TITLE: How prove this $\lim_{n\to\infty}\int_{0}^{\frac{\pi}{2}}\sin{x^n}dx=0$
QUESTION [6 upvotes]: show that
$$\lim_{n\to\infty}\int_{0}^{\dfrac{\pi}{2}}\sin{x^n}dx=0$$
I have see this similar problem
$$\lim_{n\to\infty}\int_{0}^{\dfrac{\pi}{2}}\sin^n{x}dx=0$$
poof:
$\forall \xi>0,0<\delta<\xi/2$,and there is $N$,such $0<\sin^n{\pi/2-\delta}<\xi/\pi(n\ge N)$
then we have
$$\int_{0}^{\pi/2}\sin^n{x}dx=\left(\int_{0}^{\pi/2-\delta}+\int_{\pi/2-\delta}^{\pi/2}\right)\sin^n{x}dx=I_{1}+I_{2}$$
then 
$$|I_{1}|\le\left(\sin{\pi/2-\delta}\right)^n(\pi/2-\delta)<\xi/\pi\cdot\pi/2=\xi/2$$
and
$$|I_{2}|\le\left(\pi/2-(\pi/2-\delta)\right)=\delta<\xi/2$$
and This problem have many other methods,
But for this $$\lim_{n\to\infty}\int_{0}^{\dfrac{\pi}{2}}\sin{x^n}dx=0$$
I can't prove it,Thank you

REPLY [5 votes]: One has
$$J_n:=\int_0^{\pi/2}\sin(x^n)\ dx={1\over n}\int_0^{(\pi/2)^n}\sin u\>{du\over u^{1-{1/n}}}\qquad(n\geq2)\ .$$
Looking at the graph of the integrand (the alternating bumps become ever smaller, as in an alternating series) one can see that
$$0\leq\int_0^{(\pi/2)^n}\sin u\>{du\over u^{1-{1/n}}}\leq \int_0^\pi\sin u\>{du\over u^{1-{1/n}}}\leq\int_0^\pi u^{1/n}\ du={n\over n+1}\pi^{1+{1\over n}}\qquad(n\geq2)\ .$$
As $\lim_{n\to\infty}\pi^{1+{1\over n}}=\pi$ it follows that
 $\lim_{n\to\infty}J_n=0$.