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

$\sin x^{2/3}$ remains above $1/2$ for $x$ between $[(2n+1/6)\pi]^{3/2}$ and $[2n+5/6]^{3/2}$, so the integral rises by more than $\left([2n+5/6]^{3/2}-[2n+1/6]^{3/2}\right)\pi^{3/2}/2$ during that time.
$$[2n+5/6]^{3/2}-[2n+1/6]^{3/2}=\frac{[2n+5/6]^3-[2n+1/6]^3}{[2n+5/6]^{3/2}+[2n+1/6]^{3/2}}\\
>\frac{8n^2}{2[2n+1]^{3/2}}$$
That increases as a function of $n$, so the integral does not converge.
