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I have this exercise I don't know how to approach :

Find the limit : $$\lim_{n\rightarrow\infty}\int_{n}^{n+7}\frac{\sin x}{x}\,\mathrm dx$$

I can see that with $n\rightarrow\infty$ the area under the graph of this function becomes really small as $\sin{x} \leq 1$ so $\dfrac{\sin{n}}{n}\rightarrow_{\infty}0$ but can I get something from it?

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up vote 9 down vote accepted

Hint: $\def\abs#1{\left|#1\right|}$ $$\abs{\int_n^{n+7}\frac{\sin x} x \, dx}\le \int_n^{n+7}\frac{\abs{\sin x}}{x}\, dx\le \frac 1n \int_n^{n+7}\abs{\sin x}\, dx $$

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hence $$= \frac{1}{n}(\sin(n+7)-\sin(n)) \leq \frac{1}{n} \rightarrow_{n\rightarrow\infty}0$$ ? :) – darenn Jun 5 '13 at 15:37
@darenn I think a better upper bound is $\frac{7}{n}$ since the range of $\sin$ on real numbers is $[-1,1]$. – Cameron Williams Jun 5 '13 at 17:47

Hint: By the Mean Value Theorem, there exists a $c_n \in (n,n+7)$ so that

$$\int_{n}^{n+7}\frac{\sin x}{x}\,\mathrm dx =7 \frac{\sin(c_n)}{c_n} \,.$$

What happens with $\frac{\sin(c_n)}{c_n}$ when $n \to \infty$?

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It goes to $0$, but is it enough? I've been thinkin about the MVT I just thought its not enough of a proof – darenn Jun 5 '13 at 18:00
@darenn Well if the integral is equal to something which goes to zero.... If you want a nice formal proof, just use $-\frac{1}{n} <\frac{\sin(c_n)}{c_n} < \frac{1}{n}$ and the Squeeze Theorem... – N. S. Jun 5 '13 at 18:07

Rescale the integral so that it takes the form

$$\int_0^1 dy \frac{\sin{(n+7) y}}{y} - \int_0^1 dy \frac{\sin{n y}}{y}$$

Combine and use the sine addition/subtraction rule to get that the integral is equal to

$$2 \int_0^1 dy \frac{\sin{\left(\frac{7}{2} y\right)} \, \cos{\left(n+\frac{7}{2}\right) y}}{y}$$

Integrate this by parts and this is equal to

$$\frac{2}{n+\frac{7}{2}} \sin{\left(\frac{7}{2} \right)} \, \sin{\left(n+\frac{7}{2}\right) } - \frac{2}{n+\frac{7}{2}} \int_0^1 dy \, \sin{\left(n+\frac{7}{2}\right) y } \frac{d}{dy}\left [ \frac{\sin{\left(\frac{7}{2} y\right)}}{y}\right]$$

As $n \to \infty$, the integral on the right vanishes as $O\left(1/n^2\right)$, so the behavior of the original integral is dominated by the first term, which is $O\left(1/n\right)$ in this limit.

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Srsly?${}{}{}{}$ – TonyK Jun 5 '13 at 17:59
@TonyK: were you expecting a response from me? – Ron Gordon Jun 5 '13 at 18:09
Not really. But I got one :-) – TonyK Jun 5 '13 at 18:17
You missed a chance to use Riemann Lebesgue on $$2 \int_0^1 dy \frac{\sin{\left(\frac{7}{2} y\right)} \, \cos{\left(n+\frac{7}{2}\right) y}}{y}!!$$ – Pedro Tamaroff Aug 19 '13 at 22:16

$$ \left\vert\int_{n}^{n + 7}{\sin\left(x\right) \over x}\,{\rm d}x\right\vert < \int_{n}^{n + 7}{{\rm d}x \over x} = \ln\left(1 + {7 \over n}\right) \to 0\quad \mbox{when}\quad n \to \infty $$

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Really small times seven equals really small.

You have stated that, for any positive $\epsilon$, you have $|\sin x/x| < \epsilon$ for for sufficiently large $x$.

Thus, for sufficiently large $n$,

$$ \left| \int_{n}^{n+7} \frac{\sin x}{x} ~dx \right| < \int_n^{n+7} \epsilon ~ dx $$

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No I don't think that there is a lot that can be got out of it. For example the result is not yet sufficient condition for convergence of the integral from $0$ to $\infty$. However, the Riemann improper integral of the first kind \begin{equation} \int_0^\infty \textrm{sinc}(x/\pi) dx \end{equation} where $\textrm{sinc}(x) = \Bigg\{ \begin{eqnarray} & \frac{\sin(\pi x)}{\pi x} & , \ \ x \neq 0 \\ &1&, \ \ x = 0 \end{eqnarray}$ converges as we know. Define \begin{equation} a_n = \int_0^n \textrm{sinc}(x/\pi) dx \ . \end{equation} Now $\{a_n\}_{n=1}^\infty$ is a Cauchy-sequence. The integral under the limit can be expressed as follows. \begin{eqnarray} \int_n^{n+7} \frac{\sin(x)}{x} dx & = & \int_n^{n+7} \textrm{sinc}(x/\pi) dx = \int_0^{n+7} \textrm{sinc}(x/\pi) dx - \int_0^n \textrm{sinc}(x/\pi) dx \\ \\ \\ \\ & = & a_{n+7} - a_n \rightarrow 0 \end{eqnarray} as $n \rightarrow \infty$.

Without the knowledge that the integral from $0$ to $\infty$ converges we can proceed as follows. We first show that $a_n$ is a Cauchy-sequence. We may assume $m \leq n$. Otherwise we can apply $|a_n-a_m| = |a_m-a_n|$ and interchange the order of $m$ and $n$ in the first term of the following calcualtion. We have \begin{eqnarray} |a_n-a_m| & = & \bigg| \int_0^n \textrm{sinc}(x/\pi)dx - \int_0^m \textrm{sinc}(x/\pi)dx \bigg| = \bigg| \int_m^n \textrm{sinc}(x/\pi)dx \bigg| \\ & = & \bigg| \int_m^n \frac{\sin(x)}{x}dx \bigg| = \bigg| \bigg|_m^n \frac{-\cos(x)}{x} - \int_m^n -\frac{-\cos(x)}{x^2} dx \bigg| \\ & = & \bigg|-\frac{\cos(n)}{n}+\frac{\cos(m)}{m}-\int_m^n \frac{\cos(x)}{x^2} dx \bigg| \\ & \leq & \bigg|-\frac{\cos(n)}{n} \bigg| + \bigg| \frac{\cos(m)}{m} \bigg| + \bigg| \int_m^n \frac{\cos(x)}{x^2} dx \bigg| \\ & \leq & \bigg|\frac{\cos(n)}{n} \bigg| + \bigg| \frac{\cos(m)}{m} \bigg| + \int_m^n \bigg| \frac{\cos(x)}{x^2} \bigg| dx \\ & \leq & \frac{|\cos(n)|}{|n|} + \frac{|\cos(m)|}{|m|} + \int_m^n \frac{|\cos(x)|}{|x^2|} dx \\ & \leq & \frac{1}{n} + \frac{1}{m} + \int_m^n \frac{1}{x^2} dx \leq \frac{1}{n} + \frac{1}{m} + \bigg|_m^n -\frac{1}{x} \\ & \leq & \frac{1}{n} + \frac{1}{m} - \frac{1}{n} + \frac{1}{m} \leq \frac{1}{n} + \frac{1}{m} + \frac{1}{n} + \frac{1}{m} = \frac{2}{n} + \frac{2}{m} \rightarrow 0 \end{eqnarray} as $m, \ n \rightarrow \infty$. Hence $a_n$ is a Cauchy-sequence. We can then calculate as above after the definition of $a_n$ and obtain the desired result \begin{equation} \lim_{n \rightarrow \infty} \int_n^{n+7} \frac{\sin(x)}{x} dx = 0 \ . \end{equation} The existence of the integral from $0$ to $\infty$ can be now established by calculating the distance $|a_M-a|$ where $a$ is the limit of the Cauchy-sequence. For the next integer $m$ from $M$ we have \begin{equation} |a_M-a| = |a_M-a_m+a_m-a| \leq |a_M-a_m| + |a_m-a| \end{equation} the second term goes to $0$ by definition of limit and the first can be estimated as you thought in your question. Hence \begin{equation} a = \lim_{M \rightarrow \infty} a_M = \lim_{m \rightarrow \infty} \int_0^M \textrm{sinc}(x/\pi) dx = \int_0^\infty \textrm{sinc}(x/\pi) dx \ . \end{equation} Hopefully you got some perspective to the integration of $\frac{\sin(x)}{x}$.

Then the exercise. We calculate using the previous estimate \begin{eqnarray} \Bigg|\int_n^{n+7} \frac{\sin(x)}{x} dx \Bigg| \leq \frac{1}{n+7} + \frac{1}{n} - \frac{1}{n+7} + \frac{1}{n} = \frac{2}{n} \rightarrow 0 \ , \end{eqnarray} as $n \rightarrow \infty$. Note that we succeed to get rid of the $7$ in the question. Another approach is to estimate \begin{eqnarray} \sup_{x \in [0,7]} \Bigg|\frac{\sin(x+n)}{x+n} - 0 \Bigg| \leq \sup_{x \in [0,7]} \frac{1}{n} = \frac{1}{n} \rightarrow 0 \ , \end{eqnarray} as $n \rightarrow \infty$. Hence $\frac{\sin(x+n)}{x+n}$ converges uniformly to $0$. Hence \begin{eqnarray} \lim_{n \rightarrow \infty} \int_n^{n+7} \frac{\sin(x)}{x} dx = \lim_{n \rightarrow \infty} \int_0^{7} \frac{\sin(x+n)}{x+n} dx = \int_0^7 \lim_{n \rightarrow \infty} \frac{\sin(x+n)}{x+n} dx = \int_0^7 0 dx = 0 \ , \end{eqnarray} where the order of integration and limit is changed by uniform convergence of functions defined on the compact set $[0,7]$.

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