I want to prove $\displaystyle\int_0^{\infty} \frac{\sin x}x \,\mathrm{d}x = \frac \pi 2$, and $\displaystyle\int_0^{\infty} \frac{|\sin x|}x \,\mathrm{d}x \to \infty$.

And I found in wikipedia, but I don't know, can't understand. I didn't learn differential equation, laplace transform, and even inverse trigonometric functions.

So tell me easy, please.


marked as duplicate by Start wearing purple, Jared, Namaste, Nick Peterson, Adriano Jul 18 '13 at 3:41

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  • 1
    $\begingroup$ Why do you think there is an "easy" way to compute this? See math.stackexchange.com/questions/5248 for many possible solutions. (The second question follows from estimating the integral over each half-period of $\sin$ and comparing with a harmonic series. This may also require more than your background.) $\endgroup$ – mrf Jul 11 '13 at 14:25
  • $\begingroup$ math.stackexchange.com/questions/13344/… $\endgroup$ – Guy Fsone Jan 2 '18 at 10:30

About the second integral: Set $x_n = 2\pi n + \pi / 2$. Since $\sin(x_n) = 1$ and $\sin$ is continuous in the vicinity of $x_n$, there exists $\epsilon, \delta > 0$ so that $\sin(x) \ge 1 - \epsilon$ for $|x-x_n| \le \delta$. Thus we have: $$\int_0^{+\infty} \frac{|\sin x|}{x} dx \ge 2\delta\sum_{n = 0}^{+\infty} \frac{1 - \epsilon}{x_n} = \frac{2\delta(1-\epsilon)}{2\pi}\sum_{n=0}^{+\infty} \frac{1}{n + 1/4} \rightarrow \infty $$

  • $\begingroup$ Oh! Thnx! XD It is the answer that I want! $\endgroup$ – Fourier Jul 11 '13 at 14:42
  • $\begingroup$ @CrMT If it's the answer that you wanted, you should select it as the correct answer (the check mark under the voting arrows). You even gain +2 points for it :) $\endgroup$ – Ataraxia Jul 29 '13 at 14:24
  • $\begingroup$ @Ataraxia I did thanks! $\endgroup$ – Fourier Aug 3 '13 at 10:03

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