# How to show the function $f(x) = x^2 \sin(1/x)$ has integrable derivative?

Consider the function

$$f(x) = \begin{cases} x^2\sin(1/x) & \text { if } x \neq 0 \\ 0 & \text{ otherwise} \end{cases}$$

has integrable derivative on $$(-1, 1)$$.

I found

$$f'(x) = \begin{cases} 2x\sin(1/x) - \cos(1/x) & \text { if } x \neq 0 \\ 0 & \text{ otherwise,} \end{cases}$$

but I have no clue how to show that it is integrable on $$(-1, 1)$$. Can someone please help me?

Your computations are correct and they show that $$f'$$ is bounded and it has a single discontinuity point. Therefore, it Riemann-integrable (and this would still be true if it had a countable set of discontinuity points).
• @zhw. I agree. It's easy to prove directly from the definion (together with the fact that continuous $\implies$ Riemann-integrable) that if a bounded function only has finitely many points of discontinuity, then it is Riemann-integrable. – José Carlos Santos Dec 11 '18 at 18:56