TITLE: for $p>0$, when does this integration:$\int_0^{\infty} x^pe^{-x^8\sin^2x}dx$ converge?
QUESTION [5 upvotes]: for $p>0$，find the value range of $p$, which makes this integration: $\displaystyle\int_0^{\infty} x^pe^{-x^8\sin^2x}dx$ converge.
I tried to divide $(0,\infty)$ into $(n\pi,(n+1)\pi)$, but i met difficulty estimating $\displaystyle\int_0^{\pi} (x+n \pi)^pe^{-(x+n \pi)^8\sin^2x}dx$
My teacher says it is $O(n^{p-4})$, but i think it is wrong.

REPLY [6 votes]: HINT:
Using $2x/\pi\le\sin(x)\le x$ for $x\in [0,\pi/2]$, we can assert that 
$$\begin{align}
\int_0^{\pi/2} (x+n\pi)^pe^{-(x+n\pi)^8\sin^2(x)}\,dx&\le (\pi/2+n\pi)^p\int_0^{\pi/2} e^{-(n\pi)^8(2x/\pi)^2}\,dx\\\\
&\le (\pi/2+n\pi)^p\int_0^\infty e^{-4n^8\pi^6x^2}\,dx\\\\
&=\frac{(n+1/2)^p\pi^p}{4\pi^{5/2}n^4}\\\\
&=O\left(n^{p-4}\right)
\end{align}$$
as $n\to \infty$.
Can you finish now?