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Integration of powers of ln x

  1. May 26, 2008 #1
    1. The problem statement, all variables and given/known data

    Find
    [tex]\int ln x^{2009}[/tex]
    2. Relevant equations

    The integration by parts!

    3. The attempt at a solution

    I have been trying to discover a series which might have helped:
    I have taken out the integrals of :
    1) ln x = x ln x -x
    2) [tex]ln x^{2}[/tex]=x ([tex]ln x^{2}[/tex])-2x ln x +2x
    3)[tex]ln x^{3}[/tex]=x[tex]ln x^{3}[/tex]-3x[tex]ln x^{2}[/tex]+6x ln x-6x
    4)[tex]ln x^{4}[/tex]=x[tex]ln x^{4}[/tex]-4x[tex]ln x^{3}[/tex]+12x [tex]ln x^{2}[/tex]-12x ln x +12x


    But I am unable to find a series. Please help!!
     
  2. jcsd
  3. May 26, 2008 #2
    Please note that all the powers here are not on x but on (ln x)
     
  4. May 26, 2008 #3

    Vid

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  5. May 26, 2008 #4
    Hey, have u found out the pattern urself?
     
  6. May 26, 2008 #5
    I got it as a series doing it just as Vid said. By integrating by parts you can find a recurrence relation that [itex]I_n = \int (\ln x)^n dx[/itex] must satisfy and from that you can figure out the series. So keep working, it's not too bad!
     
  7. May 26, 2008 #6

    Defennder

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  8. May 27, 2008 #7

    Vid

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    Maybe I'm missing something, but that sub got me Int( (ln(x))^2009)dx = 1/2009Int(e^(u^(1/2009))*u^(1/2009)du. The first sub got me Int(u^(2009)*e^(u))du. The second one is much easier to get to and to solve.
     
  9. May 27, 2008 #8

    Defennder

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    Here's what I have:

    [tex]u = (ln x)^{2009} \ du=2009\frac{{lnx}^{2008}}{x}[/tex]
    [tex]dv=1 \ v = x[/tex]
    [tex]\int (lnx)^{2009} dx = x(lnx)^{2009} - 2009\int (lnx)^{2008} dx[/tex]

    And so on...

    Actually I think it's the same thing, only with different variables.
     
  10. May 27, 2008 #9

    Vid

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    Yea, my u was for a change of variable followed by a by parts. Yours was just a by parts so it's the same. I had thought you meant to do a change of variable with u = ln(x)^2009 which is pretty messy.
     
  11. May 27, 2008 #10
    or maybe you can substitute u=ln(x), which would leave you with integrals in terms of e^(u) which, i think, will be easier to work with as compared to ln(x)
     
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