6
$\begingroup$

Find $\lim_{n\to \infty} \int_n^{n+1} {\sin x \over x} dx$

I thought about defining $\space F(x) = \int_0^x {\sin t \over t} dt \space$ and then the limit is $\space \lim_{n\to \infty} (F(n+1) - F(n))$, so the answer is 0? It doesn't seem ok

$\endgroup$
7
  • 4
    $\begingroup$ $| \sin x/x|\le 1/|x|$. $\endgroup$ Commented Dec 26, 2015 at 11:07
  • $\begingroup$ Why do you conclude $=0$ ? What do you know about the function $F$ ? $\endgroup$
    – user65203
    Commented Dec 26, 2015 at 11:09
  • $\begingroup$ yes the result is zero $\endgroup$ Commented Dec 26, 2015 at 11:12
  • 2
    $\begingroup$ i think this is a reason $\endgroup$ Commented Dec 26, 2015 at 11:15
  • 3
    $\begingroup$ If $a_n$ is the $n$'th term of your sequence, then $|a_n|\le 1/n$. With your method, you'd have to show $F$ has a limit at $\infty$. $\endgroup$ Commented Dec 26, 2015 at 11:18

4 Answers 4

17
$\begingroup$

We have $$\left \vert \int_n^{n+1} \dfrac{\sin(x)}xdx\right \vert \leq \int_n^{n+1} \left \vert \dfrac{\sin(x)}x\right \vert dx \leq \int_n^{n+1}\dfrac{dx}x = \log\left(1+\dfrac1n\right)$$ Hence, the integral converges to $0$.

$\endgroup$
12
$\begingroup$

The function ${\sin x} \over {x}$ is integrable on $(0,\infty)$. This means that the sum

$$ \sum_{n=0}^\infty \int _n ^{n+1} \frac{\sin x}{x} dx = \int _0 ^{\infty} \frac{\sin x}{x} dx $$ converges. And so, the $n^{th}$ term of this series must go to zero (by the $n^{th}$ term test).

$\endgroup$
12
$\begingroup$

You can use the mean value theorem for integrals. $$\int_{n}^{n+1}{\frac{\sin x}{x}}\,dx=\frac{\sin \varepsilon}{\varepsilon}\; \quad\text{where} \;\varepsilon\in [n;n+1]$$

If $n\to\infty\;$ then $\; \varepsilon\to\infty \;$. Hence $$\lim_{n\to\infty}\int_{n}^{n+1}{\frac{\sin x}{x}}\,dx=\lim_{\epsilon\to\infty}\frac{\sin \varepsilon}{\varepsilon}=0$$

$\endgroup$
2
  • $\begingroup$ wow, it's so awesome when there are so many answers... $\endgroup$
    – Stabilo
    Commented Dec 26, 2015 at 11:42
  • 4
    $\begingroup$ nice! Really like your answer! $\endgroup$
    – SiXUlm
    Commented Dec 26, 2015 at 17:54
3
$\begingroup$

EDIT: Similar to Leg's answer

It is known: $ -1 \le \sin(x) \le 1 \ \forall \ x \in \mathbb{R} $

$ \Rightarrow \int_n^{n+1} {\sin x \over x} dx \le \ \int_n^{n+1} {1 \over x} dx = \ln(n+1) - \ln(n) = \ln(1+\frac{1}{n}) \xrightarrow{n \to \infty} 0 $

$ \Rightarrow \int_n^{n+1} {\sin x \over x} dx \ge \ -\int_n^{n+1} {1 \over x} dx = -\ln(n+1) + \ln(n) = - \ln(1-\frac{1}{n}) \xrightarrow{n \to \infty} 0 $

Hence: $\lim\limits_{n\to \infty} \int_n^{n+1} {\sin x \over x} dx = 0 $

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .