Evaulate $\lim_{n \to \infty} \int_0^\infty ne^{-nx} \sin(1/x)dx$

Now, I know that the question was answered here: $\int_0^\infty ne^{-nx}\sin\left(\frac1{x}\right)\;dx\to ?$ as $n\to\infty$

But I'm looking for more of a measure theoretical approach, so if someone can point me in the right direction, I would be grateful. (i.e. use of Dominated Convergence Theorem or Monotone Convergence Theorem)

My rough idea has been the following:

We can't use MCT because the functions are not pointwise increasing. We can't use DCT since there there is no function to bound $$ne^{-nx} sin(1/x)$$.

But, it seems that on $$[\ln 2, \infty)$$, $$e^{-x} \geq ne^{-nx}$$, and on $$[\ln (\frac{n+1}{n}, \ln \frac{n}{n-1}]$$, $$ne^{-nx} \geq me^{-mx}$$ for all $$n, m \in \mathbb{N}$$ (I don't have proof of this, and it seems hard to prove).

So we can use DCT on integral restricted such intervals. That is, we can use DCT on the integral $$\int_{\mathbb{R}} ne^{-nx} \sin(\frac{1}{x}) \chi_{[\ln(n+1/n), \ln (n/n-1)]}dx$$ And on these intervals, the integrals are all 0.

Since these intervals partitions $$[0, \infty)$$, it follows our integral is 0.

I don't know if the idea is correct, and even if it was, it wouldn't be an elegant solution. So I'm hoping someone could point me in the right direction for a clean solution.

Thanks!

EDIT:

If we let $$u = nx$$, we get the integral

$$\int^\infty_0 e^{-u} \sin(\frac{n}{u}) du$$.

We could apply DCT to this to get 0?

Another EDIT:

But $$\sin(\frac{n}{u})$$ is not convergent, so we cannot apply DCT.

• The problem is now the integrand doesn't converge to zero pointwise. (Which is why I shamefully deleted my comment.) – Liam Mar 15 '20 at 2:24
• Note that the convergence is uniform to 0 on intervals of the form $[\delta, \infty)$, so you can really rephrase your question to be on an interval like $[0,1]$ – operatorerror Mar 15 '20 at 3:24
• The oscillation of $\sin(1/x)$ near $0$ and the lack of a pointwise bound on this interval pose problems though. – operatorerror Mar 15 '20 at 3:56