Why is the characteristic function of the sign function $2i/t$?

Question

A book I'm reading says:

Let $\theta(x)$ be the Heaviside function... Since $\operatorname{sgn}(x)=2\theta(x)-1$, its characteristic function will be: $$\chi_{\operatorname{sgn}(x)}(t)=\frac{2i}{t}$$

I understand why $\operatorname{sgn}(x)=2\theta(x)-1$, but I don't follow the claim about the characteristic function. If the CDF of $X$ is the Heaviside function, then $X$ is a.s. zero, and the sign of $X$ is a.s. zero, and the characteristic of a random variable that is a.s. zero is $1$. How does $2i/t$ come into it?

Answer 1

The word characteristic function is misleading.

Here $\theta(x)$ is not the CDF of a random variable, we are looking at the Fourier transform of $\theta(x)$ not of $d\theta(x)=\delta(x)dx$.

This Fourier transform exists only in the sense of distributions, it will be $PV(i/t)+\pi \delta(t)$ ($PV$ for principal value).

Answer 2

Integrating by parts viz. Riemann–Stieltjes,$$\int_{\Bbb R}e^{itx}\operatorname{sgn}x\mathrm dx=-\frac{1}{it}\int_{\Bbb R}e^{itx}2\delta(x)\mathrm dx=\frac{2i}{t},$$but this still requires the question to intend convergence in a distributional sense.