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)
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! $
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.)
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?But where do you show that "repeatedly differentiating under the integral is permitted"?