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
$$
\boxed{v\in C(\mathbb R)\cap C^\infty(\mathbb R\setminus\{0\})}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 327 / 1 / b / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-327/1/b/ii/solution