Heaviside step function fourier transform and principal values 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.

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