It follows from the question to judge whether $\int_0^{+\infty}x^p\sin(x^q)dx$, where $q\neq 0$, converges. I'm now dealing with the case that $q>0$, and I find it's necessary that $p>-q-1$, and it suffices if $p<q-1$ is added. However, I have difficulty figuring out whether the latter is necessary.
The case $p=q-1$ is easy, for $\int_1^{+\infty}x^{q-1}\sin(x^q)dx=\lim\limits_{x\to+\infty}\frac{1}{q}(\cos(x^q)-\cos1)$ the limit does not exist, but I don't see any effective way to analysis the case $q-1<p<0$.
I'd be grateful if anyone could provide some hints, ways that seems work or a complete solution.
Best Answer
HINT: what about a change of variable? Let $y=x^q$, then $dx=\tfrac{1}{q}y^{\tfrac{1}{q}-1}\,\mathrm d y=\,\mathrm d x$ and so $$ \int_{1}^{\infty }x^p\sin (x^q)\,\mathrm d x=\frac1{q}\int_{1}^{\infty }y^{\frac{p+1}q-1}\sin (y)\,\mathrm d y,\quad\text{ when } q>0 $$ And when $q<0$ we must change the infinity by a zero.