TITLE: Application of Dominated Convergence Theorem?
QUESTION [12 upvotes]: Find with proof the following limit:
$$\lim_{n \to \infty} \int_{-\infty}^{\infty} \frac{(\sin(x))^n}{x^2}dx$$
I want to use the DCT but I cannot seem to dominate $f_{n}(x)=\frac{(\sin(x))^n}{x^2}$ by an integrable function. Any help would be appreciated.

REPLY [14 votes]: Put $f_n(x):=\frac{(\sin x)^n}{x^2}\mathbf 1_{x\neq 0}$. We have for $n\geq 2$ and $x\in\mathbb R$
\begin{align*}|f_n(x)|&=\frac{|(\sin x)^n|}{x^2}\mathbf 1_{|x|\geq 1} +\frac{|(\sin x)^n|}{x^2}\mathbf 1_{|x|< 1}\mathbf 1_{x\neq 0}\\
&\leq \frac 1{x^2}\mathbf 1_{|x|\geq 1}+x^{n-2}\mathbf 1_{|x|< 1}\mathbf 1_{x\neq 0}\\
&\leq\frac 1{x^2}\mathbf 1_{|x|\geq 1}+\mathbf 1_{|x|< 1}=:g(x)
\end{align*}
which is an integrable function. Since $f_n(x)\to 0$ if $x\notin \frac{\pi}2+\pi \mathbb Z$, a set of measure $0$, we can apply the dominated convergence theorem to get 
$$\lim_{n\to\infty}\int_{-\infty}^{+\infty}\frac{(\sin x)^n}{x^2}dx=0.$$