Does $\int_0^\infty \sin(x^{2/3}) dx$ converges? My Try: 
We substitute $y = x^{2/3}$. Therefore, $x = y^{3/2}$ and $\frac{dx}{dy} = \frac{2}{3}\frac{dy}{y^{1/3}}$
Hence, the integral after substitution is: 
$$ \frac{3}{2} \int_0^\infty \sin(y)\sqrt{y} dy$$
Let's look at:
$$\int_0^\infty \left|\sin(y)\sqrt{y} \right| dy  = \sum_{n=0}^\infty \int_{n\pi}^{(n+1)\pi}\left|\sin(y)\right| \sqrt{y} dy \ge \sum_{n=0}^\infty \sqrt{n\pi} \int_{n\pi}^{(n+1)\pi}\left|\sin(y)\right| dy \\= \sum_{n=1}^\infty \sqrt{n\pi} \int_{n\pi}^{(n+1)\pi}\sqrt{\sin(y)^2}$$

I will try to look at the problem from a different perspective and to understand exactly why it will not converge.


*

*The first thing we should know is that if the integral converges the terms must get smaller after each "iteration".

*That means that the function keeps decreasing towards zero while going to infinity.

*If the function is decreasing towards infinity that means that the derivative of the function and its limit towards infinity should be negative (the derivative can be zero only if the value of the function has reached zero, because that would mean that the function remains at zero forever thus nothing else is added)
So lets get our hands dirty:
$$ \\ \large
{f(x) = \sin x^{2/3}} \\
\large
\frac {df(x)} {dx} = \cos x^{2/3}  \times \frac {2} {3} {x^{-{1/3}} } = \frac {2 \cos {x^{\frac {2} {3}}}} {3 \sqrt[3] {x}} \\
\large \lim_{x\to\infty} \frac {df(x)} {dx} = \lim_{x\to\infty} \frac {2 \cos {x^{\frac {2} {3}}}} {3 \sqrt[3] {x}} = 0 \\
AND \\
\lim_{x\to\infty} {f(x)} = {[-1,+1]}
\\$$
So as you can see the function is not decreasing towards infinity, it is in fact maintaining the same value. And the value of $f(x)$ towards infinity isn't $0$ but is in fact a value in $[-1,+1]$. Thus we can conclude the terms of the function aren't decreasing and the function isn't converging.
