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Is $$\frac{\mathrm d}{\mathrm dx} \frac{\sin x}{x} = \frac{\cos x}{x} - \frac{\sin x}{x^2}$$ Lebesgue integrable?

In other words, is $$ \int_{\mathbb{R}} \left| \frac{\mathrm d}{\mathrm dx} \frac{\sin x}{x} \right| \mathrm dx = \int_{\mathbb{R}} \left| \frac{\cos x}{x} - \frac{\sin x}{x^2} \right|\mathrm dx < \infty? $$

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  • $\begingroup$ What did you try...? $\endgroup$ – Pedro Tamaroff Aug 22 '14 at 1:06
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\begin{align} \int_{\pi/2}^{3\pi/2} \frac{d}{dx} \frac{\sin x}{x} \, dx = \left[ \frac{\sin x} x \right]_{\pi/2}^{3\pi/2} = \frac{-1}{3\pi/2} - \frac 1 {\pi/2} = -\frac{4}{3\pi}. \end{align} If the derivative were negative everywhere in this interval, the integral of the absolute value would be just minus that, thus a positive number. The absolute value makes it a bigger positive number because the intervals on which the signs of the derivatives were different no longer partially cancel each other out.

Do the same on the interval from $3\pi/2$ to $5\pi/2$, then on $5\pi/2$ to $7\pi/2$, and so on, and you'll see a familiar series, which is a lower bound on the integral.

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  • $\begingroup$ Conclusion: The integral of interest is bounded below by a harmonic series, hence diverges. Thanks. $\endgroup$ – LucasSilva Aug 22 '14 at 1:57

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