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 :)

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 $$\lim \int{f_n}=0$$
