<s>[INST] How $\int_1^\infty \sin \left( \frac{2}{x^{5/3}} \right)dx $ absolute converge? I have the integral $$\displaystyle{\int_1^\infty {\sin \left( \frac{2}{x^{\frac{5}{3}}} \right)}dx }$$
I know it is converge. but when I check absolute converge I did the following: according to the trigonometric identity :$\cos(2\theta)=1-2\sin^2(\theta)$ I got
$$\left|\sin\left(\frac{2}{x^{\frac{5}{3}}}\right)\right| \geq \sin^2\left(\frac{2}{x^{\frac{5}{3}}}\right)=\frac{1}{2} -\frac{1}{2} \cos\left(\frac{4}{x^{\frac{5}{3}}}\right)$$ 
$\displaystyle{\int_1^\infty \frac{1}{2}dx }$ is diverge so why I cant conclude that $\displaystyle{\int_1^\infty {\sin \left( \frac{2}{x^{\frac{5}{3}}} \right)}dx }$ also diverge?
 [/INST] Put $y=\frac 2 {x^{5/3}}$. We get $\int_1^{\infty}| \sin (\frac 2 {x^{5/3}})|dx= 2^{-2/5}\frac  3 5 \int_0^{2} |\sin y| y^{-8/5}dy$. Now use the fact that $\frac {\sin y} y \to 1$ as $ y \to 0$ and  $\int_0^{2} y^{-3/5}dy <\infty$ to see that the given inetgral is absolutely convergent. 
</s>