Prove the derivative of $x^2 \sin (1/x^2)$ is not (Lebesgue) integrable on $[0,1]$ Prove the derivative of $x^2 \sin (1/x^2)$ is not Lebesgue integrable on $[0,1]$.
Note at $x=0$, the value of the function is defined to be $0$. 
Here 'not integrable' means that the integral value approximated by simple functions from the above is not the same as that by the ones from the below.
Should I use some powerful theorem to prove this?
I don't think this question is a hard one but don't know what kind of approach I should take.
 A: (This solution is similar to the one above, but there you also need to show that the length of $I_n$ doesn't go to 0 fast enough -- otherwise of course the sum wouldn't diverge.)
We need to show that the following is not integrable:
$$f'(0) = \cases{
2x \sin\left( \frac{1}{x^2}\right) - \frac{2}{x}\cos\left(\frac{1}{x^2}\right), \quad x \in(0,1], \\
0, \quad x=0. 
}$$
Because a continuous bounded function on an interval is measurable, and therefore integrable by Lebesgue's theorem, we know that $2x \sin\left(\frac{1}{x^2}\right)$ is integrable. The sum of integrable functions will be integrable, so $f'$ not integrable is equivalent to $- \frac{2}{x}\cos\left(\frac{1}{x^2}\right)$ not being integrable; which implies by definition that its absolute value is also not integrable.
Define
$$g(x):= \left\lvert \frac{2}{x}\cos\left(\frac{1}{x^2}\right)\right\rvert.$$
Observe that for $y \in [2k\pi - \pi/4, 2k\pi +\pi/4]$, $\cos(y) \geq \frac{1}{\sqrt{2}}$. For integers $k \geq 1$, define
$$ \overline x(k) := \frac{1}{\sqrt{2k\pi -\pi/4}}$$
$$ \underline x(k) := \frac{1}{\sqrt{2k\pi +\pi/4}} $$
Therefore, $g(x) \geq \frac{\sqrt{2}}{\overline x(k)}$ for $x \in [\underline x(k), \overline x(k)]$. If $I_k:= [\underline x(k), \overline x(k)]$, then 
$$g \geq \sum_{k=1}^K \frac{\sqrt{2}}{\overline x(k)} \chi_{I_k} $$
for arbitrary $K$. Therefore 
$$ \int g \geq \sum_{k=1}^K \sqrt{2}\left(1-\frac{\underline x(k)}{\overline x(x)}\right) = \sum_{k=1}^K \sqrt{2}\left(1-\sqrt{\frac{2k\pi -\pi/4}{2k\pi + \pi/4} }\right).$$
By applying the inequality $\sqrt{1-a} \leq 1 - \frac{a}{2}$, for $a\in[0,1]$, we get:
$$ \int g \geq \sum_{k=1}^K \sqrt{2}\left(\frac{\pi/4}{2k\pi + \pi/4}\right)\geq \sum_{k=1}^K \frac{\sqrt{2}}{4}\left(\frac{1}{2(k+1)}\right).$$
This sum diverges for $K \rightarrow \infty$, so we are done.
A: We need only look at the term $\frac{1}{x}\cos(\frac{1}{x^2})$ which occurs in the expression for the derivative of the given function. The derivative isn't integrable because of the $\frac{1}{x}$ factor (the other $x\sin(\frac{1}{x^2})$ term in the derivative is obviously integrable being continuous on $[0,1]$). You can easily construct a minorising sequence $s_k$ of simple functions whose integral blows up as $k\to\infty$. Consider this construction taking for simplicity only the positive part of $f$. Since $\cos(1/x^2)\ge 1/2$ on $I_n:= [1/(2\sqrt{(n+2/3)\pi}),1/(2\sqrt{(n+1/3)\pi})]$ (pls check and correct if required but the principle is valid), so $\frac{1}{x}\cos(\frac{1}{x^2})\ge \sqrt{(n+1/3)\pi}=:a_n$ on $I_n$. Then if $s_k = \sum_{n=1}^k a_n\chi_{I_n}$, then the integral of $s_k$ blows up as $k\to\infty$. 
