I would like to compute the Fourier transform of the principal value distribution. Let $H$ denote the Heaviside function.

Begin with the fact that $$2\widehat{H} =\delta(x) - \frac{i}{\pi} p.v\left(\frac{1}{x}\right).$$ Rearranging gives that the principal value distribution is, up to a constant $$\delta(x) - 2\widehat{H}.$$ If we take the Fourier transform of this, we get $$1- 2H(-x) ,$$ which seems wrong.

First, why does this method produce nonsense?

Second, what is a good way to do this computation?

  • $\begingroup$ You are claiming that the Fourier transform of the Hilbert transform is the Delta function plus principal value, which is wrong. $\endgroup$ – Chee Han Jun 8 '15 at 9:25
  • $\begingroup$ @CheeHan Sorry, I was using $H$ to denote the Heaviside function. I have clarified this in the question. $\endgroup$ – Potato Jun 8 '15 at 16:34
  • $\begingroup$ Interesting, I never actually compute the Fourier transform of the Heaviside function so I couldn't recognised that expression at all. My hunch is that since you are considering $H$ as a distribution, you are "free" to choose whatever value at 0 since it is defined almost everywhere. Wikipedia gives me that $H(x)=1/2(1+\text(sgn)(x))$ en.wikipedia.org/wiki/Heaviside_step_function#Zero_argument , I must say I don't know how they got that but I think you should be able to figure it out (: $\endgroup$ – Chee Han Jun 9 '15 at 3:33

There are two ways to do it as far as I know, but the better way to do it is probably from definition (The other way is using conjugate Poisson kernel, see for example wikipedia: Hilbert transform)

I am going to do it formally, but you could easily justify the calculation below. Since $p.v(1/x)$ is a tempered distribution, by definition,

\begin{align} \widehat{p.v(\frac{1}{x})}(\varphi) & \colon = p.v(\frac{1}{x})(\hat\varphi)\\ & =\int_{\mathbb{R}}\frac{\hat\varphi(\xi)}{\xi}d\xi\\ & =\int_{\mathbb{R}}\frac{1}{\xi}\Big(\int_{\mathbb{R}}\varphi(x)e^{-2\pi ix\cdot\xi}dx\Big)d\xi\\ & =\int_{\mathbb{R}}\varphi(x)\Big(\int_{\mathbb{R}}\frac{1}{\xi}e^{-2\pi ix\cdot\xi}d\xi\Big)dx\\ & =\int_{\mathbb{R}}\varphi(x) F(x)dx \end{align}

Computing $F(x)$ will then give you the Fourier transform of $p.v.(1/x)$ (as a tempered distribution), which you should get that $F(x)=-\pi i\text{ sgn}(x)$, where $\text{sgn}(x)$ is the usual sign function. The interchange of order of integration is justified by splitting the range of integration and apply some convergence theorem as usual.

  • $\begingroup$ Interesting. It looks like the method I sketch gets the right answer after all. Thank you. $\endgroup$ – Potato Jun 8 '15 at 16:39
  • $\begingroup$ This is a brute force computation so it's easier; a cleaner method would be to use conjugate Poisson kernel but that requires some knowledge on Poisson integral. (However my friend once told me the calculation isn't that bad so one should still be able to figure out the intermediate step using conjugate Poisson kernel if the lecturer gave enough information on it.) $\endgroup$ – Chee Han Jun 9 '15 at 3:39

Another solution

The distribution $\mathrm{pv} \frac{1}{x}$ satisfies $x \, \mathrm{pv} \frac{1}{x} = 1.$ Therefore, $$ 2\pi \, \delta(\xi) = \mathcal{F} \{ 1 \} = \mathcal{F} \{ x \, \mathrm{pv} \frac{1}{x} \} = i \frac{d}{d\xi} \mathcal{F} \{ \mathrm{pv} \frac{1}{x} \} $$

Thus, $ \mathcal{F} \{ \mathrm{pv} \frac{1}{x} \} = -i \pi \, \operatorname{sign}(\xi) + C $ for some constant $C$.

But $\mathrm{pv} \frac{1}{x}$ is odd so its Fourier transform must also be odd, and since $-i \pi \, \operatorname{sign}(\xi)$ is odd while $C$ is even, we must have $C=0.$


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