# For which values of the parameter $\alpha$ function $f(x) = \frac{1}{x^{\alpha}}\left|sin\frac{1}{x}\right|$ is Lebesgue Integrable?

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?

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.

edit:

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.

• 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 $(2\pi n)^{\alpha}$ May 8 '20 at 10:18
• @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 :) May 8 '20 at 10:54