Consider the function $F(x)=x^2\sin(1/x^2)$ for $x\not=0$, and $F(0)=0$. Show that $F^{'}(x)$ exists for every x, but that $F^{'}(x)$ is not integrable on [-1,1].
So the part I am struggling with here is:
1) Showing that our derivative will exist for every x. I can show obviously take the derivative to get $F^{'}(x) = 2x\sin(1/x^{2})-\frac{2}{x}\cos(1/x^{2})$ for $x\not=0$ and $F^{'}(0)=0$, but I'm not sure that this actually shows existence.
2) Is the issue with integrability over this interval the fact that at $x=0$, we have $-\frac{2}{x}\cos(1/x^{2})$ undefined? I'm confused since our original derivative is piecewise-defined to handle a similar issue with the parent function.
Any pointers or corrections are appreciated!
The issue with integrability is not that $F'$ is undefined at $x = 0$. As you claim and is in fact easy to show, $F'(0) = 0$.
Clearly $F'$ is unbounded in the vicinity of $x = 0$ due to the term $\frac{2}{x} \cos \frac{1}{x^2}$, although this by itself does not preclude the existence of the Lebesgue integral. For example $x \mapsto 1/\sqrt{x}$ is unbounded but Lebesgue integrable on $[0,1]$.
Nevertheless, $x \mapsto \frac{2}{x} \cos \frac{1}{x^2}$ fails to be Lebesgue integrable since using a change of variables $x = 1/\sqrt{u}$ we have
$$\int_0^1\frac{2}{x} \left|\cos \frac{1}{x^2}\right| \, dx = \int_1^\infty\frac{|\cos u|}{u} \, du > \sum_{k=1}^\infty\int_{\pi k}^{\pi (k+1)}\frac{|\cos u|}{u} \, du \\ > \sum_{k=1}^\infty\frac{1}{\pi(k+1)}\int_{\pi k}^{\pi (k+1)}|\cos u| \, du = \sum_{k=1}^\infty\frac{2}{\pi(k+1)} = +\infty$$
Recall that a function $f$ is Lebesgue integrable over $E$ if and only if $\int_E |f| < \infty$.