How can one proves that $ \lim\limits _{n\rightarrow \infty}\int \limits_{0}^{\infty}\frac{\sin (x^{n})}{x^{n}}dx=1 $? I am trying to prove that $$ \lim _{n\rightarrow \infty}\int \limits_{0}^{\infty}\frac{\sin (x^{n})}{x^{n}}dx=1 .$$
My Attempt: Since $$ \lim _{a\rightarrow 0}\frac{\sin a}{a}=1 $$ and for each $ x\in [0,1) $, $$ \lim _{n\rightarrow \infty}x^{n}=0 $$ we can obtain $$ \lim _{n\rightarrow \infty}\frac{\sin (x^{n})}{x^{n}}=1 .$$
Also since for each $ a\in \mathbb{R} $, $ |\sin a|\leq |a| $, we have that for each $ x\in [0,1) $ and $ n\in \mathbb{N} $, $ \left |\frac{\sin (x^{n})}{x^{n}}\right |\leq 1 $.
Then by dominated convergence theorem, $$ \lim _{n\rightarrow \infty}\int \limits_{0}^{1}\frac{\sin (x^{n})}{x^{n}}dx=\int \limits_{0}^{1}1dx=1. $$
So, I have to show that $$ \lim _{n\rightarrow \infty}\int \limits_{1}^{\infty}\frac{\sin (x^{n})}{x^{n}}dx=0. $$
But, I still haven't managed to show it. So, could anyone give me some help ?
Any hints/ideas are much appreciated.
Thanks in advance for any replies. 
 A: Hint:Observe that: $\dfrac{\sin(x^n)}{x^n} \leq \dfrac{1}{x^n} = g_n(x)$, and use the DCT again to conclude.
A: I know this has been answered long time ago, but it might be helpful for others, so I am posting this. Your integral can be calculated in closed form
$$
I(n)=\int_0^\infty dx\frac{\sin(x^n)}{x^n}=\frac{2^{-2+\frac1n}\sqrt{\pi}\Gamma\left(\frac{1}{2n}\right)}{n\Gamma\left(\frac32-\frac{1}{2n}\right)},
$$
where $\Gamma(u)$ is the Gamma function. From this we get the asymptotic expansion
$$
I(n)\sim1-\frac{\gamma-1}{n}+O(n^{-2}),\quad n\to\infty,
$$
where $\gamma$ is the Euler constant. So the answer to your question is $\lim_{n\to\infty}I(n)=1$, and you can take this computation as the proof.
A: From this answer on this post integral of $\int \limits_{0}^{\infty}\frac {\sin (x^n)} {x^n}dx$  we have 
$$\int \limits_{0}^{\infty}\frac {\sin (x^n)} {x^n}dx=\cos\left(\frac{\pi}{2n}\right)\Gamma \left(\frac{1}{n}\right)\frac{1}{n-1}= \cos\left(\frac{\pi}{2n}\right)\frac{1}n\Gamma \left(\frac{1}{n}\right)\frac{n}{n-1}\\=\cos\left(\frac{\pi}{2n}\right)\Gamma \left(\frac{1}{n}+1\right)\frac{n}{n-1} $$
Thus, 
$$\lim_{n\to \infty}\int \limits_{0}^{\infty}\frac {\sin (x^n)} {x^n}dx=\lim_{n\to \infty}\cos\left(\frac{\pi}{2n}\right)\Gamma \left(\frac{1}{n}+1\right)\frac{n}{n-1} =1$$
