Fourier transform of the principal value distribution

Question

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?

Answer 1

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 $C=0.$

Answer 2

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.

Answer 3

I thought it might be instuctive to present a rigorous and direct approach to determining the Fourier Transform of $\text{PV}\frac1x$. To that end, we now proceed.

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First, we define the Fourier Transform for an $L^1$ function, $f$ as

$$\mathscr{F}\{f\}(k)=\int_{-\infty}^\infty f(x) e^{ikx}\,dx$$

Next, we define the Principal Value of the distribution, $d(x)=\left(\text{PV}\frac1x\right)$ as

$$\langle d,\phi\rangle= \lim_{\delta\to 0^+}\int_{|x|\ge \delta}\frac{\phi(x)}{x}\,dx$$

where $\phi\in \mathbb{S}$. Therefore, we have for any $\phi\in \mathbb{S}$

$$\begin{align} \langle \mathscr{F}\{d\},\phi\rangle &=\langle d, \mathscr{F}\{\phi\}\rangle\\\\ &=\lim_{\delta\to 0^+}\int_{|x|\ge\delta}\frac1x \int_{-\infty}^\infty \phi(k) e^{ikx}\,dk\,dx\\\\ &=\lim_{\delta\to 0^+} \lim_{L\to\infty}\int_{-\infty}^\infty \phi(k) \int_{\delta\le |x|\le L}\frac{e^{ikx}}{x}\,dx\,dk\\\\ &=\lim_{\delta\to 0^+} \lim_{L\to\infty}\int_{-\infty}^\infty \phi(k) \int_{\delta\le |x|\le L} \frac{i\sin(kx)}{x}\,dx\,dk \end{align}$$

Note that $\left| \int_{\delta\le |x|\le L} \frac{\sin(kx)}{x}\,dx\right|\le 4$ for all $0\le \delta<L$ and all $k$. Hence, applying the dominated convergence theorem we have

$$\begin{align} \langle \mathscr{F}\{d\},\phi\rangle&=\lim_{\delta\to 0^+} \lim_{L\to\infty}\int_{-\infty}^\infty \phi(k) \int_{\delta\le |x|\le L} \frac{i\sin(kx)}{x}\,dx\,dk\\\\ &=\int_{-\infty}^\infty \phi(k) \lim_{\delta\to 0^+} \lim_{L\to\infty}\int_{\delta\le |x|\le L} \frac{i\sin(kx)}{x}\,dx\,dk\\\\ &=\int_{-\infty}^\infty \phi(k) \left(i\pi \text{sgn}(k)\right) \end{align}$$

We conclude, therefore, that the Fourier Transform of the tempered distribution $d(x) = \left(\text{PV}\frac1x\right)$ is

$$\mathscr{F}\{d\}(k)=i\pi \text{sgn}(k)$$

And we are done!