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Wolfram|Alpha tells me that $\int|\sin(x)| = -\cos(x)\text{sgn}(\sin(x))$ (which happens to also be its derivative), but I don't understand how this is possible, because the resulting function jumps back to $-1$ at every $\pi$, although the $|\sin(x)|$ never goes below $0$. Also, an integral should always be continuous, and this one isn't.

It seems like the integral is missing $2\lfloor\frac{x}{\pi}\rfloor +1$. Is Wolfram|Alpha wrong, or is there something I'm not aware of?

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  • $\begingroup$ $\text{sgn}(\sin(\pi))=0$, so $-\cos(\pi)\text{sgn}(\sin(\pi))=0$, not $-1$. $\endgroup$ – vadim123 Oct 18 '14 at 22:32
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The indefinite integrals Wolfram Alpha gives aren't guaranteed to be valid over the entire line, just on some open set. In this case, it's chosen something that's valid over a single period. If you ask for a definite integral it should still give the correct answer, though.

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