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!
 A: $\sin(x) \le 1$ 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.
A: That's good work for the first part,
For $I_2$ you can see that $\sin(x) = o_{+\infty}(x)$ therefore $\dfrac{x-\sin(x)}{x^{7/2}} \sim_{+\infty} \dfrac{x}{x^{7/2}} = \dfrac{1}{x^{5/2}} $.
Moreover $\int_1^{+\infty} \dfrac{1}{x^{5/2}} \mathrm{d}x$ converges (Riemann's rule with $5/2>1$). And $\forall x\geq1, x-\sin(x)\geq 0$ so you can use the theorem of comparison for positive functions and it gives you the convergence of $I_2$.
A: Since $ \frac{x-\sin{x}}{x^{\frac{7}{2}}}\underset{x\to 0}{\sim}\frac{1}{6\sqrt{x}} $, there is no problem at $ 0 $. And since $ \frac{x-\sin{x}}{x^{\frac{7}{2}}}\underset{x\to +\infty}{\sim}\frac{1}{x^{\frac{5}{2}}} $ there is no problem at $ +\infty $ either. Thus our integral converges.
Let's try to find its value using Fubini theorem and the two following identities : \begin{aligned}\left(\forall x\in\mathbb{R}^{*}\right),\ \frac{x-\sin{x}}{x^{3}}=\frac{1}{2}\int_{0}^{1}{\left(1-y\right)^{2}\cos{\left(xy\right)}\,\mathrm{d}y}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ &\left(1\right)\\ \int_{0}^{+\infty}{\cos{\left(x^{2}\right)}\,\mathrm{d}y}=\frac{\sqrt{\pi}}{2\sqrt{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ &\left(2\right)\end{aligned}
The first equality can be easily proven, doing some integrations by parts, the second one (Fresnel Integral) is a well known identity that I'm sure has been proven a lot of times in this website.
\begin{aligned}\int_{0}^{+\infty}{\frac{x-\sin{x}}{x^{\frac{7}{2}}}\,\mathrm{d}x}&=2\int_{0}^{+\infty}{\frac{x^{2}-\sin{\left(x^{2}\right)}}{x^{6}}\,\mathrm{d}x}\\ &=\int_{0}^{+\infty}{\int_{0}^{1}{\left(1-y\right)^{2}\cos{\left(x^{2}y\right)}\,\mathrm{d}y}\,\mathrm{d}x}\\ &=\int_{0}^{1}{\left(1-y\right)^{2}\int_{0}^{+\infty}{\cos{\left(x^{2}y\right)}\,\mathrm{d}x}\,\mathrm{d}y}\\ &=\int_{0}^{1}{\frac{\left(1-y\right)^{2}}{\sqrt{y}}\int_{0}^{+\infty}{\cos{\left(\mu^{2}\right)}\,\mathrm{d}\mu}\,\mathrm{d}y}\\ &=\sqrt{\frac{\pi}{2}}\int_{0}^{1}{\frac{\left(1-y\right)^{2}}{2\sqrt{y}}\,\mathrm{d}y}\\ &=\sqrt{\frac{\pi}{2}}\int_{0}^{1}{\left(1-\nu^{2}\right)^{2}\,\mathrm{d}\nu}\\ \int_{0}^{+\infty}{\frac{x-\sin{x}}{x^{\frac{7}{2}}}\,\mathrm{d}x}&=\frac{8}{15}\sqrt{\frac{\pi}{2}}\end{aligned}
