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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.


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.


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.

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

2 Answers 2

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 $b$.

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

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I'm having trouble following your second line. –  Ron Gordon Apr 1 '13 at 15:23

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