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I'm stumped at the last part of this problem. Could anyone advise, please? Thank you.

Here is my proof. Define $f: \mathbb{R} \to \mathbb{R}$ by $$ f(x) = \left\{\begin{aligned} &\frac{\sin nx}{nx}&&, x \neq0 \\ &1 &&, x=0 \end{aligned} \right.$$

Hence, $$f(x) = \sum^{\infty}_{k=0}(-1)^k\frac{(nx)^{2k}}{(2k+1)!} , \forall x \in \mathbb{R} \mbox{ and } f \in {\mathscr R[0,b]}.$$

$$\int^{b}_{0} \frac{\sin nx}{nx}dx = \int^{b}_{0}f(x)dx=\sum^{\infty}_{k=0}\int^{b}_{0}(-1)^k\frac{(nx)^{2k}}{(2k+1)!}dx= \sum^{\infty}_{n=0}(-1)^k\frac{(n)^{2k}b^{2k+1}}{(2k+1)(2k+1)!}$$

$$\lim_{n \to \infty} \sum^{\infty}_{k=0}(-1)^k\frac{(n)^{2k}b^{2k+1}}{(2k+1)(2k+1)!}= ?$$

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    $\begingroup$ You have used $n$ for two different things in the sum. Call the summation index $k$ or so. But I don't think the Taylor series will help you crack this one. Consider a substitution in the integral to move the dependence on $n$ somewhere else. $\endgroup$ – Daniel Fischer Nov 24 '13 at 14:36
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Put $u = nx$. Then your integral becomes $$\int_0^{nb} \frac{1}{n} \frac{\sin u}{u} du = \frac{1}{n}\int_0^{nb} \frac{\sin u}{u} du .$$

Taking the limit as $n \to \infty$ and noting that if $\lim a_n = a$, $\lim b_n = b$ then $\lim a_nb_n = ab$ we have $$\int_0^b \frac{\sin x}{x} dx = \left(\lim_{n\to \infty} \frac{1}{n}\right) \times \frac{\pi}{2} = 0.$$

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  • $\begingroup$ Thank you.How does $\frac{\pi}{2}$ come into the picture? $\endgroup$ – Alexy Vincenzo Nov 24 '13 at 15:02
  • $\begingroup$ @AlexyVincenzo $\int \limits_0^{+\infty} \frac{\sin (x)}x\mathrm dx=\frac \pi 2$. $\endgroup$ – Git Gud Nov 24 '13 at 15:04

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