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Now, I know that the question was answered here.

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 $$\int_{\mathbb{R}} ne^{-nx} \sin\left(\frac{1}{x}\right) \chi_{[\ln(n+1/n), \ln (n/n-1)]}dx$$ 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

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

We could apply DCT to this to get $0$?

Another EDIT:

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

Felix Marin
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Phil
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  • 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 02:24
  • 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 03:24
  • 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 03:56

2 Answers2

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This is not a very measure theoretic approach, but more of an application of simple facts about oscillatory integrals.

As pointed out by the OP, an application of dominated convergence on the integral $$\int^\infty_0 e^{-y}\sin\Big(\frac{n}{y}\Big)\,dy\tag{1}\label{one}$$ is not possible. The main issue is that although sequence $\phi_n(y)=e^{-y}\sin(n/y)$ is uniformly integrable, it does not converge to $0$, not even in probability.

One can, however, control the oscillatory nature of the function $\sin(n/y)$ about the origin by switching to hyperbolic scale; this will distribute the oscillations more evenly and bring things to the realm of the Riemnann-Lebesgue theorem. In other words, after making the change of variable $u=\frac1y$ in \eqref{one}, we obtain $$\int^{\infty}_0 ne^{-ny}\sin(1/y)\, dy=\int^{\infty}_0e^{-y}\sin\Big(\frac{n}{y}\Big)\,dy=\int^\infty_0\frac{e^{-1/u}}{u^2}\sin(nu)\,du$$ Notice that $g(u)=\frac{e^{-1/u}}{u^2}\in L^+_1(0,\infty)$. Then, an application of the Riemann-Lebesgue theorem (or alternatively, an application of Fejér's lemma) yields

$$\int^\infty_0\frac{e^{-1/u}}{u^2}\sin(nu)\,du\xrightarrow{n\rightarrow\infty}0$$

Mittens
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Too long for a comment

It would be interesting to get the asymptotics of the integral at $n\to\infty$ $$I(n)=\int^\infty_0 e^{-u} \sin\left(\frac{n}{u}\right) du\overset{u=s\sqrt n}{=}=-\Im\sqrt n\int_0^\infty e^{-\sqrt n(s+\frac is)}ds$$ Deforming the contour in the complex plane, we go through the extremum point of $\displaystyle f(s)=s+\frac is$, which is $s_0=e^{\frac{\pi i}4}$

Near this point $$f(s)=f(s_0)+\frac 12f''(s_0)(s-s_0)^2+...=\sqrt2(1+i)+\frac{1-i}{\sqrt2}(s-s_0)^2+...$$ and the asymptotics of our integral appears to be $$I(n)\sim-\Im\sqrt n\,e^{-\sqrt{2n}(1+i)}\int_{-\infty}^\infty e^{-\sqrt{\frac n2}(1-i)z^2}dz=-\Im\,\sqrt n\,e^{-\sqrt{2n}(1+i)}\sqrt{\frac \pi{\sqrt n\,e^{-\frac{\pi i}4}}}$$ $$\boxed{\,\,I(n)\sim\sqrt{\pi\sqrt n}\,\,e^{-\sqrt{2n}}\,\sin\left(\sqrt{2n}-\frac\pi 8\right)\,\,}$$ what provides a reasonable agreement with numeric checks (for example, WA - here and here).

Svyatoslav
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