For arbitrary $r_n\in\mathbb{R}$,show that $$ \lim_{n\rightarrow\infty} \int_{0}^{+\infty} e^{-x}\sin^{n}(x+r_n)\, dx=0 $$ By Lebesgue Dominated Convergence Theorem, since $$ |e^{-x}\sin^{n}(x+r_n)|<e^{-x}\in\mathbb{L}^1(0,+\infty)$$ We have $$ \lim_{n\rightarrow\infty} \int_{0}^{+\infty} e^{-x}\sin^{n}(x+r_n) dx= \int_{0}^{+\infty} e^{-x}\lim_{n\rightarrow\infty} \left(\sin^{n}(x+r_n)\right) dx $$ However, the limit of $\lim_{n\rightarrow\infty} \sin^{n}(x+r_n)$ doesn't exist.

  • $\begingroup$ Hint : The limit $\lim_{n\rightarrow\infty} sin^{n}(x+r_n)$ does not exist for all $x$, but it exists for almost every $x$. $\endgroup$ Nov 5, 2020 at 15:08
  • $\begingroup$ @TheSilverDoe, I am skeptical that the limit exists for almost every $x$, if $r_n$ is chosen appropriately. Thinking that the peaks of $\sin^n x$ had with $\sim n^{-1/2}$, it will be enough to consider $r_n=c\sqrt{n}$ for some small $c$. $\endgroup$ Nov 5, 2020 at 17:31

1 Answer 1


The integral is bounded above in absolute value by

$$\int_0^\infty e^{-x}|\sin^n(x+r_n)|\,dx = \sum_{k=0}^{\infty} \int_{k\pi}^{(k+1)\pi}e^{-x}|\sin^n(x+r_n)|\,dx $$ $$\tag 1 \le \sum_{k=0}^{\infty} e^{-\pi k}\int_{k\pi}^{(k+1)\pi}|\sin^n(x+r_n)|\,dx.$$

But note $|\sin^n x|$ is $\pi$-periodic. Thus

$$\int_a^{a+\pi}|\sin^n x|\,dx =\int_0^\pi|\sin^nx|\,dx$$

for any $a.$ Therefore

$$\int_{k\pi}^{(k+1)\pi}|\sin^n(x+r_n)|\,dx = \int_{k\pi+r_n}^{(k+1)\pi+r_n}|\sin^n(y)|\,dy = \int_0^\pi |\sin^n(y)|\,dy$$

for all $k,n.$ Now the last integral $\to 0$ as $n\to \infty$ by the DCT. So the limit of $(1)$ equals

$$\lim_{n\to \infty}\left (\int_0^\pi |\sin^n(y)|\,dy\right)\cdot \sum_{k=0}^{\infty} e^{-\pi k}=0.$$

This gives the desired result.


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