real analysis - For $alpha>0$, $int_{0}^{+infty}frac{t-sin{t}}{t^alpha},dt$ converges iff $alphain (2,4)$

I can't prove that for $\alpha>0$, $I_\alpha=\int_{0}^{+\infty}\frac{t-\sin{t}}{t^\alpha}\,dt$ converges iff $\alpha\in (2,4)$. Here's my attempt:

1) An easy remark but important: as $\sin{t}\le t$, we're dealing with a positive function.

2) I splited the integral to $\int_{0}^{1}\frac{t-\sin{t}}{t^\alpha}\,dt$ and $\int_{1}^{+\infty}\frac{t-\sin{t}}{t^\alpha}\,dt$ and use the inequality $\frac{t^3}{3!}-\frac{t^5}{5!}\le t-\sin{t}\le \frac{t^3}{3!}$ but it didn't work.

3) This is the only thing that lead me to a part of the answer: I noticed that $\int_0^{+\infty}\frac{dt}{t^{\alpha-1}}$ is always divergent. Thus, if $I_\alpha$ converges, $\int_0^\infty\frac{\sin{t}}{t^\alpha}\,dt$ diverges. But we know that $\int_1^\infty\frac{sin{t}}{t^\alpha}\,dt$ converges for any $\alpha>0$. Thus $\int_0^\infty\frac{\sin{t}}{t^\alpha}\,dt$ diverges iff $\int_0^1\frac{\sin{t}}{t^\alpha}\,dt$ diverges. Since $\sin{t}\sim t$ near $0^+$, both positive, $\int_0^1\frac{\sin{t}}{t^\alpha}\,dt$ diverges iff $\alpha>2$. Hence we showed that $(I_\alpha\,\text{converges})\Rightarrow\alpha>2$

Could you please help me? Thank you in advance!

Answer

The idea to split the integral is fine.

First, note that the integral $I_1$ as given by

$$I_1=\int_0^1 \frac{t-\sin(t)}{t^\alpha}\,dt$$

converges for $\alpha<4$ (and diverges for $\alpha \ge 4)$ since the integrand is $O\left(t^{3-\alpha}\right)$ as $t \to 0$.

Next, note that the integral $I_2$ as given by

$$\int_1^\infty \frac{t-\sin(t)}{t^\alpha}\,dt$$

converges for $\alpha >2$ (and diverges for $\alpha \le 2$) since the integrand is $O\left(t^{1-\alpha}\right)$ as $t\to \infty$.

Putting it together, we have that the integral of interest $I=I_1+I_2$ converges for all $\alpha \in (2,4)$ and diverges elsewhere.