Representation of Heaviside function's Fourier transform

Question

I've seen here that the Fourier transform of Heaviside function $\Theta(t)$ is

$$ \Theta(\omega) = \frac{1}{i\omega} + \pi \delta(\omega) \tag{1}$$

But in some physics texts and here I've seen the following representation:

$$ \Theta(\omega) = \frac{1}{\alpha + i\omega} \tag{2}$$

where $\alpha$ is infinitesimal parameter.

Why and in what sense is this truly representation of Heaviside function?

Answer 1

Note that

$$\frac{1}{\alpha+i\omega}=\frac{\alpha}{\alpha^2+\omega^2}-\frac{i\omega}{\alpha^2+\omega^2}\tag{1}$$

Taking the limit $\alpha\rightarrow 0^{+}$ of (1) gives

$$\lim_{\alpha\rightarrow 0^+}\frac{1}{\alpha+i\omega}=\lim_{\alpha\rightarrow 0^+}\frac{\alpha}{\alpha^2+\omega^2}+\frac{1}{i\omega}\tag{2}$$

The real part of (2) is a scaled nascent Dirac delta function:

$$\lim_{\alpha\rightarrow 0^+}\frac{\alpha}{\alpha^2+\omega^2}=\pi\delta(\omega)\tag{3}$$

Combining (1), (2) and (3) gives

$$\lim_{\alpha\rightarrow 0^+}\frac{1}{\alpha+i\omega}=\pi\delta(\omega)+\frac{1}{i\omega}\tag{4}$$

So the two expressions in your question are indeed identical if you consider the limit $\alpha\rightarrow 0^+$.

Answer 2

I cannot write a comment, so writing an answer.

The function theta in the original question (and thus its Fourier transforms) is actually not the usual Heaviside step function, as $$\Theta(1)=0,$$ $$\Theta(-1)=1.$$ David C. Ullrich had already meantioned it, but I think it still causes confusion.

The answer asked here is answered in detail in Graduate Mathematical Physics by James J. Kelly, page 137.