= Solution
\b[No.] Consider the continuous even <function>
$$
f(x)=\begin{cases}
0,&x=0,\\
1/\log(1/|x|),&0<|x|\leq e^{-1},\\
1,&|x|\geq e^{-1}.
\end{cases}
$$
The formulas agree at $|x|=e^{-1}$, and $f(x)\to0$ at zero. Part (i) defines its <even-weight principal-value distribution>. However,
$$
\boxed{\int_0^{e^{-1}}\frac{|f(x)|}{x}\,dx
=\int_1^\infty\frac{ds}{s}=\infty,\qquad s=\log(1/x).}
$$
If this were a <regular distribution> represented by $g\in L^1_{\mathrm{loc}}(\mathbb R)$, testing on either side of zero would force $g=f/x$ <almost everywhere> there, by uniqueness of the <locally integrable function> representing a <regular distribution>. Changing the value at zero cannot remove the displayed divergence, contradicting <local integrability> of $g$.
More generally, the exact condition is
$$
\boxed{\operatorname{pv}(f/x)\text{ is regular}\iff
\int_0^r\frac{|f(x)|}{x}\,dx<\infty\text{ for some }r>0.}
$$
Necessity is the preceding argument; sufficiency follows because $f/x$ is then a <locally integrable function> and its truncated integrals converge absolutely. The value $f(0)=0$ alone provides no such rate of decay.
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