Showing that a finction is not integrable

Hello!

I've got what looks like a simple problem but I just can't get it right.

Here it goes: Show that $\displaystyle f(x)=d/dx (x^2\sin x^{-3}), x>0$, is not in $\displaystyle L^1(0,1)$.

My first thought was to simply compute the integral

$\displaystyle \int_{0}^{1}|f(x)|dx$ and show that it diverges - however I had no luck doing that.

Then I thought that I might simply estimate it from below with something that I know is infinite but no luck there either, all I got was:

$\displaystyle \int_{0}^{1}|2x\sin x^{-3}-3x^{-2}\cos x^{-3}|dx \geq \int_{0}^{1}(|2x\sin x^{-3}|-|3x^{-2}\cos x^{-3}|)dx $

Thanks!

Re: Showing that a finction is not integrable

You don't need to compute the integral. Just notice that $\displaystyle \sin (x^{-3})\overset{0}{\sim} x^{-3}$ and the integral $\displaystyle \int_0^1\frac 1xdx$ is divergent.