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Integral Calculation

  1. Sep 22, 2007 #1
    Hi guys,recently I came across the following integral and need assistance in solving the problem.

    The crux of the problem is calculating the definite integral of log(sin(x)*sin(x)) over
    the interval ( 0, Pi/2).
    Sorry, I made a mistake in typing the integrand.
    It should be (Log(sin(x)))^2 instead.
     
    Last edited: Sep 22, 2007
  2. jcsd
  3. Sep 22, 2007 #2
    I think you should post this in the calculus group or try yahoo answers.
     
    Last edited: Sep 22, 2007
  4. Sep 22, 2007 #3

    HallsofIvy

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    Hardly a number theory question!

    The problem is [tex]\int_0^{2\pi} log(sin^2(x))dx[/tex]?

    It might help to reduce it to [tex]2\int_0^{2\pi} log(sin(x))dx[/itex]
    If not, I tend to suspect the integral cannot be written in terms of elementary functions.
     
  5. Sep 22, 2007 #4

    mathwonk

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    a little research reveals the related problem, where apparently the definite integral is calculable in spite of the difficulty of the indefinite one.

    the integral of ln(cos) from 0 to x, is called Lobachevsky's integral f(x), and it satisfies the functional equation f(x) = 2f(pi/4 + x/2) - 2f(pi/4 -x/2) - xln(2).

    hence the definite integral from 0 to pi/2 equals pi/2 ln(2).
     
  6. Sep 22, 2007 #5

    arildno

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    ABSOLUTE VALUE SIGN!! :mad:
     
  7. Sep 22, 2007 #6
    sin^2 is always positive?
     
  8. Sep 22, 2007 #7

    arildno

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    Sure it is; but sin(x) is not in the interval given.
    So, when the "2" is re-placed in front of the logarithm, instead of being an exponent within it, you must put sin(x) within the absolute value sign.
     
  9. Sep 22, 2007 #8

    Gib Z

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    Halls typoed the interval, it was meant to be pi/2 not 2pi. So in the correct interval, absolute value signs are not needed, but for what arildno read off, they were, no were all just a little bit confused here =] Mathwonk's got some good books, because I can't find Lobachevsky integral anywhere =[ But if the function equation holds, then mathwonks value is correct.
     
  10. Sep 22, 2007 #9

    mathwonk

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    see joseph kitchen, calculus.
     
  11. Sep 23, 2007 #10
    I am sorry , the integrand should be (log(sin(x)))^2 and the interval is (Pi/2, 0) .


    -Mathslover
     
  12. Sep 23, 2007 #11
    Use a numerical formula (i.g, Gauss's).
     
  13. Sep 24, 2007 #12
    I would like to find the definite integral of (log(sin(x)))^2 under the interval (pi/2,0)
    Numerical integration can only give a numerical answer

    I would like to find the above integral in terms of well-known constants


    -Mathslover
     
  14. Oct 1, 2007 #13
    Reading through "Ramanujan's notebook Part 2" and "Collected Papers of Ramanujan ",
    I chanced upon an entry which solved my problem beautifully.
    I just wish that much more can be learned from Ramanujan's work.





    -mathslover
     

    Attached Files:

  15. Oct 6, 2007 #14
    Devilishly Clever!!! Can the argument be extended to deal with higher powers?
     
  16. Oct 6, 2007 #15
    The beautiful point about this calculation is that it is applicable for all positive integers.


    Define I(n)=Int((log(sin(x)))^n, {x=0 to pi/2}) then it can be shown that


    I(0) = pi/2
    I(1) = -(pi/2)*log(2)
    I(2) = (pi^3)/(24) + (pi/2)*(log(2))^2

    and I(3) is a function of log(2) , pi and Zeta(3)

    -mathslover
     
  17. May 16, 2009 #16
    This is not very difficult. As [tex]log(sin^2(x))=2 log(sin(x))[/itex]
    In the same way as [tex]log(x^2)=2 log x[/itex]
     
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