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Integral Proof

  1. Feb 7, 2009 #1
    1. The problem statement, all variables and given/known data
    If n is a positive integer, prove that [tex]\int_{0}^{1}(\ln{x})^ndx=(-1)^n\cdot n![/tex]
    2. Relevant equations



    3. The attempt at a solution
    I'm assuming that since ln(0) is undef and [tex]\mathop{\lim}\limits_{x \to 0^+}\ln{x}=- \infty[/tex] i need to rewrite the integral as [tex]\mathop{\lim}\limits_{t \to 0^+}\int_{t}^{1}(\ln{x})^ndx=(-1)^n\cdot n![/tex]. But I have no idea how to integrate that since n is a variable...
     
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  3. Feb 7, 2009 #2

    Dick

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    This looks like a job for induction. Can you show it's true for n=1? To do the induction step integrate by parts with u=(ln(x))^(n+1) and dv=dx.
     
  4. Feb 7, 2009 #3
    so i proved n=1, and assumed it's true for n=k
    now for n=k+1, I get:
    [tex]\mathop{\lim}\limits_{t \to 0^+}\int_{t}^{1}[/tex]ln(x)(k+1)dx which i need to prove equals [tex](-1)^k^+^1\cdot (k+1)![/tex] Now i know that ln(x)k+1=ln(x)kln(x).

    so if i integrate by parts using f=ln(x)k and g'=ln(x), i go around in circles...

    am i missing something?
     
  5. Feb 7, 2009 #4

    Dick

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    I told you what parts to use. To integrate f*dg=(ln(x))^(k+1)*dx take f=(ln(x))^(k+1), dg=dx. So g=x. What's g*df?
     
  6. Feb 7, 2009 #5
    i'm not familiar with that notation for integration by parts, but if i understand you correctly g*df=[tex]\frac{\ln{x}^k \cdot (k+1)}{x} \cdot 1[/tex]
     
  7. Feb 7, 2009 #6
    so then i get
    ln(x)k+1*x-∫ln(x)k(k+1)dx
     
  8. Feb 7, 2009 #7

    Dick

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    g=x. So I get x*(k+1)*ln(x)^k*(1/x) dx. Do you see how the x's cancel? I was trying to imitate your notation for integration by parts. Integral(f*dg)=f*g-Integral(g*df).
     
  9. Feb 7, 2009 #8

    Dick

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    Ok then. Put in your induction hypothesis for the integral of ln(x)^k.
     
  10. Feb 7, 2009 #9
    can i do that with the (k+1) inside the integral?
    or do i do parts again, and set
    f'=ln(x)^k f=(-1)kk!
    g=k+1 g'=0

    that gets me
    ln(x)k+1*x-[((-1)kk!)(k+1)]+∫0dx
     
  11. Feb 7, 2009 #10
    Physicists don't like proofs by induction.

    [tex]\int_{0}^{1}x^{p}dx = \frac{1}{p+1}[/tex]

    Differentiate both sides n times w.r.t. the parameter p:

    [tex]\int_{0}^{1}x^{p} \log^{n}(x)dx = (-1)^{n}\frac{n!}{(p+1)^{n+1}}[/tex]

    Put p = 0 to get the desired result.
     
  12. Feb 7, 2009 #11

    Dick

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    (k+1) IS a CONSTANT.
     
  13. Feb 7, 2009 #12
    oh duh thanks
     
  14. Feb 7, 2009 #13
    so i have ln(x)k+1x-(k+1)(-1)kk!
    which equals
    ln(x)k+1x-(-1)k(k+1)!

    so somehow ln(x)k+1x needs to get my (-1)k to (-1)k+1
     
  15. Feb 7, 2009 #14

    Dick

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    I'm a physicist, and I don't have any particular problems with induction. But sure, you can do it that way. Note the problem says 'prove that'. Physicists also HATE proofs.
     
  16. Feb 7, 2009 #15

    Dick

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    (-1)*(-1)^k=(-1)^(k+1), doesn't it? The term x*ln(x)^(k+1) vanishes at x=1. And it's limit is 0 as x->0. You can use l'Hopital and another induction argument to show that, if you don't already know it.
     
  17. Feb 7, 2009 #16
    yeah if i just show the l'hopital for the t-->0 that should be sufficient.

    THANKS!!!!!!! :D :D
     
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