Representation of Heaviside function's Fourier transform 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?
 A: 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 $\alpha\rightarrow 0^+$.
A: 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.
