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How can i solve this

  1. Jun 30, 2009 #1
    How can i solve this:

    [tex]\int Log(Sin(x))dx[/tex]

    ? And if someone knows a good material on this please share :). Thanks
     
  2. jcsd
  3. Jun 30, 2009 #2

    Mark44

    Staff: Mentor

    A good place to start would be integration by parts, with u = ln(sin x), dv = dx. I'm assuming you are working with the natural log, ln, rather than base-10 log.
     
  4. Jun 30, 2009 #3
    Pretty sure I end up with [tex]\int x*Cot(x) dx[/tex] and then ... ?
     
  5. Jun 30, 2009 #4

    Mark44

    Staff: Mentor

    You could do integration by parts again, but neither possible partition seems to get you anywhere. One choice-- u = cot x, dv = xdx gets you a harder integral, and the other -- u = x, dv = cot x dx takes you right back to the original integral being equal to itself, which is no help. If this were a definite integral, you might be able to approximate the integrand with a few terms of a Taylor's series.

    I'm stumped.
     
  6. Jun 30, 2009 #5
    You end up with:

    [tex]xln(sin(x)) - \int x*Cot(x) dx[/tex]

    Try using [tex]Cot(x)=\sqrt{\csc^2 (x) - 1}\ [/tex]

    and substitute t=csc(x).
     
  7. Jun 30, 2009 #6
    Wolframalpha doesn't give an elementary integral...
     
  8. Jun 30, 2009 #7
    You can compute the definite integral from zero to pi quite easily, but I don't think the indefinite integral can be expressed in terms of elementary functions.

    In case of the definite integral, you substitute x = 2 t, then t goes from zero to pi/2, and you write Log[sin(2 t)] as Log(2) + Log[sin(t)] + Log[cos(t)]. By symmetry the two latter integrals are half the original integral. There is a factor 2 from dx = 2 dt, so you find that the integral I satisfies the equation:

    I = pi Log(2) + 2 I ------->

    I = - pi Log(2)
     
  9. Jul 1, 2009 #8
    Ty Count Iblis :),nice approach .
    I need it undefined though. So I'm sure it can't be done with "ortodox" methods. I tried even with FT , but i don't have good hold of it. So i was wondering if you know something else?
    Also I think the result should be complex since Sin(x) can take a negative value.
     
    Last edited: Jul 1, 2009
  10. Jul 1, 2009 #9
    FT won't help if you're looking for a indefinite integral....right?
    I'm not sure if it can be expressed with just elementary functions.
     
  11. Jul 1, 2009 #10
    I guessed so. So I'll try [0,y] so it wont go complex
     
  12. Jul 1, 2009 #11
    For a non complex result, wouldn't you need sin(x) to remain positive?
    That would mean you have to restrict yourself to [tex][0,\pi][/tex], and not to [tex][0,y][/tex], wouldn't it? In which case, the definite integral has been shown by Count Ilbis.
     
  13. Jul 1, 2009 #12
    Yes.True, forgot to mention. But for 0->[tex]\pi[/tex]/3 it won't work for example.

    The idea is I'm missing some course,so i xero-copied from a colleague.Yet the only thing i got for this part from his courses (between 2 mickey mouse figures) is this integral defined once from 0 to [tex]\pi[/tex]/4 which gives : -(G/2 + [tex]\pi[/tex]*Log[2]/4) . Which is wow , i mean what is G first.
    The problem is their all on holiday , couldn't contact the ones that know. So i'm curious what is the method and what to learn
    Yeah.. know sounds dumb, but it happens :|
     
    Last edited: Jul 1, 2009
  14. Jul 1, 2009 #13
    You can express the indefinite integral in terms of the Barnes G-function, http://arxiv.org/abs/math/0308086" [Broken]
     
    Last edited by a moderator: May 4, 2017
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