Heaviside step function fourier transform and principal values

Question

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.

Answer 1

$$ \begin{align} \lim_{\varepsilon \to 0} \int_0^\infty e^{i\omega t - \varepsilon t} \mathrm{d}t &=\lim_{\varepsilon \to 0} \frac{1}{\varepsilon - i\omega} \\ &=\lim_{\varepsilon \to 0} \frac{1}{\varepsilon + \omega^2/\varepsilon} + \lim_{\varepsilon \to 0} \frac{\omega i}{\varepsilon^2 + \omega^2} \\ &=\pi \delta(\omega) + \mathrm{P} \frac{i}{\omega} \end{align} $$ where the last step uses the limiting representations of the delta function and the Cauchy principal value.