# Heaviside step function fourier transform and principal values

I found the following answer on 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 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 evaluation, but I have no idea why I have

$$\mathrm{P} \frac{i}{\omega} \quad \text{instead of just} \quad \frac{i}{\omega}$$

In the given expression.

-
Do you feel comfortable with $i/( \omega + i0^+)$? –  Fabian Jan 3 '13 at 20:24

Because in distribution theory (implied by the $\delta$ distribution) we want the functions to be locally integrable while $\omega \mapsto \frac 1{\omega}$ isn't.
So $\displaystyle\int_a^b \frac {dx}x$ is not well defined while $\quad P\displaystyle\int_a^b \frac {dx}x$ will be defined for different signs of $a$ and $b$.
\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.