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How do I show $$\lim_{n \to \infty} \int_0^\infty \frac{n}{n^2+x}\sin\left(\frac{1}{x}\right)\, dx = 0\,\,?$$ I've tried splitting into the cases where $x \leq 1$ and $x \geq 1$ but I am having trouble finding bounds so that I can apply the dominated convergence theorem.

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  • $\begingroup$ What is the difficulty in finding bounds for the $x\leq 1$ case? The $x\geq 1$ case? $\endgroup$ – Michael Jul 14 '20 at 20:28
  • $\begingroup$ Look at function values. First factor goes to 0 regardless of x, second factor is bounded between -1 and 1. Product therefore goes to 0 at each individual $x$. Integral of the 0 function is 0. $\endgroup$ – mathreadler Jul 14 '20 at 20:40
  • $\begingroup$ Wait, is this allowed to do, to put limit inside of integral like I did..? In what cases? $\endgroup$ – mathreadler Jul 14 '20 at 20:46
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Edit: The second half of this is nonsense. See the comments below...

Say the integrand is $f$. If $0<x\le1$ then $|f(x)|\le 1$, while if $x\ge1$ then $|f(x)|\le 1/x^2$, since $|\sin(t)|\le|t|$.

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  • $\begingroup$ For the case $x \geq 1$, is there a way to easily see the bound you have given? I so far have $|f(x)| \leq \frac{n}{n^2x+x^2}$. I'm having trouble seeing why the bound you have given holds for every $n$. $\endgroup$ – Nino Jul 14 '20 at 20:49
  • $\begingroup$ Hi David. I hope you're doing well and staying safe and healthy. How is $n/(n^2+x)$ bounded by $1/x$ for $x\ge1$? Certainly it is bounded by $1/2\sqrt x$, which is a bound fit for purpose. $\endgroup$ – Mark Viola Jul 14 '20 at 20:54
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    $\begingroup$ @nina Apply the AM_GM and find the bound $1/2x^{3/2}$. $\endgroup$ – Mark Viola Jul 14 '20 at 20:57
  • $\begingroup$ @MarkViola Good question. Wish I had a good answer... $\endgroup$ – David C. Ullrich Jul 14 '20 at 21:52
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[15px,#ffd]{\lim_{n \to \infty}\int_{0}^{\infty} {n \over n^{2} + x}\,\sin\pars{1 \over x}\,\dd x} \,\,\,\stackrel{x\ \mapsto\ 1/x}{=}\,\,\, \lim_{n \to \infty}\bracks{{1 \over n}\int_{0}^{\infty} {\sin\pars{x} \over \pars{x + 1/n^{2}}x}\,\dd x} \\[5mm] = &\ \lim_{n \to \infty}\bracks{{\pi n \over 2} - n\int_{0}^{\infty}{\sin\pars{x} \over x + 1/n^{2}}\,\dd x} \\[5mm] = &\ \lim_{n \to \infty}\pars{{\pi n \over 2} - n\braces{\operatorname{Ci}\pars{1 \over n^{2}}\sin\pars{1 \over n^{2}} + {1 \over 2}\bracks{\pi -2\operatorname{Si}\pars{1 \over n^{2}}}\cos\pars{1 \over n^{2}}}} \end{align} $\ds{\operatorname{Ci}}$ and $\ds{\operatorname{Si}}$ are the Cosine and Sine Integrals Functions, respectively.

As $\ds{z \to 0}$, $\ds{\quad\operatorname{Ci}\pars{z} \sim \gamma + \ln\pars{z} - {1 \over 4}\,z^{2}\quad}$ and $\ds{\quad\operatorname{Si}\pars{z} \sim z - {1 \over 18}\,z^{3}\quad}$ from this link.


Therefore, \begin{align} &\bbox[15px,#ffd]{\lim_{n \to \infty}\int_{0}^{\infty} {n \over n^{2} + x}\,\sin\pars{1 \over x}\,\dd x} = -\lim_{n \to \infty}{\gamma - 2\ln\pars{n} \over n} \\[5mm] = &\ -\lim_{n \to \infty}{\bracks{\gamma - 2\ln\pars{n + 1}} - \bracks{\gamma - 2\ln\pars{n}} \over \pars{n + 1} - n} = 2\lim_{n \to \infty}\ln\pars{1 + {1 \over n}} = \bbox[15px,#ffd,border:1px solid navy]{0} \end{align}
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By the self-adjointness of the Laplace transform

$$\int_{0}^{+\infty}\frac{\sin x}{x}\cdot \frac{1}{nx+\frac{1}{n}}\,dx =\frac{1}{n} \int_{0}^{+\infty}\left(\frac{\pi}{2}-\arctan(s)\right)e^{-s/n^2}\,ds $$ where the RHS is more manageable than the LHS since no oscillating functions are involved.
We have $$ \int_{0}^{n^3}\left(\frac{\pi}{2}-\arctan(s)\right)\,ds =n^3\arctan\frac{1}{n^3}+\frac{1}{2}\log(1+n^6)=O(\log n)$$ $$ \int_{n^3}^{+\infty} e^{-s/n^2}\,ds = n^2 e^{-n}=O(1)$$ hence the wanted limit is clearly zero.

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