By popular demand: integral of sin floor 1/x

By popular demand: integral of sin floor 1/x
You asked for it and Dr Peyam delivers! Here we calculate the integral from 0 to 1 of sin floor 1/x where floor is the integer part function. For this we split the sum up into smaller intervals and use the laws of sin to write this in terms of the sum of sin(n)/n and cos(n)/n, for which we have more explicit formulas. Please enjoy this calculus extravaganza which is a must see for all mathematics lovers!
Link to blackpenredpen video: • using complex numbers ...
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[@Mephisto707]

That's not the floor symbol, it's an undecided mixture of floor and ceiling!

[@saj_f0]

In some textbooks they use this notation.

[@ANTONIOMARTINEZ-zz4sp]

I think that in this case, a more convenient way to split up the interval (0,1) should be the following: (1,1/π),(1/π,1/2π),(1/2π,1/3π), etc. So, the interval (1,1/π) and those of the forns of the form (1/2nπ,1/(2n+1)π), n ≥ 1, would have value equal to zero for the floor of sin(1/x) while the intervals of the form (1/(2n-1)π,1/2nπ), n ≥ 1, would have value equal to (-1) for the floor of sin(x) and value (-1/π)*[(1/(2n-1))-(1/2n)], n ≥ 1, for the integral of the interval. So, the result of the integrals for the intervals with non-zero value would be: (-1/π)*(1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... ). In short, this is (-1/π)*(Ln(2)). @drpeyam, please let me know whether my reasoning is correct.

[@geekjokes8458]

floor of sin or sin of floor?

[@drpeyam]

Oops hahaha 😂

[@marcelob.5300]

I love floor sins.

[@Happy_Abe]

Which video does he calculate the cos sum? The one in the description is just for the sin

[@depressedguy9467]

@@Happy_Abe u can see blackpenredpen

[@DihinAmarasigha-up5hf]

Is the answer to the integral in the thumbnail (-ln 2)/pi ?

[@csilval18]

No, it's 0. It's not Reimman integrable pretty sure. But Lebesgue integration gives you 0

[@DihinAmarasigha-up5hf]

@csilval18 but can't you just substitute 1/x = u then break up the u domain such that the floor of sin(u) either become 0 or -1 then evaluate the resulting series by some prefered method (series is a result of the 1/u^2 that you get)

[@csilval18]

Nooo. The thumbnail is wrong. Solved it myself, proving the integral of the floor of sin (1/x) is 0. (It's non zero in a numerable subset of the interval, so the lebesgue integral is 0). But opened the video ans the integral is different :(

[@drpeyam]

Sorry :/

[@worldnotworld]

wtf

[@depressedguy9467]

Second