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  1. #1
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    integral

    Please help, I'm stuck...

    Prove the equality: int[(t^n)(e^-t)] from 0 to infinity = n!, for n=0,1,2,...

    (Evaluate int(e^(-xt)) dt from 0 to infinity and consider whether repeatedly differentiating under the integral is permitted)
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  2. #2
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    Quote Originally Posted by lsutigers View Post
    Please help, I'm stuck...

    Prove the equality: int[(t^n)(e^-t)] from 0 to infinity = n!, for n=0,1,2,... Mr F says: Proof by induction would be a simple approach (combined with integration by parts).

    (Evaluate int(e^(-xt)) dt from 0 to infinity and consider whether repeatedly differentiating under the integral is permitted) Mr F says: Is this a seperate question? Or is it related to the above?
    ..
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  3. #3
    Super Member Random Variable's Avatar
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    I'll give it a try.


    We know that $\displaystyle \int^{\infty}_{0} e^{-at} \ dt = \frac {1}{a} $

    $\displaystyle (\text{-}1)^{n}\frac {\partial^{n}}{\partial a^{n}} \int^{\infty}_{0} e^{-at} \ dt = (\text{-}1)^{n}\int^{\infty}_{0} \frac {\partial^{n}(e^{-at})}{\partial a^{n}} \ dt $ $\displaystyle = \int^{\infty}_{0}t^{n}e^{-at} \ dt $

    and $\displaystyle (\text{-1})^{n}\frac {\partial^{n}}{\partial a^{n}}\frac{1}{a} = \frac{n!} {a^{n-1}}$

    so $\displaystyle \int^{\infty}_{0}t^{n}e^{-at} \ dt = \frac{n!}{a^{n-1}}$

    now let a =1

    then $\displaystyle \int^{\infty}_{0}t^{n}e^{-t} \ dt = n! $
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  4. #4
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    But where do you show that "repeatedly differentiating under the integral is permitted"?

    I think I would be inclined to use "proof by induction", avoiding differentiating under the integral.

    If t= 0, $\displaystyle \int_0^\infty e^{-t}dx= 1= 0!$

    Assume that for some k, [tex]\int_0^\infty t^ke^{-t}dx= k!. Then do$\displaystyle \int_0^\infty t^{k+1}e^{-t}dt$ using integration by parts: let $\displaystyle u= t^{k+1}$, $\displaystyle dv= e^{-t}dt$ so that [tex]du= (k+1)t^k dt[tex] and $\displaystyle v= -e^{-t}$. Then the integral becomes $\displaystyle \left[-t^{k+1}e^{-t}\right]_0^\infty+ (k+1)\int_0^\infty t^k e^{-t}dt$[tex]= (k+1)k!= (k+1)!

    (Ahhh! I see that mr. fantastic already suggested that.)
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  5. #5
    Super Member Random Variable's Avatar
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    But where do you show that "repeatedly differentiating under the integral is permitted"?
    I think it's permitted because $\displaystyle f(t, a) = e^{-at} $ and the first n derivatives of $\displaystyle f(t,a)$ are continuous in both t and a. Am I mistaken?
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