Show that $\lim_{n\to\infty}\int_0^\infty\frac{\sin(x^n)}{x^n}dx = 1$

Question

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.

Answer 1

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 $n\geq 2$, $x>1$.

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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 $n\geq 2$.

Answer 2

$$|\sin(x^n)|\le1\therefore \frac{|\sin(x^n)|}{x^n}\le\frac{1}{x^n}$$ $$\int_1^\infty\frac{\sin(x^n)}{x^n}dx\le\int_1^\infty x^{-n}dx=\left[\frac{x^{1-n}}{n-1}\right]_\infty^1=\frac{1}{n-1}\{n>1\}$$ 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.

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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 $x$

Answer 3

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 $$I_n=-\frac 1n\left(1-\frac{\pi ^2}{8 n^2}+O\left(\frac{1}{n^4}\right)\right)\,\left(-n+(\gamma -1)-\frac{12(1-\gamma) +6 \gamma ^2+\pi ^2}{12 n}+O\left(\frac{1}{n^2}\right)\right)$$ $$I_n=1+\frac{1-\gamma }{n}+\frac{24+12 (\gamma -2) \gamma -\pi ^2}{24 n^2}+O\left(\frac{1}{n^3}\right)$$