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:
$$\int_0^\infty \left|\sin(y)\sqrt{y} \right| dy  = \sum_{n=0}^\infty \int_{n\pi}^{(n+1)\pi}\left|\sin(y)\right| \sqrt{y} dy \ge \sum_{n=0}^\infty \sqrt{n\pi} \int_{n\pi}^{(n+1)\pi}\left|\sin(y)\right| dy \\= \sum_{n=1}^\infty \sqrt{n\pi} \int_{n\pi}^{(n+1)\pi}\sqrt{\sin(y)^2}$$
 A: Maple writes the indefinite integral as
$$
\int \sin(x^{2/3})\,dx = \frac{-3x^{1/3}\cos(x^{2/3})}{2}+
\frac{3\sqrt{\pi}\;C\left(\displaystyle \frac{\sqrt{2} x^{1/3}}{\sqrt{\pi}}\right)}{2\sqrt{2}}
$$
where $C$ is the Fresnel C function.  The term with the Fresnel function does converge, but the first term oscillates wildly as $x \to \infty$, so the integral diverges.
This suggests an alternate way to do it.  There should be an integration by parts, where we get $\frac{-3x^{1/3}\cos(x^{2/3})}{2}$ plus an integral that can be seen to converge.
A: 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 $-\infty$)
A: the convergence of $\int_0^\infty \sin(x^{2/3}$ at the lower limit $x = 0$ is not a problem. the trouble is at the upper limit $x = \infty$
to handle the upper limit, i will make a change of variable $x = t^{3/2}, dx = 3/2 t^{1/2} dt.$
then 
$\int_0^\infty \sin x^{2/3} dx = \frac{3}{2} \int_0^\infty t^{1/2} \sin t \ dt$
taking the idea from GEdgar's answer
$$ \int t^{1/2} \sin t \ dt  = -t^{1/2}\cos t + \dfrac{1}{2}t^{-1/2}\sin t - \dfrac{1}{4}\int t^{-3/2}\sin t \ dt \tag 1$$
the last two terms are alright at $t = \infty,$ and the first term has no limit at $t = \infty.$
A: $\sin x^{2/3}$ 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.
A: 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.
A: 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.
