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Jul 28 at 21:09 comment added Anixx @StevenStadnicki look here, it addresses the issue: mathoverflow.net/questions/475902/… I thought I gave you this link already...
Jul 28 at 21:07 comment added Steven Stadnicki I wasn't one of the downvotes, but I feel like the point in my initial comment still holds strongly.
Jul 28 at 11:05 comment added Anixx Why are the downvotes?
Jul 28 at 0:32 comment added Anixx @StevenStadnicki as you can see, if only at one point the function takes value $2\omega/\pi$, the whole integral is $2$. But on the interval $[0,1)$ there are $\omega_1$ such points. So, the whole integral should be $2\omega_1$.
Jul 26 at 21:18 comment added Anixx @StevenStadnicki also here: philosophy.stackexchange.com/a/112762/796
Jul 26 at 21:10 comment added Anixx @StevenStadnicki see more here: math.stackexchange.com/questions/4627326/… Also, notice, these infinite values have non-zero derivations while finite constants have derivations $0$
Jul 26 at 21:07 comment added Anixx @StevenStadnicki they are "constants" but they are infinite constants. For instance, $2\omega/\pi$ is the derivative of the function $\text{sgn}(x)$ at 0, at this point the function makes a jump of 2, so $\int_{-1}^1\left( {\begin{cases}2\omega/\pi,&{\text{if }}x=0\\ 0,&{\text{if }}x\ne 0\end{cases}}\right)\,dx=2$
Jul 26 at 21:01 comment added Steven Stadnicki Maybe I'm confused, but aren't all of the values inside your example integrals 'constants'? I would expect, for instance, that $\int_0^1\omega\ dt=\omega$ for any 'reasonable' definition of integration...
Jul 26 at 20:53 history edited Anixx CC BY-SA 4.0
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Jul 26 at 19:30 history edited Anixx CC BY-SA 4.0
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Jul 26 at 19:09 history edited Daniele Tampieri CC BY-SA 4.0
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Jul 26 at 17:17 review Low quality posts
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Jul 26 at 16:58 history answered Anixx CC BY-SA 4.0