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.



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.

  • $\begingroup$ The problem is now the integrand doesn't converge to zero pointwise. (Which is why I shamefully deleted my comment.) $\endgroup$
    – Liam
    Commented Mar 15, 2020 at 2:24
  • $\begingroup$ 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 $[0,1]$ $\endgroup$ Commented Mar 15, 2020 at 3:24
  • $\begingroup$ The oscillation of $\sin(1/x)$ near $0$ and the lack of a pointwise bound on this interval pose problems though. $\endgroup$ Commented Mar 15, 2020 at 3:56


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