1. ## integral proofing

let f(x,t)=xe^(-xt).show that the integral I(x)=∫f(x,t)dt (integration from 0 to infinite)exists for all x>=0 . is x->I(x) continuous on [0,infinite)
what should i use here to prove the integral exist ???can someone give me the detail expanlation???

2. ## Re: integral proofing

Originally Posted by cummings123321
let f(x,t)=xe^(-xt).show that the integral I(x)=∫f(x,t)dt (integration from 0 to infinite)exists for all x>=0 . is x->I(x) continuous on [0,infinite) what should i use here to prove the integral exist ???can someone give me the detail expanlation???
$\int_0^\infty {xe^{ - xt} dt} = \lim _{b \to \infty } \left( {\left. { - e^{ - xt} } \right|_{t = 0}^{t = b} } \right)=\lim _{b \to \infty } \left( {1 - e^{-xb} } \right) = ?$

3. ## Re: integral proofing

thank you,then the limit is 1 so the integral exists,but how about the continuous part, for now x->I(x) ,I(X)=1 it seems ,it is continuous?? i am not sure

4. ## Re: integral proofing

Originally Posted by cummings123321
thank you,then the limit is 1 so the integral exists,but how about the continuous part, for now x->I(x) ,I(X)=1 it seems ,it is continuous?? i am not sure
Frankly, I have the same question about $I(x)$.
We know that $I(x)\ge 0$. But I have no idea how it fits into the question.

5. ## Re: integral proofing

Isn't $I(x)=1$ for $x>0$ and $I(x)=0$ for $x=0$? So it would be continuous on $(0,\infty)$ and discontinuous at 0, right?

- Hollywood