How can I prove $\int_{0}^{+\infty }\frac{x-sinx}{\sqrt{x^{7}}}dx$ converges/diverges? I need to determine if the following integral converges/diverges:
$$\int_{0}^{+\infty }\frac{x-\sin x}{\sqrt{x^{7}}}dx$$

We can write the integral as:
$$\int_{0}^{+\infty }\frac{x-\sin x}{\sqrt{x^{7}}}dx=\int_{0}^{a }\frac{x-\sin x}{\sqrt{x^{7}}}dx+\int_{a}^{+\infty }\frac{x-\sin x}{\sqrt{x^{7}}}dx=I_{1}+I_{2}$$
+For $I_{1}=\int_{0}^{a}\frac{x-\sin x}{\sqrt{x^{7}}}dx$.
Since $\sin x= x-\frac{x^{3}}{3!} +O(x^{3})$, I know that $I_{1}$ is convergent.
+For $I_{2}=\int_{a}^{+\infty }\frac{x-\sin x}{\sqrt{x^{7}}}dx$.
I don't know how to use $x-\sin x$ for other ideas such as the comparison test. The only thing I can exploit is $x - \sin x>0 , x>0$.

Can you give me some hints? Thank you very much!

$\sin(x) \le 1$ for all $x \in \mathbb{R}$, so $$ x - \sin(x) \le x+ 1$$
and you can see that $I_2 \le \int_a^\infty \frac{x+1}{x^{7/2}}dx$
Split the fraction and you can conclude that $I_2$ converges.
