For $\alpha>0$, $\int_{0}^{+\infty}\frac{t-\sin{t}}{t^\alpha}\,dt$ converges iff $\alpha\in (2,4)$ 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!
 A: 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. 
A: Split the integral
$$\int_{0}^{+\infty}f_\alpha(t)\,dt = \int_{0}^{1}f_\alpha(t)\,dt + \int_{1}^{+\infty}f_\alpha(t)\,dt$$
where $f_\alpha(t) = \frac{t-\sin{t}}{t^\alpha}$.
For small $t$ (i.e. $t \in [0, 1]$), we have that 
$$f_\alpha(t) =-\frac{\frac{t^3}{6} + O(t^5)}{t^\alpha} \sim -t^{3-\alpha}.$$
Then:
$$-\int_{0}^{1} t^{3-\alpha}\,dt = -\int_{0}^{1} \frac{1}{t^{\alpha-3}}\,dt$$
is convergent if $$\alpha-3 < 1 \Rightarrow \alpha < 4.$$
For large $t$ (i.e. $t > 1$), we have that $$f_\alpha(t) \sim \frac{t}{t^{\alpha}} = t^{1-\alpha}.$$
In this case:
$$\int_{1}^{+\infty} t^{1-\alpha}dt = \int_{1}^{+\infty} \frac{1}{t^{\alpha-1}}dt$$ converges when
$$\alpha-1 > 1 \Rightarrow \alpha > 2.$$
Joining the two condition, you get that the integral converges when $$\alpha \in (2, 4).$$ 
