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A Much Harder Floor Integral

  1. Mar 25, 2014 #1
    Doing integrals for just plain curiosity! So, given the integral:

    ∫dx/floor(1-logbase2(1-x)) from 0 to 1

    I have looked at the graph of the integral, and I notice what seems to be an infinite number of areas under the curve (from 0 to 1/2 the area is 1/2, from 1/2 to 3/4 the area is 1/8, etc). How can I get a general expression for the nth rectangle (perhaps a summation) and a numerical answer for the integral? Thanks for any help!
     
  2. jcsd
  3. Mar 25, 2014 #2

    HallsofIvy

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    If there are indeed an infinite number of areas then the integral will be an infinite series.

    Assuming that your values are right (I didn't check), then 1/2+ 1/8+ 1/64+ ... , all powers of 1/2 so that is a geometric series. There is a formula for sums of geometric series that you probably know.
     
  4. Mar 25, 2014 #3

    Dick

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    I think the sum is actually 1*(1/2)+(1/2)*(1/4)+(1/3)*(1/8)+(1/4)*(1/16)+(1/5)*(1/32)+... That's a little harder. It's the integral of a geometric series.
     
    Last edited: Mar 26, 2014
  5. Mar 25, 2014 #4
    Yes, but after some more inspection of the graph I observed that the heights of each rectangle start as 1 and form the harmonic sequence (1, 1/2, 1/3, 1/4, 1/5, ...) and the length form a geometric sequence (1/2, 1/4, 1/8, 1/16, ...). So I get the sum from 1 to infinity of (1/n)*(1/2^n) which comes out to ln(2) which I think is the answer...
     
  6. Mar 25, 2014 #5

    Dick

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    Why, yes, I think it is.
     
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