Fourier transform of the distribution PV $\left( \frac{1}{x} \right)$

Let $u = PV\left(\frac1x\right)$. Then $xu = 1$. Now $\hat 1 = 2\pi \, \delta$ so we have $$ \langle 2\pi \, \delta, \phi \rangle = \langle \hat 1, \phi \rangle = \langle \widehat{xu}, \phi \rangle = \langle xu, \hat\phi \rangle = \langle u, x \hat\phi \rangle = \langle u, -i \widehat{\phi'} \rangle = \langle -i \hat u, \phi' \rangle = \langle i (\hat u)', \phi \rangle $$

Thus, $i(\hat u)' = 2\pi \, \delta$ which gives $\hat u(\xi) = -i\pi \operatorname{sign}(\xi) + C$. But since $u$ is odd so is also $\hat u$ which forces $C = 0$.

First note that $xp.v.\frac{1}{x} = 1$. If $\delta$ is Dirac's delta function as a distribution then for all test functions, $\varphi$, $$\bigg\langle \hat{\delta}, \varphi\bigg\rangle = \bigg\langle \delta, \hat{\varphi}\bigg\rangle = \hat{\varphi}(0) = \int_{\mathbb{R}}\varphi (x) dx = \bigg\langle 1, \varphi\bigg\rangle$$ Thus $\hat{\delta}$ = 1.

In order to deduce the Fourier transform of $1$ in the sense of distributions consider:

$\bigg\langle \hat{1}, \varphi\bigg\rangle = \bigg\langle 1, \hat{\varphi}\bigg\rangle = \int_{\mathbb{R}}\hat{\varphi} (x) dx = \int_{\mathbb{R}}e^{2i\pi 0 x}\hat{\varphi} (x) dx = \mathcal{F}^{-1}\hat{\varphi}(0) = \varphi(0) = \bigg\langle \delta, \varphi\bigg\rangle$

Thus $\hat{1} = \delta$.

Now, using properties of derivatives of distributions and Fourier transforms of derivatives of functions:

$$ \bigg\langle \delta, \varphi\bigg\rangle = \bigg\langle \hat{1}, \varphi\bigg\rangle = \bigg\langle \mathcal{F}(xp.v.\frac{1}{x}), \varphi\bigg\rangle = \bigg\langle xp.v.\frac{1}{x}, \hat{\varphi}\bigg\rangle = \bigg\langle p.v.\frac{1}{x}, x\hat{\varphi}\bigg\rangle = \bigg\langle p.v.\frac{1}{x}, x \frac{\hat{\varphi '}}{2i\pi x}\bigg\rangle$$ (In the last equality we use the fact that $\mathcal{F}(f^{(k)}(\xi)) = (2i\pi \xi)^{(k)})\hat{f}(\xi)$.) $$= \bigg\langle p.v.\frac{1}{x}, \frac{1}{2i\pi } \hat{\varphi '}\bigg\rangle = \bigg\langle \frac{1}{2i\pi } p.v.\frac{1}{x}, \hat{\varphi '}\bigg\rangle = \bigg\langle \frac{1}{2i\pi } \mathcal{F}(p.v.\frac{1}{x}), \varphi '\bigg\rangle = \bigg\langle -\frac{1}{2i\pi } \big(\mathcal{F}(p.v.\frac{1}{x})\big)', \varphi \bigg\rangle $$ $$\Rightarrow -\frac{1}{2i\pi } \big(\mathcal{F}(p.v.\frac{1}{x})\big)' = \delta$$

$H$ is the Heaviside function that is $0$ on $(-\infty, 0)$, $\frac{1}{2}$ on $0$, and $1$ on $(0, \infty)$. The derivative of $H$ in the sense of distributions is $\delta$. Thus we have:

$$ \mathcal{F}(p.v.\frac{1}{x}) = -2i\pi H + C$$

Since $p.v.\frac{1}{x}$ is odd, $\mathcal{F}(p.v.\frac{1}{x})$ is odd, and thus $-2i\pi H + C$ is an odd distribution. Thus the function describing the distribution, $-2i\pi H(x) + C$ is odd. That means $-2i\pi H(x) + C = -(-2i\pi H(-x) + C) = 2i\pi H(-x) - C$.

If $x$ is positive, $H(x) = 1, H(-x) = 0$, and we have $C = i\pi.$

If $x$ is negative, $H(x) = 0, H(-x) = 1$, and we have $C = i\pi.$

If $x$ is 0, $H(x) = H(-x) = H(0) = \frac{1}{2}$, and we have $C = i\pi.$

Thus $\mathcal{F}(p.v.\frac{1}{x}) = -2i\pi H + i\pi$.

Note, $-2i\pi H + i\pi = -i\pi sgn$, where $sgn$ is the sign function, based on how we defined $H$.