How can we evaluate $\lim_{n\to\infty}\int_0^\infty nx^2\sin\left(\frac{1}{nx}\right)dx$

Question

If the right side of integral is finite, we can shift limit into to the integral, but how ? $\lim_{n\to\infty}\int_0^\infty nx^2\sin\left(\frac{1}{nx}\right)dx$

Answer 1

Assuming that the integral exists we have: $$ I = \int_{0}^{+\infty}nx^2\sin\left(\frac{1}{nx}\right)\,dx=\frac{1}{n^2}\int_{0}^{+\infty}x^2\sin\frac{1}{x}\,dx,$$ however $x^2\sin\frac{1}{x}$ is not integrable over $\mathbb{R}^+$ since for every $x\geq\frac{2}{\pi}$ we have $\sin\frac{1}{x}\geq\frac{2}{\pi x}$, so $ x^2\sin\frac{1}{x}\geq \frac{2x}{\pi}$.

Answer 2

I think the integrals won't converge for any $ n $ because the integral over $ x> n/\pi $ is $\infty $ while the rest is finite.