Real Analysis. Integration. Confused

Question

Prove that the function $f:[0,+\infty) \longrightarrow \mathbb{R}$ defined by $f(x) = x^{\frac{3}{2}}\sin\left(\frac{1}{x}\right)$ if $x > 0$ and $f(0) = 0$ is differentiable. Prove that $f'$ is not integrable in any interval $[0, \epsilon]$.

I did the first part of the question, but I could not do the second part. The derivative $f'$ is product and sum of continuous functions if $x>0$. Thus $0$ is the only point of discontinuity (the set of the descontinuity points of $f'$ is a null set).

I'm confused.

Answer 1

$$f'(x)=\frac32\sqrt x\,\sin\frac1x-\frac{\cos\frac1x}{\sqrt x}$$

and thus this function isn't bounded in any (right) neighbourhood of zero...and this condition is a must, not only that its set of discontinuities has measure zero.

Answer 2

$$f'\Bigl(\frac {1}{(2n+1)\pi}\Bigr )=\sqrt {(2n+1)\pi} $$

this proves that $f'$ is unbounded.

$$\lim_{n\to+\infty}f' \Bigl(\frac {1}{(2n+1)\pi}\Bigr)=+\infty $$