$$\int_{0}^{1} \frac{x-\sin x}{x^{\alpha}} dx$$ I'm stuck on this, I'm supposed to solve it from $0$ to $\infty$, from the $(1,\infty)$ case i got $\alpha>2$. Here I’m supposed to show that $\alpha < 4$ and I don’t understand why...
EDIT: How stupid I am, thanks a lot!:) Solved
This is a positive function, so you may use equivalents: $$x-\sin x\sim_0\frac{x^3}6,\quad\text{so }\quad \frac{x-\sin x}{x^{\alpha}}\sim_0\frac{x^3}{6x^\alpha}=\frac1{6x^{\alpha-3}}, $$ which converges if and only if $\;\alpha-3<1$.