What $q$ makes this integral $\int _0 ^\infty \frac { \sin (x ^q ) } { x } \, dx$ converge? I'm asked by the problem what real number $q$ makes the integral 

$$\int _0 ^\infty \frac { \sin (x ^q ) } { x }  \, dx$$ 

converges. 
Using integral by parts I showed that the integral converges for all $q > 0$ (I'm not very sure about my proof). And I think the integral will not converge if $q \le 0 $. But I have trouble showing this. Any help is appreciated.

*

*Assume $q>0$. 
Then, by the change of variable $u=x^q$, we have $\log u =q \log x$
thus $\dfrac{du}u=q\:\dfrac{dx}x$, we find that $$ \int_0^\infty
   \frac{\sin(x^q)}{x}\, dx=\frac1q\int_0^\infty \frac{\sin u}{u}\,
   du=\frac1q\int_0^\infty \frac{1-\cos u}{u^2}\, du $$ where we have
integrated by parts in the last step, the latter integral converges.

*Assume $q<0$. 
Then, by the change of variable $u=x^q$, we have $\log u =q \log x$
thus $\dfrac{du}u=q\:\dfrac{dx}x$, we find that $$ \int_0^\infty
   \frac{\sin(x^q)}{x}\, dx=-\frac1q\int_0^\infty \frac{\sin u}{u}\,
   du=-\frac1q\int_0^\infty \frac{1-\cos u}{u^2}\, du $$ where we have
integrated by parts in the last step, the latter integral converges.

*Assume $q=0$. Then the inital integral diverges.
Remark. The two first cases can be handled at the same time.
