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

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.

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marked as duplicate by O.L., Jared, amWhy, Nick Peterson, Adriano Jul 18 '13 at 3:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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.) –  mrf Jul 11 '13 at 14:25

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up vote 2 down vote accepted

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: $$\int_0^{+\infty} \frac{|\sin x|}{x} dx \ge 2\delta\sum_{n = 0}^{+\infty} \frac{1 - \epsilon}{x_n} = \frac{2\delta(1-\epsilon)}{2\pi}\sum_{n=0}^{+\infty} \frac{1}{n + 1/4} \rightarrow \infty $$

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Oh! Thnx! XD It is the answer that I want! –  Fourier Jul 11 '13 at 14:42
@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 :) –  Ataraxia Jul 29 '13 at 14:24
@Ataraxia I did thanks! –  Fourier Aug 3 '13 at 10:03

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