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.

Problem

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.

Question

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.

Solution 1:

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 $$\hat{H}(\xi) = -i\text{PV}\left(\frac{1}{\xi}\right) + \pi\delta$$

Solution 2:

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

Solution 3:

First let us prove that for any particular interval $[-a,a]$, $\def\d{\operatorname{d}\!{}}\displaystyle \int_{-\infty}^{+\infty}\d\xi\int_{-a}^{a}f(x)e^{-ix\xi}\d{}x$ converges.

As in \begin{split} \lim_{c\to+\infty}\int_{0}^{c}\d\xi\int_{-a}^{a}f(x)e^{-ix\xi}\d{}x &= \int_{0}^{+\infty}f(0)\frac{e^{ia\xi}-e^{-ia\xi}}{i\mkern1mu\xi}\d\xi \\ & \quad {} +\lim_{c\to+\infty}\int_a^b\frac{f(x)-f(0)}{ix}\bigl(1-e^{-icx}\bigr)\d{}x, \end{split} both of the RHS converge.

For any $f\in\mathcal{S}(\mathbb{R})$, it's obvious that \begin{align} \int_{0}^{+\infty}\d\xi\int_{-\infty}^{+\infty}f(x)e^{-ix\xi}\d{}x &= \lim_{a\to+\infty}\int_{0}^{+\infty}\d\xi\int_{-a}^{a}f(x)e^{-ix\xi}\d{}x\\ &= \int_0^{+\infty}f(0)\frac{2\sin a\xi}{\xi}\d\xi + \int_{-\infty}^{+\infty}\frac{f(x)-f(0)}{ix}\d{}x\\ &= \pi{}f(0) + \int_0^{+\infty}\frac{f(x)-f(-x)}{ix}\d{}x \end{align} hence \begin{align} \langle\hat{H},f\,\rangle &= \langle{H,\hat{f}}\rangle\\ &= \int_{0}^{+\infty}\hat{f}(\xi)\,\mathrm d\xi\\ &= \int_{0}^{+\infty}\mathrm d\xi\int_{-\infty}^{+\infty}f(x)e^{-ix\xi}\,\mathrm dx\\ &= \pi{}f(0) + \int_0^{+\infty}\frac{f(x)-f(-x)}{ix}\d{}x\\ &= \pi{}f(0) + \lim_{x_0\to0^+}\int_{|x|\geqslant x_0}f(x)\frac{1}{ix}\d{}x \end{align} we obtain $$\langle \hat{H},f\,\rangle=\operatorname {V.\!P.}\biggl(\frac{1}{ix}\biggr)[f\,]+\pi f(0)$$ So $$\hat{H}=\operatorname {V.\!P.}\biggl(\frac{1}{ix}\biggr)+\pi\delta$$