I must show that the functional $P\left(\frac{1}{x}\right) \in \mathscr{D}^{\prime}(\mathbb{R})$ defined by
$\left\langle P\left(\frac{1}{x}\right), \varphi\right\rangle \stackrel{\operatorname{def}}{=} P . V . \int_{-\infty}^{\infty} \frac{\varphi(x)}{x} d x=\lim _{\varepsilon \rightarrow 0}\left(\int_{-\infty}^{-\varepsilon} \frac{\varphi(x)}{x} d x+\int_{\varepsilon}^{\infty} \frac{\varphi(x)}{x} d x\right)$
is not a regular distribution. So, I believe I have an idea into how to solve this. Suppose there is a function f that is locally integrable such that
$\int f \phi=\lim _{\varepsilon \rightarrow 0}\left(\int_{-\infty}^{-\varepsilon} \frac{\varphi(x)}{x} d x+\int_{\varepsilon}^{\infty} \frac{\varphi(x)}{x} d x\right)$, we know that the first and second integrals are defined, when I calculate this
$=\lim _{\varepsilon \rightarrow 0^{+}}\left(\varphi(x) \ln \left.|x|\right|_{\varepsilon} ^{\infty}+\varphi(x) \ln \left. |x|\right|_{-\infty} ^{-\varepsilon}+\left(\int_{\varepsilon}^{\infty}+\int_{-\infty}^{-\varepsilon}\right) \varphi^{\prime}(x) \ln (x) d x\right)$
$=\lim _{\varepsilon \rightarrow 0^{+}}\left(-\varphi(\varepsilon) \ln |\varepsilon|+\varphi(-\varepsilon) \ln |-\varepsilon|+\left(\int_{\varepsilon}^{\infty}+\int_{-\infty}^{-\varepsilon}\right) \varphi^{\prime}(x) \ln (x) d x\right)$
$=\lim _{\varepsilon \rightarrow 0^{+}}\left(\ln |\varepsilon| (-\varphi(\varepsilon)+\varphi(-\varepsilon)) +\left(\int_{\varepsilon}^{\infty}+\int_{-\infty}^{-\varepsilon}\right) \varphi^{\prime}(x) \ln (x) d x\right)$
$\leq \lim _{\varepsilon \rightarrow 0^{+}}\left(\ln |\varepsilon|C+\left(\int_{\varepsilon}^{\infty}+\int_{-\infty}^{-\varepsilon}\right) \varphi^{\prime}(x) \ln (x) d x\right)$
Nontheless,if I take a limit the first term can tend to zero, but what happens to the second term of integrals? Will the whole expression tend to zero when taking the limit? Also, if this is equivalently $0$ then through contradiction since I assumed $f \in L^1_{loc}$ then I this would be a contradiction, hence, there would be no such $f \in L^1_{loc}$, and $P(\frac{1}{x})$ wouldn't not be a regular distribtion. Any help would be really appreciated.
You bounded it by $\infty$ since log blows up at 0. I don't think this works.
The distribution $P(1/x)$ is "equal to $1/x$ away from $0$." That is, if there was a function $f$ that induced $P(1/x)$, then at every point $x\neq 0$, by using test functions with support containing $x$ and avoiding $0$, we deduce by the fundamental lemma of the calculus of variations that $f(x) = 1/x$. As there is no $L^1$ extension of $1/x$ to $\mathbb R$, we're done.