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Math Help - Integration question

  1. #1
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    Integration question

    Can someone show me how

    \int_{0}^{\infty}t^{k}e^{-at}dt=\frac{k!}{a^{k+1}}\text{ where }k\geq 0 \text{ is an integer and }a>0

    I understand that you repeatedly integrate by parts, but how does everything reduce to \frac{k!}{a^{k+1}}?

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  2. #2
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    Re: Integration question

    Quote Originally Posted by downthesun01 View Post
    Can someone show me how

    \int_{0}^{\infty}t^{k}e^{-at}dt=\frac{k!}{a^{k+1}}\text{ where }k\geq 0 \text{ is an integer and }a>0

    I understand that you repeatedly integrate by parts, but how does everything reduce to \frac{k!}{a^{k+1}}?

    Thanks
    Did you repeatedly integrate it by parts and see what happend ? Show us what you have done.
    Thanks from topsquark
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  3. #3
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    Re: Integration question

    Quote Originally Posted by downthesun01 View Post
    Can someone show me how

    \int_{0}^{\infty}t^{k}e^{-at}dt=\frac{k!}{a^{k+1}}\text{ where }k\geq 0 \text{ is an integer and }a>0

    I understand that you repeatedly integrate by parts, but how does everything reduce to \frac{k!}{a^{k+1}}?

    Thanks
    \displaystyle \begin{align*} \int_0^{\infty}{ t^k \, e^{ -a \, t } \, dt } &= \frac{1}{ a^{k+1} } \int_0^{\infty}{ a^{k+1} \, t^k \, e^{-a\,t} \, dt } \\ &= \frac{1}{a^{k+1}} \int_0^{\infty}{ a \left( a\,t \right) ^k\, e^{-a\,t}\,dt} \\ &= \frac{1}{a^{k+1}}\int_0^{\infty}{ u^k\, e^{-u}\,du } \textrm{ after making the substitution } u = a\,t \implies du = a\,dt \\ &= \frac{1}{a^{k+1}} \int_0^{\infty}{ u^{(k+1) - 1} \, e^{-u}\,du } \\ &= \frac{1}{a^{k+1}}\,\Gamma{ \left( k + 1 \right) } \\ &= \frac{k!}{a^{k+1}} \textrm{ since } \Gamma{ (n)} = (n - 1)! \textrm{ if } n \in \mathbf{Z} \end{align*}
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  4. #4
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    Re: Integration question

    Here's what I have after integrating by parts 10 times:

    \frac{-t^{k}}{a}e^{-at}+\frac{kt^{k-1}}{a^2}e^{-at}-\frac{k(k-1)t^{k-2}}{a^3}e^{-at}-...+\frac{k(k-1)(k-2)(k-3)(k-4)(k-5)(k-6)(k-7)(k-8)t^{k-9}}{a^{10}}e^{-at}
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    Re: Integration question

    Thank you. Where does the gamma come from?

    Nevermind, I looked up the gamma function.
    Last edited by downthesun01; June 26th 2013 at 02:09 AM.
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  6. #6
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    Re: Integration question

    It's the Gamma Function, which is the continuous function you get if you interpolate the Factorial function so it is defined over most numbers instead of just the positive integers.
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  7. #7
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    Re: Integration question

    Quote Originally Posted by downthesun01 View Post
    Here's what I have after integrating by parts 10 times:

    \frac{-t^{k}}{a}e^{-at}+\frac{kt^{k-1}}{a^2}e^{-at}-\frac{k(k-1)t^{k-2}}{a^3}e^{-at}-...+\frac{k(k-1)(k-2)(k-3)(k-4)(k-5)(k-6)(k-7)(k-8)t^{k-9}}{a^{10}}e^{-at}
    You are on the good track !
    But in the result of each intagration by part , there is a remaining integral (to be used at the next step). Where is the last remaining integral ?
    You did it 10 times. OK. But what happend before, if k=4 for example ?
    And do not forget that the limits of the integral are t=0 and t=infinity. Where are they applied into your result ?
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