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Thread: integral proofing

  1. #1
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    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???
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  2. #2
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    Re: integral proofing

    Quote Originally Posted by cummings123321 View Post
    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???
    $\displaystyle \int_0^\infty {xe^{ - xt} dt} = \lim _{b \to \infty } \left( {\left. { - e^{ - xt} } \right|_{t = 0}^{t = b} } \right)=\lim _{b \to \infty } \left( {1 - e^{-xb} } \right) = ?$
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  3. #3
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    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
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    Re: integral proofing

    Quote Originally Posted by cummings123321 View Post
    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 $\displaystyle I(x)$.
    We know that $\displaystyle I(x)\ge 0$. But I have no idea how it fits into the question.
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  5. #5
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    Re: integral proofing

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

    - Hollywood
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