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Math Help - Please help

  1. #1
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    Please help

    limit(x^p*exp^(-x),x = infinity)=0

    G(p) := Int(x^p*exp^(-x),x = 0 .. infinity)

    show that

    G(p+1)=G(p)

    I used integration by parts by can't get G(p+1)=G(p)

    Thank you very much
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  2. #2
    Eater of Worlds
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    Are you sure you have it written correctly.

    {\Gamma}(p)=\int_{0}^{\infty}x^{p-1}e^{-x}dx

    Replace p with p+1 to get:

    {\Gamma}(p+1)=\int_{0}^{\infty}x^{p}e^{-x}dx=p!

    We can integrate by parts, let x^{p}=u,\;\ e^{-x}dx=dv

    du=px^{p-1}, \;\ v=-e^{-x}

    {\Gamma}(p+1)=-x^{p}e^{-x}|_{0}^{\infty}-\int_{0}^{\infty}(-e^{-x})px^{p-1}dx

    = p\int_{0}^{\infty}x^{p-1}e^{-x}dx=p{\Gamma}(p)

    Therefore, {\Gamma}(p+1)=p{\Gamma}(p)

    This is why I questioned G(p+1)=G(p). It is G(p+1)=pG(p). Maybe that's why you couldn't arrive at the correct answer.

    Does this help?.
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  3. #3
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    Thank you very much. I've got the same. Must be I wrote it wrong
    Great help!
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  4. #4
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    If you want to get into formal detail about convergence check here
    (I did make a mistake in the proof, that the second part when I evaluate by parts cannot be done. However, you can increase the domain in part one to correct the proof).
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  5. #5
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    Smile Thank you so much

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