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

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

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

  • 1
    $\begingroup$ 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

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 $$\int_0^b \frac{\sin x}{x} dx = \left(\lim_{n\to \infty} \frac{1}{n}\right) \times \frac{\pi}{2} = 0.$$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.