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Define $f_n:(0,\infty)\to\mathbb{R}$ by $f_n(x)=\frac{1}{x^{3/2}}\sin(\frac{x}{n})$. Compute $\lim\limits_{n\to\infty}\int\limits_0^{\infty}f_n(x)dx$.

I'm trying to use the dominated convergence theorem.

So to find a dominated function,
$|f_n(x)|=\frac{1}{x^{3/2}}|\sin(\frac{x}{n})|\leq\frac{1}{x^{3/2}}\frac{x}{n}=\frac{1}{n\sqrt{x}}\leq\frac{1}{\sqrt{x}}$

But this function cannot be used because it is not integrable on $(0,\infty)$.
I really appreciate your help in solving this question.

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    $\begingroup$ $|\sin(x)| \le \min\{|x|,1\}$. $\endgroup$ – Stephen Montgomery-Smith Aug 16 '20 at 3:20
  • $\begingroup$ don't we have $|\sin(a)|\leq|a|$ for any $a$? $\endgroup$ – Charith Aug 16 '20 at 3:24
  • $\begingroup$ Yes, and we also have $|\sin(a)| \le 1$ for any $a$. $\endgroup$ – Stephen Montgomery-Smith Aug 16 '20 at 3:24
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    $\begingroup$ @StephenMontgomery-Smith Well.. so are you suggesting to divide the $(0,\infty)$ to two portions, $(0,1)$ and $[1,\infty)$ and use the two functions appropriately ? $\endgroup$ – Charith Aug 16 '20 at 3:29
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    $\begingroup$ @PacoAdajar It is an upper bound. So it is correct as stated. $\endgroup$ – Stephen Montgomery-Smith Aug 16 '20 at 3:44
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I would try to make a substitution for the integral.

$$\int_0^\infty \frac{1}{x^{3/2}}\sin(x/n)dx = \frac{1}{\sqrt{n}}\int_{0}^\infty \frac{\sin(x)}{x^{3/2}}dx$$.

The quantity in the right integral converges since $|\sin(x)| < 1$ for all $x\in \mathbb{R}$. Thus, the limit as $n\rightarrow\infty$ is 0.

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  • $\begingroup$ Can you please explain why the integral on the right should be finite... $\endgroup$ – Charith Aug 16 '20 at 3:33
  • $\begingroup$ You haven't really explained why the right integral converges. $\endgroup$ – Stephen Montgomery-Smith Aug 16 '20 at 3:33
  • $\begingroup$ @Stephen Montgomery-Smith the reason for that is as follows: if $f \le g \le h$ on $I$ and both $\int_I \! f$ and $\int_I \! h$ converge, so does $\int_I \! g$. (The inequalities as given aren't strict, but that hardly matters.) $\endgroup$ – Paco Adajar Aug 16 '20 at 3:38
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    $\begingroup$ Yes, but $1/x^{3/2}$ doesn't converge on $[0,\infty)$. $\endgroup$ – Stephen Montgomery-Smith Aug 16 '20 at 3:39
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    $\begingroup$ Right. In that case you break it into $[0, 1]$ and $[1, +\infty)$. Around $0$, $\sin{x} \sim x$ so the integral on $[0, 1]$ might as well be $\int_0^1 \! x^{-1/2} \, dx$, which we know to converge. $\endgroup$ – Paco Adajar Aug 16 '20 at 3:40

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