I was trying to prove that the following limit
$\lim_{n\to\infty}\int^{\infty}_{1}\frac{\sin{x}}{x^{n+1}}\mathrm{d}x$
is equal to $0$. I believe that the easiest option in similar cases - and the only one I know... - is proving that that $f_{n}$ converges uniformly to $f$ on the interval of the integral.
However, this is not the case here, as for $x=1$ we have $\lim_{n\to\infty}f_{n}=1$ and for $x>1$ $\lim_{n\to\infty}f_{n}=0$.
I would be very thankful for thoughts on how this should be proven.