# integral

• Jul 28th 2009, 07:25 PM
lsutigers
integral

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)
• Jul 28th 2009, 07:52 PM
mr fantastic
Quote:

Originally Posted by lsutigers

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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• Jul 28th 2009, 08:26 PM
Random Variable
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 $\displaystyle \int^{\infty}_{0}t^{n}e^{-t} \ dt = n!$
• Jul 29th 2009, 06:27 AM
HallsofIvy
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.)
• Jul 29th 2009, 08:06 AM
Random Variable
Quote:

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?