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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.

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Yes, we can: first note that we can integrate over $\mathbb R^+$, and write $\int_0^{+\infty}=\int_0^1+\int_1^{\infty}$. For $n\geq 2$ and $x\leq 1$, $\frac{(\sin x)^n}{x^2}\leq x^{n-2}\leq 1$ and for $x\geq 1,\frac{(\sin x)^n}{x^2}\leq \frac 1{x^2}$. –  Davide Giraudo Dec 3 '11 at 19:26
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1 Answer

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 $$\lim_{n\to\infty}\int_{-\infty}^{+\infty}\frac{(\sin x)^n}{x^2}dx=0.$$

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