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I've already been able to show that $\frac{\sin(x^n)}{x^n}\rightarrow 0$ pointwise if $x>1$, and (using L'Hospital's rule) that $\frac{\sin(x^n)}{x^n}\rightarrow 1$ pointwise on $(0,1)$. But in order to show that $\lim_{n\to\infty}\int_0^\infty\frac{\sin(x^n)}{x^n}\,dx = 1$ I'd need to show that $\lim_{n\to\infty}\int_0^\infty\frac{\sin(x^n)}{x^n}\,dx = \int_0^\infty\lim_{n\to\infty}\frac{\sin(x^n)}{x^n}\,dx$.

I'm having trouble finding the right bounding to apply the dominated convergence theorem.

Any thoughts would be greatly appreciated.

Thanks.

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Hint: Break the integral in two: $\int^1_0+\int^\infty_1$.

  • Over $[0,1]$, you can use dominated convergence with $\Big|\frac{\sin x^n}{x^n}\Big|\leq 1$.

  • Over $(1,\infty]$ notice that $\Big|\frac{\sin x^n}{x^n}\Big|\leq \frac{1}{x^n}$. This leads to $$ \int^\infty_1\frac{1}{x^n}\,dx=\frac{1}{n-1}\xrightarrow{n\rightarrow\infty}0 $$ One can also use dominated convergence here with $\Big|\frac{\sin x^n}{x^n}\Big|\leq \frac{1}{x^2}$ for $n\geq 2$, $x>1$.


Also, one can combined all of the above and use dominated convergence over all $[0,\infty)$ with

$$\Big|\frac{\sin x^n}{x^n}\Big|\leq \mathbb{1}_{[0,1]}(x)+\frac{1}{x^2}\mathbb{1}_{(1,\infty]}$$

for $n\geq 2$.

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  • $\begingroup$ +1 You want the lower limit to be $1$ in the second integral. $\endgroup$ – saulspatz Aug 8 '20 at 17:42
  • $\begingroup$ Is $\left\vert\frac{sin(x^n)}{x^n} \right\vert < 1$ a relatively well known inequality? Or is there a straightforward way of showing it? $\endgroup$ – Bears Aug 8 '20 at 17:48
  • $\begingroup$ It is well known, but you can also prove it with simple calculus methods. Remember that $\lim_{t\rightarrow0}\frac{\sin x}{x}=1$. That will give you a head start. You can also use the auxiliary function $\phi(x)=x-\sin x$ and see where the maximum is and so on. $\endgroup$ – Oliver Diaz Aug 8 '20 at 17:50
  • $\begingroup$ Okay. This all makes sense. Thanks a bunch for your help. $\endgroup$ – Bears Aug 8 '20 at 17:55
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$$|\sin(x^n)|\le1\therefore \frac{|\sin(x^n)|}{x^n}\le\frac{1}{x^n}$$ $$\int_1^\infty\frac{\sin(x^n)}{x^n}dx\le\int_1^\infty x^{-n}dx=\left[\frac{x^{1-n}}{n-1}\right]_\infty^1=\frac{1}{n-1}\{n>1\}$$ now we just need to look at the other part of the integral: $$I=\int_0^1\frac{\sin(x^n)}{x^n}dx$$ we know that: $$\lim_{x\to 0}\frac{\sin(x^n)}{x^n}\to1$$ and that is the region $x\in[0,1],\,\,\frac{\sin(x^n)}{x^n}\le1$ furthermore we can show that for $n\to\infty,\frac{\sin(x^n)}{x^n}\to1$ and so taking the limit first we get: $$\lim_{n\to\infty}\int_0^1\frac{\sin(x^n)}{x^n}dx\to\int_0^1dx=1$$ combining this with the first integral calculated and taking the limit we get our desired result.


If you were wondering why we can make the assumption in the last integral use the fact that: $$u\to0,\sin(u)\approx u$$ and we have our $u=x^n$ so for large $n$ and $x\in[0,1]$ our $u\to0$ for all $x$

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Just for your curiosity.

Let $x^n=t$ to make $$I_n=\int_0^\infty\frac{\sin(x^n)}{x^n}\,dx=\frac 1n\int_0^\infty t^{\frac{1}{n}-2} \sin (t)\,dt$$ The antiderivative can be computed in terms of the gamma function and $$I_n=-\frac 1n \cos \left(\frac{\pi }{2 n}\right) \Gamma \left(\frac{1}{n}-1\right)$$ Now, using expansions $$I_n=-\frac 1n\left(1-\frac{\pi ^2}{8 n^2}+O\left(\frac{1}{n^4}\right)\right)\,\left(-n+(\gamma -1)-\frac{12(1-\gamma) +6 \gamma ^2+\pi ^2}{12 n}+O\left(\frac{1}{n^2}\right)\right)$$ $$I_n=1+\frac{1-\gamma }{n}+\frac{24+12 (\gamma -2) \gamma -\pi ^2}{24 n^2}+O\left(\frac{1}{n^3}\right)$$

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