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Math Help - Integral definition of the factorial function

  1. #1
    Super Member Bacterius's Avatar
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    Integral definition of the factorial function

    Hello,
    just out of mathematical curiosity, I found out on some website that the following statement holds for any n > 0 :

    \int_0^{\infty} \frac{x^n}{e^x} \ dx = n!

    How would one go to actually prove this statement ? Without going in too advanced mathematics if possible.

    Thanks all
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  2. #2
    MHF Contributor
    skeeter's Avatar
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    Quote Originally Posted by Bacterius View Post
    Hello,
    just out of mathematical curiosity, I found out on some website that the following statement holds for any n > 0 :

    \int_0^{\infty} \frac{x^n}{e^x} \ dx = n!

    How would one go to actually prove this statement ? Without going in too advanced mathematics if possible.

    Thanks all
    you opened a can of worms ...

    The Gamma Function
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  3. #3
    Super Member Bacterius's Avatar
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    Ah ... Well, I'll just keep this website in my favorites and come back to it in a couple of years
    That is, when I will have learnt the basics of advanced calculus
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  4. #4
    Super Member Random Variable's Avatar
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    You can integrate by parts n-times, or you can differentiate under the integral sign.

    We know that  \int_{0}^{\infty} e^{-ax} \ dx = \frac{1}{a}

    And notice that  -\int^{\infty}_{0}\frac{\partial^{n}}{\partial a^{n}} \ e^{-ax} \ dx = \int_{0}^{\infty} x^{n}e^{-ax} \ dx (1)

    Now switch the order of integration and differentiation (which in this case is allowed).

      -\frac{\partial^{n}}{\partial a^{n}}\int^{\infty}_{0}\ e^{-ax} \ dx =  -\frac{\partial^{n}}{\partial a^{n}}\frac{1}{a} = n!a^{(-1-n)} (2)

    Finally equate (1) and (2) and let a=1.
    Last edited by Random Variable; February 11th 2010 at 08:00 PM.
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