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

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

• $| \sin x/x|\le 1/|x|$. – David Mitra Dec 26 '15 at 11:07
• Why do you conclude $=0$ ? What do you know about the function $F$ ? – Yves Daoust Dec 26 '15 at 11:09
• yes the result is zero – Dr. Sonnhard Graubner Dec 26 '15 at 11:12
• i think this is a reason – Dr. Sonnhard Graubner Dec 26 '15 at 11:15
• 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 $\infty$. – David Mitra Dec 26 '15 at 11:18

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 $0$.

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).

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

• wow, it's so awesome when there are so many answers... – Stabilo Dec 26 '15 at 11:42
• nice! Really like your answer! – SiXUlm Dec 26 '15 at 17:54

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