= Solution
Because $x^{-2}$ is <Lebesgue integrable function>[integrable] on $[1,\infty)$, the <Dominated convergence theorem> applied to the defining integral proves that $v$ is <continuous> on $\mathbb R$.
For $\lambda>0$, the <change of variables formula> $t=\lambda x$ gives
$$
v(\lambda)=-i\lambda\int_\lambda^\infty\frac{e^{-it}}{t^2}\,dt,
$$
and for $\lambda<0$ the analogous formula is obtained with $e^{it}$. On either open half-line the lower endpoint stays away from zero locally, so repeated <differentiation under the integral sign> proves smoothness. Thus
$$
\boxed{v\in C(\mathbb R)\cap C^\infty(\mathbb R\setminus\{0\})}.
$$
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