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?
 A: Let $f(x) = \left|\sin \left( \dfrac{2}{x^{5/3}}\right)\right|$. Is is positive continuous on $[1,+\infty)$. Moreover, we know that $\sin h = h + o(h)$ as $h \to 0$. Since $x^{5/3} \to \infty$ as $x \to \infty$, $\dfrac{2}{x^{5/3}}\to 0$ as $x \to \infty$. Thus
\begin{align}
f(x) \sim_{x\to \infty} \dfrac{2}{x^{5/3}}
\end{align}
By the comparison theorem, $\int_1^{\infty}f(t)\mathrm{d}t$ has the same nature than $\int_1^{\infty} \dfrac{2}{t^{5/3}}\mathrm{d}t$ which is convergent. Thus $f$ is integrable and your function is absolutly integrable.
What you missed is that even if $f$ and $g$ have divergent integrals, $f-g$ can have a convergent one (for example, $f-f = 0$ is always integrable).
A: Since $|\sin t|\leq|t|$ for all real $t$,
$$
|\sin(2x^{-5/3})|\leq 2x^{-5/3},\qquad x>0
$$
As $\int^\infty_1 x^{-5/3}\,dx<\infty$, it follows that $\int^\infty_1|\sin(2x^{-5/3})|\,dx$ converges.
A: 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. 
