2
$\begingroup$

I happened upon the following integration exercise. I don't know if it would count as standard in measure theory, but it seems interesting, and I am wondering how best to approach it.

Let $r \in \mathbb{R}$. For which values of $r$ is the following function from $(0,1] \rightarrow \mathbb{R}$ (1) Riemann integrable in the improper sense or (2) Lebesgue integrable:

$y^r \sin(1/y).$

For (1), I was thinking that an effective argument will require integration by parts and L'Hopital's Rule in tandem, but I am a little rusty in using these tools (it's embarassing!); perhaps the Riemann-Lebesgue lemma will factor in too. Unfortunately, I don't have such a "knee-jerk response" in mind for (2), since Lebesgue integration is something new that I haven't fully grappled with yet. Any assistance a visitor to the site cares to give would be appreciated! Thanks.

  • 0
    I don't see why the Riemann-Lebesgue lemma should factor in. What do you mean specifically? (Especially since the RL lemma is a statement where the hypothesis requires the function to be integrable.)2012-03-02

1 Answers 1

2

Since $|y^r\sin(1/y)|\le y^r$, the integral is absolutely Riemann integrable if $r>-1$ (in the improper sense if $-1 ). For these values of $r$, it is also Lebesgue integrable.

To see what happens when $r\le-1$, it is convenient to make the change of variable $y=1/t$, which transforms the integral into $ \int_1^\infty t^{-(r+2)}\sin t\,dt. $ This is integrable in the improper Riemann sense if and only if $-2, and is not Lebesgue integrable for any $r\le-1$.

  • 0
    @ Alexandre Yes, I refere to r>-1.2012-03-02