Fourier transform of the distribution PV $\left( \frac{1}{x} \right)$ 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.
 A: 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 $C = 0$.
A: 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 $H$. 
A: First off, you're off by a minus sign in your definition of the Fourier transform. Accounting for this, you can then make a change of variables in your first $dx$ integral and combine the two $\epsilon$ limits as one:
\begin{align}
&= \int_{-\infty}^{\infty}\varphi(\xi)\lim_{\epsilon\to 0} \left( \int_{\epsilon}^{\infty} -\frac{e^{ix\xi}-e^{-ix\xi}}{x} dx\right)d\xi \\
&= \int_{-\infty}^{\infty}\varphi(\xi)\lim_{\epsilon\to 0} \left( -2i\int_{\epsilon}^{\infty} \frac{\sin x\xi}{x} dx\right)d\xi \\
&= \int_{-\infty}^{\infty}\varphi(\xi) (-2i) \left(\frac{\pi}{2}\text{sign}(\xi) \right) d\xi \\
&= \int_{-\infty}^{\infty}\varphi(\xi) \Big( -i\pi\,\text{sign}(\xi)\Big)d\xi \\
&= \langle F,\varphi \rangle,
\end{align}
where $F(\xi) = -i\pi\,\text{sign} \xi$. But now this means that $\widehat{PV(\frac{1}{x})} = -i\pi\,\text{sign} \xi$ as a distribution, since it's what you are integrating $\varphi$ against.
A: I thought it might be instructive to present an approach that the OP was attempting, but was not addressed in the currently posted answers.  Moreover, the interchange of integrals and the interchange of the limit with the integral need to be justified.  In the following we provide a way forward with justification.  To that end, we proceed.

Your approach is formally on target.  Let the distribution $f$ be given by $f(x)=\text{PV}\left(\frac1x\right)$.  Then certainly we have
$$\begin{align} \langle \mathscr{F}\{f\},\phi \rangle&=\langle f, \mathscr{F}\{\phi\}\rangle\\\\
&=\text {PV}\int_{-\infty}^\infty \frac{\mathscr{F}\{\phi\}(x)}{x}\,dx\\\\
&=\lim_{\varepsilon\to0^+\\L\to \infty}\int_{\varepsilon\le|x|\le L}\frac1x \int_{-\infty}^{\infty} \phi(k)e^{ikx}\,dk\,dx\tag1\\\\
&=\lim_{\varepsilon\to0^+\\L\to \infty} \int_{-\infty}^\infty \phi(k) \int_\varepsilon^L \frac{2i\sin(kx)}{x}\,dx\,dk\tag2\\\\
&=\int_{-\infty}^\infty \phi(k) i\pi \text{sgn}(k) \,dk\tag3\\\\
&=\langle i\pi \text{sgn}, \phi \rangle
\end{align}$$
So, we find that in distribution,
$$\mathscr{F}\{f\}=i\pi \text{sgn}$$


NOTES:
In going from $(1)$ to $(2)$, Fubini-Tonelli applies since $\phi \in \mathbb{S}$.  To see this, we note that since $\phi \in \mathbb{S}$, then $\int_{-\infty}^\infty |\phi(k)e^{ikx}|\,dk=\int_{-\infty}^\infty |\phi(k)|\,dk<\infty$.  Hence, the interated integral $\int_{\varepsilon<|x|<L}\int_{-\infty}^\infty |\phi(k)e^{ikx}|\,dk<\infty$.
In going from $(2)$ to $(3)$, the Dominated Convergence Theorem applies.  To see this, note that $\left|\phi(k)\int_\varepsilon^L \frac{2i \sin(kx)}{x}\,dx\right|\le |\phi(k)\int_0^\pi \frac{\sin(x)}{x}\,dx$ and is uniformly bounded by a function that is independent of $\varepsilon$ and $L$.  And inasmuch as $\phi\in\mathbb{S}$, $\int_{-\infty}^\infty |\phi(k)|\,dk<\infty$, the Dominated Convergence Theorem is applicable.
