Limit of a sequence defined as definite integral

What is the limit of $$a_n=\int_n^{n+\sqrt{n}}\frac{\sin x}{x}\ dx\ ?$$ What's the key thing to do here?

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Limit of the integral as n approaches ___? – Shubham May 24 '14 at 16:19
Infinity of course. – Jules May 24 '14 at 16:20

Subbing first $u=x-n$ and then $\displaystyle v=\frac{u}{\sqrt{n}}$,

the initial integral is $\displaystyle \int_0^1\frac{\sin(v\sqrt{n}+n)}{v+\sqrt{n}}$

Taking limits under the integral sign yields $0$ as an answer.

In order to justify the limit/integral permutation, note that $$|\frac{\sin(v\sqrt{n}+n)}{v+\sqrt{n}}| \leq\frac{1}{\sqrt{n}}$$

which grants uniform convergence.

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Apply the mean value theorem for integrals. Since the integrand is continuous on any interval $[n,n+\sqrt{n}]$ there exists a number $\xi_n$ between $n$ and $n + \sqrt{n}$ such that

$$0\leq\Big{|}\int_n^{n+\sqrt{n}}\frac{\sin x}{x}dx\Big{|}=\frac{|\sin \xi_n|}{\xi_n}\sqrt{n}\leq \frac{1}{\sqrt{n}}\rightarrow0$$

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And what about conditions under which we can use this theorem? – Jules May 24 '14 at 16:58
The integrand is a continuous function. – RRL May 24 '14 at 17:06

Hint: Integrate by parts, using $u=\frac{1}{x}$ and $dv=\sin x\,dx$. Estimate the integral that remains.

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Recall that if $\int_a^\infty f$ exists then $$\lim_{A\to\infty}\int_{A}^\infty f=0$$

The integral $$\int_0^\infty \frac{\sin x}{x}dx$$ exists (and equal $\frac\pi2$) so

$$a_n=\int_n^\infty \frac{\sin x}{x}dx-\int_{n+\sqrt n}^\infty \frac{\sin x}{x}dx\xrightarrow{n\to\infty}0$$

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