TITLE: Fourier transform of the distribution PV $\left( \frac{1}{x} \right)$
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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 $H$.