TITLE: Heaviside step function fourier transform and principal values
QUESTION [6 upvotes]: I found the following answer on Math.SE:
Fourier transform of unit step?
However, it is still not clear to me and maybe somebody could explain it clearer.
Problem
I have the following in my notes of a theoretical physics course:
$$
\hat{\Theta}(\omega) = \int_{-\infty}^\infty \Theta (t) e^{i\omega t} \mathrm{d} t 
= \lim_{\varepsilon \to 0} \int_0^\infty e^{i\omega t - \varepsilon t} \mathrm{d}t
= \pi \delta(\omega) + \mathrm{P} \frac{i}{\omega},
$$
where the $\mathrm{P}$ denotes the Cauchy's principal value. 
Question
I understand why I get a delta function in this computation, but I have no idea why I have $\mathrm{P} \frac{i}{\omega}$ instead of just $\frac{i}{\omega}$ in the resulting expression.

REPLY [4 votes]: First let us prove that for any particular interval $[-a,a]$, $\def\d{\operatorname{d}\!{}}\displaystyle
\int_{-\infty}^{+\infty}\d\xi\int_{-a}^{a}f(x)e^{-ix\xi}\d{}x$ converges.
As in
\begin{split}
\lim_{c\to+\infty}\int_{0}^{c}\d\xi\int_{-a}^{a}f(x)e^{-ix\xi}\d{}x &= \int_{0}^{+\infty}f(0)\frac{e^{ia\xi}-e^{-ia\xi}}{i\mkern1mu\xi}\d\xi \\
& \quad {} +\lim_{c\to+\infty}\int_a^b\frac{f(x)-f(0)}{ix}\bigl(1-e^{-icx}\bigr)\d{}x,
\end{split}
both of the RHS converge.
For any $f\in\mathcal{S}(\mathbb{R})$, it's obvious that
\begin{align}
\int_{0}^{+\infty}\d\xi\int_{-\infty}^{+\infty}f(x)e^{-ix\xi}\d{}x &= \lim_{a\to+\infty}\int_{0}^{+\infty}\d\xi\int_{-a}^{a}f(x)e^{-ix\xi}\d{}x\\
&= \int_0^{+\infty}f(0)\frac{2\sin a\xi}{\xi}\d\xi + \int_{-\infty}^{+\infty}\frac{f(x)-f(0)}{ix}\d{}x\\
&= \pi{}f(0) + \int_0^{+\infty}\frac{f(x)-f(-x)}{ix}\d{}x
\end{align}

hence

\begin{align}
\langle\hat{H},f\,\rangle &= \langle{H,\hat{f}}\rangle\\
&= \int_{0}^{+\infty}\hat{f}(\xi)\,\mathrm d\xi\\
&= \int_{0}^{+\infty}\mathrm d\xi\int_{-\infty}^{+\infty}f(x)e^{-ix\xi}\,\mathrm dx\\
&= \pi{}f(0) + \int_0^{+\infty}\frac{f(x)-f(-x)}{ix}\d{}x\\
&= \pi{}f(0) + \lim_{x_0\to0^+}\int_{|x|\geqslant x_0}f(x)\frac{1}{ix}\d{}x
\end{align}
we obtain
$$\langle \hat{H},f\,\rangle=\operatorname {V.\!P.}\biggl(\frac{1}{ix}\biggr)[f\,]+\pi f(0)$$
So
$$\hat{H}=\operatorname {V.\!P.}\biggl(\frac{1}{ix}\biggr)+\pi\delta$$