Fourier transformation of tempered distribution $p.v. \frac{1}{x}$

Question

Prove the fourier transformation of $p.v. \frac{1}{x}\in \mathcal{S}'(\Bbb{R})$ is $-i\pi \text{sgn}(y)$.

This is a classical result that shows in many textbook,my question is some detail question that I can't work out,that is :

\begin{aligned} \text { p.v. } \frac{1}{x}(\varphi) &=\text { p.v. } \frac{1}{x}(\widehat{\varphi})=\lim _{\epsilon \downarrow 0} \int_{\epsilon<|x|<1 / \epsilon} \frac{\widehat{\varphi}(x)}{x} d x \\ &=\lim _{\epsilon \downarrow 0} \int_{\epsilon<|x|<1 / \epsilon} \frac{1}{x}\left(\int_{-\infty}^{\infty} \varphi(y) e^{-2 \pi i x y} d y\right) d x\\ &=\lim _{\epsilon \downarrow 0} \int_{-\infty}^{\infty} \varphi(y)\left(\int_{\epsilon<|x|<1 / \epsilon} \frac{e^{-2 \pi i x y}}{x} d x\right) d y \\ &=\int_{-\infty}^{\infty} \varphi(y)\left(\lim _{\epsilon \downarrow 0} \int_{\epsilon<|x|<1 / \epsilon} \frac{e^{-2 \pi i x y}}{x} d x\right) d y \\ &=-i \pi \int_{-\infty}^{\infty} \operatorname{sgn}(y) \varphi(y) d y \end{aligned}

In the fourth equality,we put the limit into the integral.I can't work out this step,maybe we need to apply DCT,denote $\int_{\epsilon<|x|<1 / \epsilon} \frac{e^{-2 \pi i x y}}{x} d x = g(\epsilon,y)$,we need to bound $|f(y)g(\epsilon,y)|$ in order to apply DCT,but I don't know how.

Answer 1

You meant $$\widehat{\text {p.v.} \frac{1}{x}}(\varphi) =\text {p.v.} \frac{1}{x}(\widehat{\varphi})$$

Then for $y\ne 0$ do the change of variable $x = u/y$ in $$\lim _{\epsilon \downarrow 0} \int_{\epsilon<|x|<1 / \epsilon} \frac{e^{-2 \pi i x y}}{x} d x$$ you'll get $C \text{sign}(y)$ where $C =-i \int_{-\infty}^\infty \frac{\sin(u)}{u}du =- i\pi$

$\int_{\epsilon<|x|<1 / \epsilon} \frac{e^{-2 \pi i x y}}{x} $ converges boundedly and locally uniformly on $\Bbb{R}^*$ to $-i\pi \text{sign}(y)$, so no problem to exchange $\lim $ and $\int$ (as $\hat{\varphi}$ is $L^1$) nor with the case $y=0$.