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?

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 $C=0.$
