Let $f$ be a real-valued $C^{1}$ function on $\mathbb{R}$. Let $1 < \alpha < 2$. Show $\int^{1}_{0} \frac{f(x) - f(0)}{x^\alpha}\text{d}x$ converges.

Question

I have already shown that there exists a real $C >0$ such that, for any, $x \in [0,1]$: $|f(x) - f(0)| \leq C|x|$.

I understand that, for all $a>0$: $\int^{1}_{a} \frac{f(x) - f(0)}{x^\alpha}\text{d}x \in \mathbb{R}$ since the smooth and bounded for all $x \neq 0$. However, I can't figure out how to show that the integral doesn't blow up as $a$ approaches $0$. I tried using the Rule of L'Hospital, but that doesn't help, since $\alpha > 1$ and I can only differentiate $f$ once, since it is $C^{1}$. Any hints?

Answer 1

Use comparison test with The function $1/x^{\alpha-1}$ and the fact that $\frac{f(x)-f(0)}{x}$ converges to $f’(0)$