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


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 $\lim_{n\to\infty}J_n=0$.

  • $\begingroup$ This is certainly the way to go. Cannot we argue that $$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\int_1^{{{\left( {\frac{\pi }{2}} \right)}^n}} {\frac{{\sin u}}{u}{u^{1/n}}du = } \mathop {\lim }\limits_{n \to \infty } \frac{1}{n} \cdot \int_1^0 {\frac{{\sin u}}{u}du} = 0$$ somehow? The integrand converges almost everywhere to $u^{-1}\sin u$ (everywhere but $0$) and monotonically for $n\geq 2$. I have cut $[0,1]$ since that one is treatable with $\sin x\leq x$. $\endgroup$ – Pedro Tamaroff Jun 19 '13 at 20:56
  • $\begingroup$ But $(\pi/2)^n\longrightarrow \infty ,n\longrightarrow \infty$,so why $LHS>0$ $\endgroup$ – math110 Jun 20 '13 at 11:37
  • $\begingroup$ @math110: The first bump of the integrand $\bigl(=: f(u)\bigr)$ is upwards, the second bump downwards with smaller area, the third bump again upwards with still smaller area, and so on. It follows that the function $F(x):=\int_0^x f(u)\ du$ takes its max at $x=\pi$. $\endgroup$ – Christian Blatter Jun 20 '13 at 12:08

The integral from 0 to 1 will vanish. Then you have a bunch of integrals from $[2k\pi]^{1/n}$ to $[2(k+1)\pi]^{1/n}$. If you can show that each of these integrals becomes very small as a function of both $k$ and $n$, because of near-cancellation, then you might be able to sum over all $k$ and still get a sum, as a function of $n$, that approaches 0.

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    $\begingroup$ In other words, the part from $0$ to $1$ vanishes because $|\sin(x^n)|\leqslant x^n\to0$, the part from $1$ to $(2\pi)^{1/n}$ vanishes because $(2\pi)^{1/n}\to1$, and each part from $(2k\pi)^{1/n}$ to $((2k+1)\pi)^{1/n}$ nearly cancels out with the part from $((2k+1)\pi)^{1/n}$ to $((2k+2)\pi)^{1/n}$. $\endgroup$ – Did Jun 19 '13 at 15:47
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    $\begingroup$ Why downvote this answer? $\endgroup$ – Did Jun 19 '13 at 15:48
  • $\begingroup$ @Michael It is fine enough. +1 $\endgroup$ – AD. Jun 20 '13 at 22:35

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