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Gamma function

  1. Sep 7, 2013 #1
    1. The problem statement, all variables and given/known data
    "Show that [tex]- \int^1_0 x^k\ln{x}\,dx = \frac{1}{(k+1)^2} ; k > -1[/tex].

    Hint: rewrite as a gamma function.
    2. Relevant equations
    Well, I know that [tex]\Gamma \left( x \right) = \int\limits_0^\infty {t^{x - 1} e^{ - t} dt}[/tex].

    3. The attempt at a solution
    I've tried various substitutions, beginning with u=k*ln(x), but I'm not getting very far. To write it as a gamma function, I'd have to change the limits from (0 to 1) to (0 to infinity) and I can't find a way to do that.

    Can someone point me in the right direction?
     
  2. jcsd
  3. Sep 7, 2013 #2

    vanhees71

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    I wonder why you should use the [itex]\Gamma[/itex] function for this elementary integral. Perhaps I'm missing something, but if I'm not wrong, it should be pretty easily solvable by integration by parts.

    If you are forced to use the [itex]\Gamma[/itex] function, I'd indeed try the substitution
    [tex]x=\exp (-u/k),[/tex]
    which leads to an integral you can evaluate immediately in terms of the [itex]\Gamma[/itex] function.
     
  4. Sep 7, 2013 #3
    It does seem pretty straightforward with integration by parts, but since I'm told to use the gamma function, I'd at least like to know how to do that.
    If I use the substitution [tex]x=e^{-u/k}[/tex], I get [tex]dx = -k*e^{-u/k}\,du[/tex] The integral then becomes [tex]\int^1_0 e^{-u}*-\frac{u}{k}*-k*e^{-u/k}\,du = \int^1_0 e^{-(u + \frac{u}{k})}u\,du[/tex],
    but the problem remains that it's between 0 and 1, not 0 and infinity. How do I get to a gamma function from there?
    Thanks!
     
  5. Sep 7, 2013 #4

    LCKurtz

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    Isn't that a ##\frac {-1} k## instead of ##-k## out in front?

    If ##x=e^{-u/k}## and ##x## goes from ##0## to ##1##, how do you get ##u## going from ##0## to ##1##?
     
  6. Sep 7, 2013 #5
    Ah! That clears it up. Thanks!
     
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