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Prove that $$\underset{n\rightarrow \infty }{\lim} \ \int_{\epsilon }^{\pi} \frac{\sin(nx)}{nx}dx=0\ ;\ \epsilon>0$$ then use the result to deduce: $$\underset{n\rightarrow \infty }{\lim} \ \int_{0 }^{\pi} \frac{\sin(nx)}{nx}dx=0$$


My Attempt:

Since $\frac{\sin(nx)}{nx} \leq \frac{1}{n \epsilon} \forall x \in [\epsilon, \pi]$ (of course if we choose $\epsilon$ small enough), it converges uniformly to 0.

Solving first part is trivial, however when it comes to the second one: $$\underset{n\rightarrow \infty }{\lim} \ \int_{0 }^{\pi} \frac{\sin(nx)}{nx}dx=\underset{n\rightarrow \infty }{\lim} \ \int_{0 }^{\epsilon} \frac{\sin(nx)}{nx}dx+\underset{n\rightarrow \infty }{\lim} \ \int_{\epsilon }^{\pi} \frac{\sin(nx)}{nx}dx$$ I am stuck with the improper integral $\underset{n\rightarrow \infty }{\lim} \ \int_{0 }^{\epsilon} \frac{\sin(nx)}{nx}dx$. It's obvious that it's equal to 0 but I am facing difficulties in showing that. Help would be appreciated.

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  • $\begingroup$ @Tunk-Fey I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$ – epimorphic Jul 4 '14 at 2:51
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    $\begingroup$ Possible duplicate of Let $b>0.$ Evaluate $\lim_{n \to \infty} \int^{b}_{0} \frac{\sin nx}{nx}dx$ $\endgroup$ – Rohan Dec 31 '17 at 15:56
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Since $$\left\lvert \frac{\sin (nx)}{nx}\right\rvert\leqslant 1,$$

you have

$$\left\lvert \int_0^\epsilon \frac{\sin (nx)}{nx}\,dx\right\rvert \leqslant \epsilon,$$

and therefore

$$\limsup_{n\to\infty} \left\lvert \int_0^\pi \frac{\sin (nx)}{nx}\,dx\right\rvert \leqslant \epsilon.$$

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$$\int_0^\pi\frac{\sin(nx)}{nx}dx=\frac1n\int_0^{n\pi}\frac{\sin t}{t}dt\sim_\infty\frac1n \int_0^{\infty}\frac{\sin t}{t}dt$$ and the last integral is convergent ($0$ has a false problem and at $\infty$do an integration by parts to see the convergence).

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As $\left\lvert \frac{\sin (nx)}{nx}\right\rvert\leqslant 1$ by Bounded Convergence Theorem we may take the limit inside the integral, so

$\displaystyle\underset{n\rightarrow \infty }{\lim} \ \int_{0 }^{\pi} \frac{\sin(nx)}{nx}dx=\int_{0 }^{\pi} \underset{n\rightarrow \infty }{\lim}\frac{\sin(nx)}{nx}dx=\int_{0}^{\pi} 0 \ dx=0 $

You can use this line of reasoning with your attempt to get $\underset{n\rightarrow \infty }{\lim} \ \int_{0 }^{\pi} \frac{\sin(nx)}{nx}dx=\underset{n\rightarrow \infty }{\lim} \ \int_{0 }^{\epsilon} \frac{\sin(nx)}{nx}dx+\underset{n\rightarrow \infty }{\lim} \ \int_{\epsilon }^{\pi} \frac{\sin(nx)}{nx}dx= \ \int_{0 }^{\epsilon}\underset{n\rightarrow \infty }{\lim} \frac{\sin(nx)}{nx}dx+ \ \int_{\epsilon }^{\pi} \underset{n\rightarrow \infty }{\lim}\frac{\sin(nx)}{nx}dx=0+0=0$

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  • $\begingroup$ and the region of integration is finite is the other condition needed for the application of the Bounded Convergence Theorem $\endgroup$ – Permian Jul 3 '14 at 16:51
  • $\begingroup$ The Bounded Convergence Theorem comes from a standard course in measure and integration. $\endgroup$ – Permian Jul 3 '14 at 16:57
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You have: $$\int_{0}^{\pi}\frac{\sin(nx)}{nx}dx=\frac{1}{n}\int_{0}^{n\pi}\frac{\sin y}{y}=\frac{1}{n}\sum_{j=0}^{n-1}\int_{0}^{\pi}(-1)^j\frac{\sin y}{y+j\pi}dy,$$ so, since $x\,(1-x^2/\pi^2)\geq\sin(x)\geq 0$ when $x\in[0,\pi]$ and $x+(j+1)\pi > x+j\pi\geq 0$, $$0\leq\int_{0}^{\pi}\frac{\sin(nx)}{nx}dx\leq \frac{1}{n}\int_{0}^{\pi}\frac{\sin x}{x}\,dx\leq \frac{1}{n}\int_{0}^{\pi}\left(1-\frac{x^2}{\pi^2}\right)dx=\frac{2\pi}{3n}.$$

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  • $\begingroup$ This is a truly clever solution, I am wondering how did you find it? $\endgroup$ – SomeOne Jul 4 '14 at 11:05
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    $\begingroup$ This is the same trick used to compute the Dirichlet integral, since, by the Residue Theorem, $$\sum_{n\in\mathbb{Z}}\frac{(-1)^n}{y+n\pi}=\frac{1}{\sin y}.$$ $\endgroup$ – Jack D'Aurizio Jul 4 '14 at 18:51

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