Given the sequence of distributions: $$ x^3~ \sin (nx),~~n \in (\mathbb{N}) $$ How can i find the limit for $n \rightarrow \infty$?
I tried with the usual substitution $y=nx$, but it leads to integrals that don't converge.
Using instead the Riemann-Lebesgue theorem, i get that the limit is zero, but I don't think this is the right answer (Wolfram alpha says that, in the sense of functions, the limit is $x^3 -1$). Am I missing something? Why can't i use the Riemann-Lebesgue theorem?
We indeed obtain that the limiting distribution is $0$ because $x\mapsto x^3\varphi(x)$ is a smooth integrable function with integrable derivative. So Riemann-Lebesgue lemma applies.