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I was trying to prove that the following limit


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? – Christopher A. Wong Oct 4 '12 at 19:59
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 $f(1)=1$. – Johnny Westerling Oct 4 '12 at 20:27
@JohnnyWesterling You might add where the problem comes from. I assume your definition of integral is that of Riemann? – AD. Oct 5 '12 at 5:47
@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. – Johnny Westerling Oct 5 '12 at 14:19

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? – Johnny Westerling Oct 4 '12 at 20:11
@JohnnyWesterling Approximate the integral of $|\sin x/ x^{n+1}|$. – David Mitra Oct 4 '12 at 20:16
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 $0\leq|\sin{x}/x^{n+1}|\leq1/x^{n+1}$... – Johnny Westerling Oct 4 '12 at 20:23
@JohnnyWesterling Sorry, that wasn't what I meant to say; but I think you have the idea... – David Mitra Oct 4 '12 at 20:30
Hint: use $\triangle$ – AD. Oct 4 '12 at 20:33

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...? – Johnny Westerling Oct 4 '12 at 20:16
sure, what can you say about $f_n$ there ? – AD. Oct 4 '12 at 20:30
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? – Johnny Westerling Oct 4 '12 at 20:37
There is a problem on when to switch limits..(btw i would stick to the less confusingr $1+\varepsilon$). – AD. Oct 4 '12 at 20:49
I would try to estimate the integral, it is like the area of a thin rectangl. – AD. Oct 4 '12 at 20:53

STEP 1: NOTE THAT, SINCE $x>1$ $$\left|\frac{sin x}{x^{n+1}}\right|\leq \frac{1}{x^{n+1}}$$


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

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

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


$$\lim_{n\to\infty}\left|\int_{1}^{\infty}\frac{\sin x}{x^{n+1}}\mathrm{d}x\right|=0,$$ 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 $0$? – Johnny Westerling Nov 3 '12 at 6:15

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