<s>[INST] 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:
$$\int_0^\infty \left|\sin(y)\sqrt{y} \right| dy  = \sum_{n=0}^\infty \int_{n\pi}^{(n+1)\pi}\left|\sin(y)\right| \sqrt{y} dy \ge \sum_{n=0}^\infty \sqrt{n\pi} \int_{n\pi}^{(n+1)\pi}\left|\sin(y)\right| dy \\= \sum_{n=1}^\infty \sqrt{n\pi} \int_{n\pi}^{(n+1)\pi}\sqrt{\sin(y)^2}$$
 [/INST] By $y = x^{2/3}$, we get $$\int_0^\infty \sin(x^{2/3})dx = \frac{3}{2}\int_0^\infty \sin(y)\sqrt{y}dy$$
Consider the integral over the interval $[2n\pi, 2(n+1)\pi]$, we have
$$
\begin{align}
\int_{2n\pi}^{2(n+1)\pi} \sin(y)\sqrt{y}dy &= \int_{2n\pi}^{(2n+1)\pi}\sin(y)\sqrt{y}dy +  \int_{(2n+1)\pi}^{(2n+2)\pi}\sin(y)\sqrt{y}dy \\
&=\int_{2n\pi}^{(2n+1)\pi}\sin(y)\sqrt{y}dy + \int_{2n\pi}^{(2n+1)\pi}\sin(y+\pi)\sqrt{y+\pi}dy\\
&=\int_{2n\pi}^{(2n+1)\pi}\sin(y)\sqrt{y}dy - \int_{2n\pi}^{(2n+1)\pi}\sin(y)\sqrt{y+\pi}dy\\
&=-\int_{2n\pi}^{(2n+1)\pi}\sin(y)\frac{\pi}{\sqrt{y} + \sqrt{y+\pi}}dy\\
&\le -\int_{2n\pi}^{(2n+1)\pi}\sin(y)\frac{\pi}{\sqrt{(2n+1)\pi} + \sqrt{(2n+2)\pi}}dy\\
&=-\frac{2\pi}{\sqrt{(2n+1)\pi} + \sqrt{(2n+2)\pi}}
\end{align}$$ 
then $$\sum_{n=1}^\infty\int_{2n\pi}^{2(n+1)\pi} \sin(y)\sqrt{y}dy \leq \sum_{n=1}^\infty -\frac{2\pi}{\sqrt{(2n+1)\pi} + \sqrt{(2n+2)\pi}} =-\infty$$
so this integral does not converge. To be more convinced, see @Kyson's comment below(so this integral oscillates between $+\infty$ and $-\infty$)
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