Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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 $b$.

share|improve this answer

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

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.