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,

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

  • 2
    $\begingroup$ $|\sin(x)| \le \min\{|x|,1\}$. $\endgroup$ – 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @PacoAdajar It is an upper bound. So it is correct as stated. $\endgroup$ – Stephen Montgomery-Smith Aug 16 '20 at 3:44

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ Yes, but $1/x^{3/2}$ doesn't converge on $[0,\infty)$. $\endgroup$ – Stephen Montgomery-Smith Aug 16 '20 at 3:39
  • 3
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.