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$$\int_{1}^{\infty} \left| \frac{\sin(x)}{x}\right|dx$$ I'd like to know if this integral converges or not. I tried Wolframalpha but it didn't give me answer.

Seirios
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YD55
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    It does not, you should look for a proof that $${\rm sinc}(x) :=\frac{\sin x}x \notin L^1(\mathbb R)$$which I assume has been presented around here several times – AlexR Feb 23 '14 at 17:26
  • Oh, I didn't study that before. Thanks. – YD55 Feb 23 '14 at 17:30
  • Multi duplicate. – Did Feb 23 '14 at 17:33
  • @Did Couldn't find one ad-hoc so I wrote down a pretty standard proof for it. – AlexR Feb 23 '14 at 17:36
  • Pretty comprehensive answer there: http://math.stackexchange.com/questions/67198/does-int-0-infty-frac-sin-xxdx-have-an-improper-riemann-integral-or – Did Feb 23 '14 at 17:43

2 Answers2

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Note that $$\left.\frac{|\sin x|}{|x|}\right|_{[n, n+1]} \ge\frac1{n+1}|\sin(x)|\tag 1$$ And that $$\int_{\alpha}^{\alpha +1}|\sin(x)| dx \ge \int_{-\frac12}^{\frac12} |\sin x| dx = 2\int_0^{\frac12} \sin x dx = 2(1-\cos(\frac12)) =: C > 0\tag 2$$ So we have $$\int_1^\infty \left|\frac{\sin x}x\right| dx =\sum_{n=1}^\infty \int_n^{n+1}\left|\frac{\sin x}x\right| dx \stackrel{(1)}\ge \sum_{n=1}^\infty \frac1{n+1} \int_n^{n+1} |\sin x| dx \stackrel{(2)}\ge C \sum_{n=2}^\infty \frac1n$$ The latter is the harmonic series (minus the first term) and is well-known to diverge, or can be shown by comparison to $\ln$.

AlexR
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    @YD55 Since you so promptly accepted this answer, surely you know how to justify the inequality between the integral from $\alpha$ to $\alpha+1$ and the integral from $-1/2$ to $1/2$? – Did Feb 23 '14 at 17:45
  • Oh, I checked it by graph. Not rigorously though. – YD55 Feb 23 '14 at 17:48
  • @YD55 "By graph"? What does that mean? You need the thing to hold for every $\alpha$... And this led you to accept the answer without asking for explanations or anything? I don't get it. – Did Feb 23 '14 at 17:54
  • @Did You can argue by symmetry and the monotony of $|\sin|$, or as the solution to the minimisation problem $\min_{\alpha\in[0,2\pi)} \int_{\alpha}^{\alpha+1}|\sin|$. Alternatively you can make the subdivision into intervals of $[n\pi, (n+1)\pi]$ and receive $=2$ and a different (scaled) harmonic series. – AlexR Feb 23 '14 at 17:54
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    @AlexR Be assured that I know how to prove this inequality, in several ways. This is not the point. – Did Feb 23 '14 at 17:55
  • @Did I assume this, knowing your valuable contributions to the site :). Maybe I should switch to other intervals to make the estimate more apparent... – AlexR Feb 23 '14 at 17:57
  • If you move [-1/2,1/2] along x-axis of |sin| graph and watch the variation of area, it is kind of evident that the minimum area with bottom length 1 should be the area under [-1/2,1/2] . – YD55 Feb 23 '14 at 17:59
  • @AlexR If you ask me, your solution can stay as it is, it is up to the OP to decide what they understood from it... – Did Feb 23 '14 at 17:59
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First, for $0 < x < \frac{\pi}{2}$, $\sin x \ge \dfrac{2x}{\pi}$. Likewise for any interval $[k\pi, (k + \frac12)\pi]$ (with integer $k$), $|\sin{x}| \ge \dfrac{2(x-k\pi)}{\pi}$.

For each interval $[k\pi, (k + \frac12)\pi]$, $\dfrac{|\sin x|}{x} \ge \dfrac{2(x - k\pi)}{\pi x} \gt \dfrac{2(x-k\pi)}{(k+1)\pi} = f_k(x)$.

$f_k$ over the $k$th interval forms a triangle with base $\dfrac{\pi}2$ and height $\dfrac{2((k+\dfrac12)\pi - k\pi)}{(k+1)\pi} = \dfrac{1}{k+1}$, and area $\dfrac{\pi}{4(k+1)} = g_k$

The $f_k$ cover the left half of each "bump" in $\dfrac{|\sin x|}{x}$. This isn't a problem because the integrand is non-negative for all of $[0,\infty]$. $\lim_{n \rightarrow \infty} \sum_{k=0}^{n} g_k$ diverges as $O(\log n)$. Since $\dfrac{|\sin x|}{x} \ge f_k(x)$ over every respective interval, the integral must diverge also.

NovaDenizen
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