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

Question

I need to find the fourier transform of $f =$ PV $\left( \frac{1}{x} \right) $ which is defined as
\begin{align}
PV \left( \frac{1}{x} \right)(\varphi) = \lim_{\varepsilon\to 0} \int_{|x|>\varepsilon} \left( \frac{\varphi(x)}{x} \right) dx
\end{align}
Let $\hat{f}$ denote Fourier transform of $f$. We know that $\langle \hat f,\varphi\rangle= \langle f,\hat \varphi\rangle$ where $\varphi$ is in Schwartz class , $S(\mathbb R)$. $\\$
My attempt is as follows
\begin{align}
\langle \hat f,\varphi\rangle
= \langle f,\hat \varphi\rangle
& = \lim_{\varepsilon\to 0} \int_{|x|>\varepsilon} \left( \frac{\hat\varphi(x)}{x} \right) dx \\&
= \lim_{\varepsilon\to 0} \int_{|x|>\varepsilon} \left( \frac{1}{x} \right)\left( \int_{-\infty}^{\infty}\varphi(\xi)e^{ix\xi}d\xi\right) dx \\&
= \int_{-\infty}^{\infty}\varphi(\xi)\lim_{\varepsilon\to 0} \int_{|x|>\varepsilon} \left( \frac{e^{ix\xi}}{x} \right) dx d\xi
\\&= \int_{-\infty}^{\infty} \varphi(\xi)\lim_{\varepsilon \to 0}\left(\int_{-\infty}^{-\varepsilon}\left( \frac{e^{ix\xi}}{x}\right)dx+\int_{\varepsilon}^{\infty}\left( \frac{e^{ix\xi}}{x}\right)dx\right)d\xi.
\end{align}
I am stuck here. I intituvely expect something like Heaviside function coming out of limit process and integration because of presence of $\frac{1}{x}$.
Any help will be deeply acknowledged.

Answer 1

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 $C = 0$.