Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I found the following answer on Math.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 a 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 computation, but I have no idea why I have $\mathrm{P} \frac{i}{\omega}$ instead of just $\frac{i}{\omega}$ in the resulting expression.

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

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

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

While chaohuang's answer gave you basically everything you needed, I'd like to elaborate a little on the last part, and on generalized functions (a.k.a. distributions) in general; I apologize for the unavoidable pun.

I realize that in physics, you might treat generalized functions, like the delta distribution, in a "classical" way similar to regular functions. The truth is, however, that distributions are really only defined in the context of how they act on a set of (nicely behaved) test functions. These test functions are usually either Schwartz space functions, or compactly supported functions. Occasionally we also consider the (much larger) test space of $C^\infty$ functions. So the reason why $\hat{H}(\xi)$ involves $\text{PV}\left(\dfrac{1}{\xi}\right)$, as opposed to just $\dfrac{1}{\xi}$, is because of how it integrates against a test function in Schwartz space.

To make \begin{align} \hat{H}(\xi) &= \int_{-\infty}^{\infty}H(x)e^{-ix\xi}\;dx\\ &= \lim_{\epsilon\to0^+}\;\int_{0}^{\infty}e^{-\epsilon x}e^{-ix\xi}\;dx\\ &= \lim_{\epsilon\to0^+}\;(\epsilon+i\xi)^{-1}\\ &= \lim_{\epsilon\to0^+}\;-i(\xi-i\epsilon)^{-1}\\ &= -i(\xi-i0)^{-1} \end{align} rigorous, you need to see how it acts as a linear functional on Schwartz space: \begin{align} \hat{H}[\varphi] &= \int_{-\infty}^{\infty}\hat{H}(\xi)\varphi(\xi)\;d\xi\\ &= \lim_{\epsilon\to0^+}\;\int_{-\infty}^{\infty}-i(\xi-i\epsilon)^{-1}\varphi(\xi)\;d\xi\\ &= -i\lim_{\epsilon\to0^+}\;\int_{-\infty}^{\infty}\left(\frac{\xi+i\epsilon}{\xi^2+\epsilon^2}\right)\varphi(\xi)\;d\xi\\ &= -i\left(\lim_{\epsilon\to0^+}\;\int_{-\infty}^{\infty}\left(\frac{\xi}{\xi^2+\epsilon^2}\right)\varphi(\xi)\;d\xi + i\lim_{\epsilon\to0^+}\;\int_{-\infty}^{\infty}\left(\frac{\epsilon}{\xi^2+\epsilon^2}\right)\varphi(\xi)\;d\xi\right)\\ &=-i\lim_{\epsilon\to0^+}A(\epsilon)+\lim_{\epsilon\to0^+}B(\epsilon) \end{align} where \begin{align} A(\epsilon) &= \int_{-\infty}^{\infty}\left(\frac{\xi}{\xi^2+\epsilon^2}\right)\varphi(\xi)\;d\xi\\ &= \int_{-\infty}^{\infty}\frac{d}{d\xi}\left(\frac{1}{2}\ln(\xi^2+\epsilon^2)\right)\varphi(\xi)\;d\xi\\ &= -\frac{1}{2}\int_{-\infty}^{\infty}\ln(\xi^2+\epsilon^2)\varphi'(\xi)\;d\xi\\ \end{align} and \begin{align} B(\epsilon) &= \int_{-\infty}^{\infty}\left(\frac{\epsilon}{\xi^2+\epsilon^2}\right)\varphi(\xi)\;d\xi\\ &= \int_{-\infty}^{\infty}\frac{d}{d\xi}\left(\arctan\left(\frac{\xi}{\epsilon}\right)\right)\varphi(\xi)\;d\xi\\ &= -\int_{-\infty}^{\infty}\arctan\left(\frac{\xi}{\epsilon}\right)\varphi'(\xi)\;d\xi\\ \end{align} with $$\lim_{\epsilon\to0^+}\;A(\epsilon) = -\int_{-\infty}^{\infty}\ln(|\xi|)\varphi'(\xi)\;d\xi$$ and $$\lim_{\epsilon\to0^+}\;B(\epsilon) = -\frac{\pi}{2}\int_{-\infty}^{\infty}\text{sgn}(\xi) \varphi'(\xi)\;d\xi$$ by various Lebesgue integration theorems, and the fact that $\varphi$ and its derivatives are rapidly decaying. In particular, whereas the area under $1/\xi$ is infinite to either side of the singularity (making the integration of $\varphi/\xi$ ill-defined around $\xi=0$), the function $\ln|\xi|$ has finite integral near $\xi=0$, and so $\lim_{\epsilon\to0^+}\;A(\epsilon)$ is well-defined.

An easy computation then gives us that $$\lim_{\epsilon\to0^+}\;A(\epsilon) = \lim_{\epsilon\to0^+}\; \int_{\mathbb{R}\setminus(-\epsilon,\epsilon)}\frac{1}{\xi}\varphi(\xi)\;d\xi =: \text{PV}\left(\frac{1}{\xi}\right)[\varphi]$$ by definition of the Cauchy principal value distribution. Similarly, we find that $$\lim_{\epsilon\to0^+}\;B(\epsilon) = 2\left(\frac{\pi}{2}\varphi(0)\right):= \pi\delta[\varphi]$$ by definition of the delta distribution.

So in conclusion, we have finally arrived at the fact that $$\hat{H}(\xi) = -i\text{PV}\left(\frac{1}{\xi}\right) + \pi\delta$$

share|cite|improve this answer

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|cite|improve this answer

Your Answer


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.