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

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    That gives us $1/n$, which obviously tends to 0. Thus, I believe we can write that $\int^{\infty}_{1}\frac{\sin{x}}{x^{n+1}}dx\leq\int^{\infty}_{1}\frac{dx}{x^{n+1}}$. Now, do we need to bound our integral by something from below or is that sufficient?2012-10-04
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    @JohnnyWesterling Approximate the integral of $|\sin x/ x^{n+1}|$.2012-10-04
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    Sorry, but I am not exactly sure what do you mean by "approximate" in this case... What I do see is that for any $x\geq{1}$ and $n\geq{1}$, we can write $0\leq|\sin{x}/x^{n+1}|\leq1/x^{n+1}$...2012-10-04
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    @JohnnyWesterling Sorry, that wasn't what I meant to say; but I think you have the idea...2012-10-04
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    Hint: use $\triangle$2012-10-04
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    Hm... Do you mean estimating our integral by using sums? If yes, then I would ask for some hints on how this could be done...2012-10-04
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    no it is a triangle :)2012-10-04
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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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    Well, our sequence is indeed uniformly convergent on $[1+\varepsilon,\infty)$. Now we would have to check our integral on $[1,1+\varepsilon]$. Well, in this case I believe we should consider $\varepsilon\to{1}^{+}$... is that the correct direction...?2012-10-04
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    sure, what can you say about $f_n$ there ?2012-10-04
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    Well, we know that its integral on this interval would be equal to $F_{n}(\varepsilon)-F_{n}(1)$ where $F_{n}$ is the antiderivative of $f_{n}$, which must be continuous on $[1,1+\varepsilon]$; thus, we can conclude that $\lim_{\varepsilon\to{1}^{+}}(F(\varepsilon)-F(1))=0$. Is that correct?2012-10-04
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    There is a problem on when to switch limits..(btw i would stick to the less confusingr $1+\varepsilon$).2012-10-04
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    I would try to estimate the integral, it is like the area of a thin rectangl.2012-10-04
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    Oh, sorry... I should have $F_{n}(1+\varepsilon)$ and then $\varepsilon\to{0}^{+}$. Yes, I can certainly see the geometric interpretation of that, but I guess we would have to somehow switch to sums... And to be honest, I am not sure how to do this properly in our case.2012-10-04
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    _Estimates the thing for fixed $\varepsilon$..2012-10-04
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    Doesn't the dominated convergence theorem apply easily? Am I mising something?2012-10-04
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    @bogus - not really, the sequence converges pointwise to a quite unpleasant branched function in our case. AD. - Ok... here it goes: for some small $\varepsilon$, for any $n\in\mathbb{N}$ the sequence $f_{n}$ will be decreasing on $[1,1+\varepsilon]$. Thus, the integral will always be smaller than the rectangle: $[(1+\varepsilon)-1]\frac{\sin{1}}{1^{n+1}}=\varepsilon\sin{1}$2012-10-04
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    @Johnny: the sequence converges to $\sin(1)$ at $x=1$, $0$ for $x >1$. That is, it converges to $0$ a.e. But from rereading the OP's question, it appears he hasn't been taught the DCT yet.2012-10-04
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    @bogus Based on the term "uniform convergence" in the OP I assumed the problem is from an earlier course.2012-10-05
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    @JohnnyWesterling That is a good try - but was not what I meant either, also, the statement $f_n$ will be decreasing on $[1,1+\varepsilon]$ would need a proof (in particular note that $x\mapsto\sin x$ increases on $[-\pi/2,\pi/2]$).2012-10-05
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    We may in fact use a very crude estimate here, the sequence $f_n$ is bounded. Use that together with the triangle inequality, that is also what you need to proceed with Robert's hint.2012-10-05
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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