I wanted to prove this statement. I know how to show that $F(x)=x^2\sin(1/x) (x\ne 0)$ and $F(0)=0$ is AC on [-1,1]. But I want to generalize it to $x^2\sin(x^{-p})$ where $p\in(1,2)$.
I tried to show that $F'$ is in $L^1([-1,1])$ which worked out. But to conclude that it is AC, I need $$F(x)=F(-1)+\int_{-1}^x F'(t)dt$$ which I'm struggling to prove.
Any help would be appreciated!
Suppose $x > 0$. Then,
$$ \int_0^x F'(t) \, dt = \int_0^\varepsilon F'(t) \, dt + \int_\varepsilon^x F'(t) \, dt $$
Since we can apply FTC on $[\varepsilon, 1]$, the right-hand side is equal to
$$ \int_0^\varepsilon F'(t) \, dt + F(x) - F(\varepsilon) $$
Because $F'$ is integrable, the integral in the middle tends to $0$ as $\varepsilon \to 0$. Also, $F(\varepsilon)$ tends to $F(0)$. Hence, in the limit you get $F(x) - F(0)$.
Edit: I didn't pay attention to the fact that $x^{-5/4}$ is not defined for negative values of $x$. You should consider $F$ only on $[0, 1]$. I changed my proof accordingly