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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.

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    Are you familiar with the monotone convergence theorem for integrals?2012-10-04
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    The problem that I see with applying it here is that our function $f$, to which $f_{n}$ converges, is a branched function... and I guess even that wouldn't be problem, if it wasn't the fact that $f(1)=1$.2012-10-04
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    @JohnnyWesterling You might add where the problem comes from. I assume your definition of integral is that of Riemann?2012-10-05
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    @AD. Oh, its just a set of problems I have, not even in English, and I don't think it has some online source either, but it got into my hands and I thought of solving it. The question doesn't ask for more than what I wrote, and does not set forth any assumptions.2012-10-05

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