Existence of limit over integral

Question

Does the following limit exist?

$\lim _{n\ \rightarrow \ \infty }\int ^{\infty }_{1}\frac{\sin^{n} \ ( x)}{x^{2}} \ d( x)$

With the Theorem of Lebesgue (dominating convergence) there is a function g ∈ $L^{1}$ with g > $\frac{\sin^{n} \ ( x)}{x^{2}}$ (for example $\frac{2}{x^{2}}$). But I don't find the function f with $\lim _{n\ \rightarrow \ \infty } \ \frac{\sin\ ^{n} \ x}{x^{2}}$. I cannot use any of the convergence theorems.

Since it looks as if I cannot use them, I would guess that there is no limit, but I cannot prove it.

Answer 1

We want to find the limit of $\frac{\sin^n(x)}{x^2}$ for $n\to\infty$ for almost every $x\in[0,\infty)$.

For $|\sin(x)|<1$ it is $\lim_{n\to\infty}\sin^n(x) = 0$. Thus $\lim_{n\to\infty}\frac{\sin^n(x)}{x^2} = 0$.

As $|\sin(x)|<1$ for almost every $x\in[0,\infty)$. The point wise limit is the zero function.

Answer 2

I think I have found the answer as follows:

$\displaystyle \lim _{n\ \rightarrow \infty } sin^{n} x\ =\ 0$ nearly everywhere since for all x in $\displaystyle \mathbb{R} \setminus \left\{( 2k-1)\frac{\pi }{2}\right\}$ is $\displaystyle sinx\ < \ 1$