly $\frac{2}{s}$. Thus by setting $a=0$ and $s=i\omega$, then \begin{align} \mathscr{F}[\operatorname{sgn}(t)]=\frac{2}{i\omega}. \end{align}

Answer 3

Echoing @Omnonmnomnm's point: the notion of "integral" in Fourier transforms very often needs to be an "extension by continuity" (in a topology that may vary, depending on context), so is in any case absolutely not any kind of Riemann, Lebesgue, or other "integral". Nevertheless, if/when we know what kind of limit the integrand is, and in what sense we want the outcome to have a sense... it is often possible, in real-life situations, to attach useful meaning to "divergent" integrals or sums which in a strict, late 19th-century interpretation, would be meaningless.

That is, "integral" is not literal integral, but is a linear map from one topological vector space to another, sometimes just to scalars. It should be continuous in relevant topologies... though the relevant topologies are not always obvious, and not always classical, not always related to "pointwise convergence", etc.

So it's not that one "defines" the Fourier transform of $1$ and $\delta$, but that one determines what these are, in the extended sense of Fourier transform, and then @Omn.'s computation is what you want.