How do you find this limit $\lim_{n\to +\infty} \int _{\frac{1}{n}}^{n} \frac{|\sin x|^n}{x^{\alpha}}\,dx $ I don't know how to solve the limit
$$\lim_{n\to +\infty} \int _{\frac{1}{n}}^{n} \frac{|\sin x|^n}{x^{\alpha}}\,dx $$
for each $\alpha>1$.
My attempt:
$\displaystyle f_n(x)=\frac{\chi_{[\frac{1}{n},n]}(x) |\sin x|^n}{x^{\alpha}}$
If $x>0$ : $0<|f_n(x)|<\frac{1}{x^\alpha}$ and $f_n \to 0 $ pointwise
$\lim_{n\to +\infty} \int _{\frac{1}{n}}^{n} \frac{|\sin x|^n}{x^{\alpha}}\,dx =\lim_{n\to +\infty} \int _{0}^{+\infty} \frac{|\sin x|^n}{x^{\alpha}}\,dx < \lim_{n\to +\infty} \int _{0}^{+\infty} \frac{1}{x^{\alpha}}\,dx $ but $\alpha>1$ !
Any help is appreciated :)
 A: I think it should converge to zero if $\alpha>1/2$.
Very roughly, the $\sin^nx$ factor becomes $n/\pi$ spikes of width $1/\sqrt{n}$.  So the integral turns into $\frac1{\sqrt{n}}\sum_{k=1}^n k^{-\alpha}=O(n^{1/2-\alpha})$
A: Let be: $$f_n(x)=\chi_{[\frac{1}{n},1]}\frac{|sin(x)|^n}{x^\alpha}+\chi_{[1,n]}\frac{|sin(x)|^n}{x^\alpha}$$ and let be$$\chi_{[\frac{1}{n},1]}\frac{|sin(x)|^n}{x^\alpha}=g_n(x)$$ $$\chi_{[1,n]}\frac{|sin(x)|^n}{x^\alpha}=h_n(x)$$
so we have:
$$g_n(x)<\chi_{[0,1]}sin(1)^nn^\alpha\to0$$$$\int{\chi_{[0,1]}sin(1)^nn^\alpha}=sin(1)^nn^\alpha\to0=\int 0$$
and
$$h_n(x)<\frac{1}{x^\alpha}\chi_{[1,\infty]}$$
so for the first and second theorem of dominated convergence$$\lim\int{f_n}=\lim\int{g_n+h_n}=\lim(\int{g_n}+\int{h_n})=\int{\lim h_n}+\int{\lim g_n}=\int{\lim f_n}$$ so $$\lim \int{f_n}=0$$
A: Observe that
$$\lim_{n\to\infty}\frac{|\sin x|^n}{x^\alpha}=0$$
for almost every $x>0$. Indeed, the above limit is zero for every $x>0$ where $|\sin x|<1$, which is true except on $\{n+\frac\pi2\,:\,n\in\mathbb{N}\}$ which has measure zero. Our aim is then to use the dominated convergence theorem to show that
$$\lim_{n\to\infty}\int_{1/n}^n\frac{|\sin x|^n}{x^\alpha}\ dx=0.$$
It suffices now to find a suitable dominating function. If $x\in(0,1]$, then for all $n>\alpha$, $\frac{|\sin x|^n}{x^\alpha}\leq\frac{|\sin x|^\alpha}{x^\alpha}$. For $x>1$, note that $\frac{|\sin x|^n}{x^\alpha}\leq\frac1{x^\alpha}$. So if we set
$$g(x):=\begin{cases}
\frac{|\sin x|^\alpha}{x^\alpha}&\text{if }0<x\le1,\\
\frac1{x^\alpha}&\text{if }x>1
\end{cases}$$
we find that $\frac{|\sin x|^n}{x^\alpha}\le g(x)$ for all $x>0$ and $n>\alpha$ (note that it is acceptable for the dominating function to dominate all but finitely many of the functions). Note that $g$ is continuous on $(0,1)$ and $\lim_{x\to0^+}g(x)=1$, so there exists $M>0$ such that $g(x)\le M$ for all $x\in(0,1]$. Hence
$$\int_0^\infty g(x)\ dx=\int_0^1g(x)\ dx+\int_1^\infty\frac1{x^\alpha}\ dx\leq M+\frac1{\alpha-1}<\infty$$
and so the dominated convergence theorem applies, completing the proof.
