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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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    @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

3 Answers 3

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Hint: What is $\int_1^\infty \dfrac{dx}{x^{n+1}}$ ?

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    Oh, OK :) Well, in order to get rid of the absolute value, we could simply use $-1/x^{n+1}$ on our left. I mean, then we have $\int^{\infty}_{1}\frac{-1}{x^{n+1}}dx\leq\int^{\infty}_{1}\frac{\sin{x}}{x^{n+1}}dx\leq\int^{\infty}_{1}\frac{dx}{x^{n+1}}$. Since both left and right side tend to $0$, we can simply use the squeeze theorem... Is it that, or did you mean something more sublime? :)2012-10-04
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Your idea almost works, to proceed with it you might first consider the sequence $f_n$ on $[1+\varepsilon,\infty)$ for fixed $\varepsilon>0$.

Do you see the next step?

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    @AD. Well, we can use the fact that its bounded, as I mentioned in under Robert's answer, by $1/x^{n+1}$; integrating it on $[1,1+\varepsilon]$, we get $\frac{1-\frac{1}{(1+\varepsilon)^{n}}}{n}$. I don't really see how we can use the triangle inequality here, unless we rewrite the $\sin{x}$ using one of the trigonometric identities, but again, I don't see where this will bring us to.2012-10-05
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STEP 1: NOTE THAT, SINCE $x>1$ $\left|\frac{sin x}{x^{n+1}}\right|\leq \frac{1}{x^{n+1}}$

STEP 2: TAKING INTEGRAL

$\int_{1}^{\infty}\left|\frac{\sin x}{x^{n+1}}\right| dx\leq \int_{1}^{\infty}\frac{dx}{x^{n+1}}=\lim_{t\to\infty}\left(\frac{x^{-n}}{-n}\right)\Big|_{1}^t=\frac{1}{n}.$

STEP 3: TAKING LIMIT WHEN $n\to\infty$

$0\leq\lim_{n\to\infty}\left|\int_{1}^{\infty}\frac{\sin x}{x^{n+1}}\mathrm{d}x\right|\leq \lim_{n\to\infty}\int_{1}^{\infty}\left|\frac{\sin x}{x^{n+1}}\right| dx\leq\lim_{n\to \infty}\frac{1}{n}=0$

Therefore

$\lim_{n\to\infty}\left|\int_{1}^{\infty}\frac{\sin x}{x^{n+1}}\mathrm{d}x\right|=0,$ and so $\lim_{n\to\infty}\int_{1}^{\infty}\frac{\sin x}{x^{n+1}}\mathrm{d}x=0$

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    $\sin{x}$ can indeed take on negative values, while for $x\geq{1}$ the denominator can't. Why would say then in step three that this integral is always greater or equal to $0$?2012-11-03