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We know that $\int_0^\infty \left(\frac{\sin x}{x}\right)^2 dx=\int_0^\infty \frac{\sin x}{x} dx=\frac{\pi}{2}.$

How do I show that $\int_0^\infty \left\vert\frac{\sin x}{x}\right\vert dx$ converges?

3 Answers 3

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It doesn't. Using the convexity of $1/x$,

$\int_0^\infty \left\vert\frac{\sin x}{x}\right\vert \,\mathrm{d}x=\sum_{k=0}^\infty\int_{k\pi}^{(k+1)\pi}\left\vert\frac{\sin x}{x}\right\vert \,\mathrm{d}x>\sum_{k=0}^\infty\int_{k\pi}^{(k+1)\pi}\frac{\left\vert\sin x\right\vert}{(k+1/2)\pi} \,\mathrm{d}x=\frac{2}{\pi}\sum_{k=0}^\infty\frac{1}{k+1/2}\;,$

which diverges since the harmonic series diverges.

  • 0
    @Mielu: You're right, that would have been easier :-) I'm using the convexity by replacing the argument of $1/x$ in the sum of the contributions at $(k+1/2)\pi-\Delta$ and $(k+1/2)\pi+\Delta$ (which have the same value of $\lvert\sin x\rvert$) by the average value $(k+1/2)\pi$.2011-03-07
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It doesn't. This function is a sequence of bumps of decreasing size. The $n^{\text{th}}$ "bump" bounds area on the order of $\frac{1}{n}$ (there are a number of ways to show this), but $\sum_{n=1}^\infty \frac{1}{n} = \infty$.

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$\int_0^\infty \frac{\sin x}{x} dx$ is in fact my most favorite example of a function which is Riemann integrable but not Lesbegue integrable. (just to tease people which think that Lesbegue is so much better than Riemann; in fact this integral might appear more frequent in applications than any integral which is Lesbegue but not Riemann integrable)

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    @Meyer: I agree and I don't want to give the impression that I don't see the beauty of Lesbegues (e.g., the dominated convergence theorem), but Lesbegues doesn't solve all problems and the sinc integral stays an improper integral...2011-03-07