# showing that $\lim_{n \to \infty} \int_{\mathbb R} \frac{\sin^n(x)}{x^2} \, dx= 0$

I want to use the dominated convergence theorem to show that: $$\lim_{n \to \infty}\int_{\mathbb R} \frac{\sin^n(x)}{x^2} \,dx =0$$

I want to make sure my approach is correct. Since $|\sin x| \le 1$, we have that $\forall x \neq 0$:

$$\bigg | \frac{\sin^n(x)}{x^2} \bigg | \le \frac{1}{x^2}$$

So I define $g(x) = x^{-2} ~~\forall x \neq 0$ and $g(x) = 0$ if $x = 0$. Then $g$ is integrable and $|f_n(x)| \le g$, where

$$f_n(x) = \frac{\sin^n(x)}{x^2}$$

So therefore, by DCT, as $f_n \to 0$ pointwise, we can interchange the limit and integral:

$$\lim_{n \to \infty}\int_{\mathbb R} \frac{\sin^n(x)}{x^2} \,dx = \int_{\mathbb R} \lim_{n \to \infty} \frac{\sin^n(x)}{x^2} \, dx = \int0\,dx = 0$$

I'm just wondering if this is the correct approach given the way I construct $g$. My thinking is that $1/x^2$ is integrable everywhere except for a set of measure zero $(-\varepsilon, \varepsilon)$

My thinking is that $1/x^2$ is integrable everywhere except for a set of measure zero $(-\epsilon, \epsilon)$.
That's not a set of measure zero, because it has measure $2\epsilon > 0$. Your dominating function is not integrable, because $$\int_0^{\infty} \frac 1 {x^2} \, dx \ge \int_0^1 \frac 1 {x^2} \, dx = \infty.$$ You need to handle a neighborhood of zero as a special case when you're finding an integrable majorant.
Here is an approach to fixing the proof. Using the fact that $|\sin x| \le |x|$ for all $x$, we actually have the estimate
$$\frac{|\sin^n x|}{x^2} \le \left\{\begin{array}{cc}1/x^2 & |x| \ge 1\\ 1 & |x| < 1\end{array}\right.$$
provided that $n \ge 2$. Now the function on the right actually is integrable on $\mathbb{R}$.