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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- Aug 26th 2006, 02:57 PM #1

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- Aug 26th 2006, 03:44 PM #2
Are you sure you have it written correctly.

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

Replace p with p+1 to get:

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

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

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

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

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

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

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?.

- Aug 26th 2006, 04:08 PM #3

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- Aug 26th 2006, 04:55 PM #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).

- Aug 26th 2006, 06:21 PM #5