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:
$$\int_0^\infty \left|\sin(y)\sqrt{y} \right| dy  = \sum_{n=0}^\infty \int_{n\pi}^{(n+1)\pi}\left|\sin(y)\right| \sqrt{y} dy \ge \sum_{n=0}^\infty \sqrt{n\pi} \int_{n\pi}^{(n+1)\pi}\left|\sin(y)\right| dy \\= \sum_{n=1}^\infty \sqrt{n\pi} \int_{n\pi}^{(n+1)\pi}\sqrt{\sin(y)^2}$$

While typing, I noticed that @GEdgar already noted this, but here it goes anyway.
Integrating by parts, we find that
$$
\begin{align}
\int \sin(x^{2/3})\,dx &=\int -\frac{3}{2}x^{1/3}\frac{d}{dx}\cos(x^{2/3})\,dx \\
&= -\frac{3}{2}x^{1/3}\cos(x^{2/3})+\int \frac{1}{2}x^{-2/3}\cos(x^{2/3})\,dx.
\end{align}
$$
Next, show that
$$
\int_0^{+\infty} x^{-2/3}\cos(x^{2/3})\,dx
$$
converges.
