Fourier transform of the principal value distribution 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?
 A: 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 $C=0.$
A: 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. 
