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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?

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    $\begingroup$ in the limit $\alpha \rightarrow 0$ both expressions are equivalent as can be shown by decomposing (2) in real and imaginary part and using some represetnations of $\delta(x)$ and the principal value $P$ $\endgroup$ – tired Jun 24 '15 at 12:49
  • $\begingroup$ No, it does not say there that the Fourier transform of the Heaviside function is what you say it says! That's the Fourier transform of something that's called $H_a$ there. $H_a$ is not the Heaviside function. $\endgroup$ – David C. Ullrich Jun 24 '15 at 17:04
  • $\begingroup$ @DavidC.Ullrich Fair enough. It does say so here, that's why I asked in what sense it is true. I think I now understand this. $\endgroup$ – Minethlos Jun 24 '15 at 18:57
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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 $\alpha\rightarrow 0^+$.

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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.

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  • $\begingroup$ I checked Kelly's book - he gets a different result because his definition of Fourier transform is $\int f(t) e^{i\omega t} dt$ while the source I quoted use $\int f(t) e^{-i\omega t} dt$. $\endgroup$ – Minethlos Nov 20 '15 at 15:04

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