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.

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

  • $\begingroup$ Oh, it's the simple :) Thanks @ Back2Basic $\endgroup$ – ASB Feb 9 '15 at 0:10
  • $\begingroup$ You are welcome, TG. $\endgroup$ – DeepSea Feb 9 '15 at 1:00

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, $$\lim_{n\to \infty}\int \limits_{0}^{\infty}\frac {\sin (x^n)} {x^n}dx=\lim_{n\to \infty}\cos\left(\frac{\pi}{2n}\right)\Gamma \left(\frac{1}{n}+1\right)\frac{n}{n-1} =1$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.