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Why is it zero ?

  1. Apr 8, 2009 #1
    hi to all ...
    it is just an attempt to understand ....
    the problem was to prove that
    gamma ( z+ 1) = z * gamma(z) using the integral definition of gamma function ...
    when i used the integration by parts i get the following :

    gamma ( z+ 1 ) = ( x^z) * e^(-x) (0 ,∞) + ⌠ z x^(z-1) e^(-x ) dx
    where the limits of integration ( 0 ,∞)

    the integration already gives z * gamma (z) , then ( x^z) * e^(-x) │( 0 ,∞) = 0

    i dont have the evidense that this term is zero ...

    have u ?
     
    Last edited: Apr 8, 2009
  2. jcsd
  3. Apr 8, 2009 #2

    Gib Z

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    Homework Helper

    What does the " | " actually denote? What values do you get when you sub in 0 and take the limit as x goes to infinity ? It should be quite straight forward. Remember exponential terms dominate algebraic terms in limits.
     
  4. Apr 8, 2009 #3
    thank u ... " | " denote nothing other than ( when or at ) , we can neglect it..
    when we sub in zero we get zero , since
    ( 0^z) * e^(-0) = 0 * 1 = 0

    but the problem still hold when we try that limit as x goes to infinity , where it gives
    ( ∞ / ∞ ) or ( 0 / 0) if we try l'Hôpital's rule , in fact i want to understand or get the answer of the question :
    why do exponential terms dominate algebraic terms in limits ?
    i think that you put the exact description of what my question originally is , when you said : Remember exponential terms dominate algebraic terms in limits
     
    Last edited: Apr 8, 2009
  5. Apr 8, 2009 #4

    Dick

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    Take the log of L=x^z*e^(-x), log(L)=z*log(x)-x. x grows faster than z*log(x). Use l'Hopital to prove it. So the limit of the log(L) goes to negative infinity. That means L goes to zero.
     
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