Let $f(x)$ non-negative measurable function on $E$ $$ f(x) = \frac{1}{x^{\alpha}}\left|sin\frac{1}{x}\right| $$

I'm trying to figure out for which $\alpha$ function $f(x) = \frac{1}{x^{\alpha}}\left|sin\frac{1}{x}\right|$ is Lebesgue Integrable on $E = \left(0, 1 \right]$

What have I done?

  1. $0 < \alpha < 1$
    $f(x) \leq \frac{1}{x^{\alpha}}$, and I know that for $0 < \alpha < 1$ function $\frac{1}{x^{\alpha}}$ is Lebesgue Integrable and $\int_{0}^{1}\frac{1}{x^{\alpha}} = \frac{1}{1 - \alpha}$. Hence $f(x)$ Lebesgue Integrable
  2. $\alpha$ < 0
    In this case $f(x) = x^{\beta}\left|sin\frac{1}{x}\right|$ where $\beta > 0$. $f(x) \leq x^{\beta}$, $x^{\beta}$ Riemann integrable function. Hence $f(x)$ Lebesgue Integrable.
  3. $\alpha = 0$
    In this case $f(x) = \left|sin\frac{1}{x}\right|$ function limited. Hence $f(x)$ Lebesgue Integrable
  4. $\alpha \geq 1$ I'm stuck on this case now

How can I prove $f(x)$ is Lebesgue Integrable or not for above case?


1 Answer 1


My attempt:

As $f$ is Lebesgue integrable if and only if $f^+$ is integrable and $f^-$ is integrable .

(Denote $f^+=f$ when $f>0$ and $f^+=0$ when $f\le 0$; $f^-=-f$ when $f<0$ and $f^-=0$ when $f\ge 0$)

So we just need to show the positive part of $f$ is integrable.

First, we choose the part where $f>0$:

when $x\in(\frac{1}{\pi+2\pi n},\frac{1}{2\pi n})(n\in\mathbb{N}^+),1/x\in(2\pi n,\pi)$, $sin(1/x)\in(0,1)$.

Then we get a series of interval: $\{(\frac{1}{\pi+2\pi n},\frac{1}{2\pi n})\}$, in which $f(x)>0$, we denote that series as $\{I_n\}$.

After that, we can do the tricky job: on each interval $I_n$, $$\int_{I_n}fdm\ge \frac{1}{2}*(\frac{1}{2\pi n}-\frac{1}{2\pi n+\pi})*(2\pi n)^\alpha = \frac{1}{2}*\frac{\pi}{(2\pi n)(2\pi n+\pi)}*(2\pi n)^{\alpha} = \frac{1}{2}*\frac{n^{\alpha-1}}{2n+1}*(2\pi)^{\alpha-1}$$

(Note:Just think about the usual Sine plot, there is lots of "arch"s, and within every "arch", there is a triangle, which area is $\frac{1}{2}*\pi*1$. Similarly, we can get the triangle in every "arch" of $\frac{1}{x^\alpha}|sin\frac{1}{x}|$, which area is bigger than $\frac{1}{2}*\frac{\pi}{(2\pi n)(2\pi n+\pi)}*(2\pi n)^{\alpha}$, then the inequality is easy to see.)

We can find that when $\alpha\ge1$, then first part $\frac{n^{\alpha-1}}{2n+1}\ge\frac{1}{2n+1}$, then $\int_{I_n}fdm\ge\frac{1}{2n+1}$. So, we sum these $\int_{I_n}fdm$, which is unbounded, thus when $\alpha\ge1$, $f$ is not integrable.


It seemed that i mistake the problem. If $f(x)=\frac{1}{x^\alpha}|sin\frac{1}{x}|$, then $f\ge0$ all the time. So $f=f^+ + f^-$, and the conclusion is right as well.

  • $\begingroup$ Could you explain this $\int_{I_n}fdm\ge \frac{1}{2}*(\frac{1}{2\pi n}-\frac{1}{2\pi n+\pi})*(2\pi n)^\alpha = \frac{1}{2}*\frac{\pi}{(2\pi n)(2\pi n+\pi)}*(2\pi n)^{\alpha} = \frac{1}{2}*\frac{n^{\alpha-1}}{2n+1}*(2\pi)^{\alpha-1}$. As I understand $\frac{1}{2\pi n} - \frac{1}{2\pi n + \pi}$ = length of intervale $I_{n}$, but what are $\frac{1}{2}$ and $(2\pi n)^{\alpha}$ $\endgroup$
    – user717043
    May 8, 2020 at 10:18
  • $\begingroup$ @user717043 well, just think about the $sin x$ plot, the area of every "arch" is bigger than the triangle in the "arch". And the area of the triangle is $\frac{1}{2}*\text{length of interval}*\text{the height of the arch}$. In the solution above, i use the minimum "height" of the "arch". And this is one of my homework undo, so i'm not sure whether it's correct :) $\endgroup$
    – robothead
    May 8, 2020 at 10:54

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