Show that derivative exists, but derivative is not integrable Let $f(x) = \begin{cases} x^{2} \cdot \sin(1/x^{2}) & x \ne 0 \\ 0 & x = 0\end{cases}$. How do I show that $f'(x)$ exists everywhere but $f'$ is not integrable. We have that $f'(x) = \begin{cases} 2x\cdot \sin(1/x^{2}) -\frac{2}{x} \cos(1/x^{2}) & x \ne 0 \\ 0& x = 0\end{cases}$. Certainly $f'(x)$ is defined for all $x \in \mathbb{R}$, but I suppose that $f'(x)$ blows up at $x \rightarrow 0$. Does this then imply that $f'$ is not integrable on $[-1,1]$. I think I need help rigorously proving this claim. 
 A: A Riemann integrable function must be bounded. But you can pick a sequence $x_{n}\rightarrow 0$ such that $f'(x_{n})\rightarrow\infty$, so it is not Riemann integrable.
But actually it is not even Lebesgue integrable, the problem is the function $x^{-1}\cos(1/x^{2})$. If it were, we have
\begin{align*}
\int_{0}^{1}\dfrac{1}{x}|\cos(1/x^{2})|dx&=\int_{1}^{\infty}\dfrac{|\cos(u^{2})|}{u}du\\
&\geq\sum_{n=1}^{\infty}\int_{(2n\pi)^{1/2}}^{(2n\pi+\pi/3)^{1/2}}\dfrac{1}{u}\cos(u^{2})du\\
&\geq\dfrac{1}{2}\sum_{n=1}^{\infty}\int_{(2n\pi)^{1/2}}^{(2n\pi+\pi/3)^{1/2}}\dfrac{1}{u}du\\
&\geq\dfrac{1}{2}\sum_{n=1}^{\infty}\dfrac{1}{(2n\pi+\pi/3)^{1/2}}\left((2n\pi+\pi/3)^{1/2}-(2n\pi)^{1/2}\right)\\
&=\dfrac{\pi}{6}\sum_{n=1}^{\infty}\dfrac{1}{(2n\pi+\pi/3)^{1/2}}\dfrac{1}{(2n\pi+\pi/3)^{1/2}+(2n\pi)^{1/2}}\\
&\geq\dfrac{\pi}{12}\sum_{n=1}^{\infty}\dfrac{1}{2n\pi+\pi/3}\\
&=\infty,
\end{align*}
a contradiction.
A: $Rf'(0)=\lim_{h \rightarrow 0} \frac{h^2 \sin(\frac{1}{h^2})-0}{h}=\lim_{h \rightarrow 0} h \sin(\frac{1}{h^2})=0$ as $-h \le h \sin({1}{h^2}) \le h.$ (Sandwich/squeez law).
$Lf'(0)=\lim_{h \rightarrow 0} \frac{h^2 \sin(\frac{1}{h^2})-0}{-h}=\lim_{h \rightarrow 0} -h \sin(\frac{1}{h^2})=0$ 
So $f'(0)=0=g(0)$ exists. However, for $x\ne 0$  as found by fou $g(x)=f'(x)=2x \sin(\frac{1}{x^2}) -\frac{2}{x} \cos(\frac{1}{x^2})$. This means $g(o^{\pm})=\pm \infty$.
$g(x)=f'(x)$ has infinite jump at $x=0$, hence it is not integrable in the interval $(-\epsilon, \epsilon)$.
