# 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.

• I guess it would also help to change variable to $u= x^n$. – Rogelio Molina Feb 8 '15 at 8:13
• What about $\sin(x^n)\leq1$ for any $n$. – servabat Feb 8 '15 at 8:16

Hint:Observe that: $\dfrac{\sin(x^n)}{x^n} \leq \dfrac{1}{x^n} = g_n(x)$, and use the DCT again to conclude.
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.
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, $$\lim_{n\to \infty}\int \limits_{0}^{\infty}\frac {\sin (x^n)} {x^n}dx=\lim_{n\to \infty}\cos\left(\frac{\pi}{2n}\right)\Gamma \left(\frac{1}{n}+1\right)\frac{n}{n-1} =1$$