integral

**lsutigers** · July 28th 2009, 07:25 PM

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)

**mr fantastic** · July 28th 2009, 07:52 PM

Originally Posted by lsutigers

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?

..

**Random Variable** · July 28th 2009, 08:26 PM

I'll give it a try.

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

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

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

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

now let a =1

then $\int^{\infty}_{0}t^{n}e^{-t} \ dt = n!$

**HallsofIvy** · July 29th 2009, 06:27 AM

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

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

(Ahhh! I see that mr. fantastic already suggested that.)

**Random Variable** · July 29th 2009, 08:06 AM

But where do you show that "repeatedly differentiating under the integral is permitted"?

I think it's permitted because $f(t, a) = e^{-at}$ and the first n derivatives of $f(t,a)$ are continuous in both t and a. Am I mistaken?

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