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Thread: Integration problem

  1. #1
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    Integration problem

    Can anyone help me to get $\displaystyle S(x)$ from

    $\displaystyle
    S^{'} (x)=x^{-b} e^{ -a x}
    $
    $\displaystyle x>0$

    If I tryed to use integration by parts, by I can't figure put, what to call $\displaystyle f$ and $\displaystyle g$. And Maple does not give me anything to go with.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Mourits View Post
    Can anyone help me to get $\displaystyle S(x)$ from

    $\displaystyle
    S^{'} (x)=x^{-b} e^{ -a x}
    $
    $\displaystyle x>0$

    If I tryed to use integration by parts, by I can't figure put, what to call $\displaystyle f$ and $\displaystyle g$. And Maple does not give me anything to go with.
    If it's a definite integral (from $\displaystyle 0$ to $\displaystyle \infty$), then the answer is $\displaystyle a^{b-1}\Gamma(1-b)$, and I can show that, but I'm not sure how to solve the indefinite integral to find $\displaystyle S(x)$. Wolfram Alpha gives this answer, but I don't know how to get there at the moment.
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  3. #3
    Super Member Deadstar's Avatar
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    Quote Originally Posted by redsoxfan325 View Post
    If it's a definite integral (from $\displaystyle 0$ to $\displaystyle \infty$), then the answer is $\displaystyle a^{b-1}\Gamma(1-b)$, and I can show that, but I'm not sure how to solve the indefinite integral to find $\displaystyle S(x)$. Wolfram Alpha gives this answer, but I don't know how to get there at the moment.
    I think your definite integral answer is the one he is after but... As for the indefinite one.

    Express $\displaystyle e^{-ax}$ as a power series and integrate etc...

    Should end up with...

    $\displaystyle x^{b+1} \sum_{n=0}^{\infty} \frac{(-ax)^n}{(b+n+1)n!}$

    But I'm not sure how to complete it from there!
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  4. #4
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    Supposing we are actually interested in the definite integral... in case Mourits needs to see how to arrive at that answer.

    By letting $\displaystyle x=t$ and $\displaystyle a=s$, we can see this is just a Laplace transform:

    $\displaystyle \int_0^{\infty} x^{-b} e^{-ax} \, dx = \int_0^{\infty} t^{-b} e^{-st} \, dt = \mathcal{L} \{ t^{-b} \}$

    Using a Laplace transform table, we get

    $\displaystyle \mathcal{L} \{ t^{-b} \} = \frac{\Gamma(-b+1)}{s^{-b+1}} = \frac{\Gamma(1-b)}{a^{1-b}} \qquad \mbox{for }b<1 \mbox{ and } a>0$
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  5. #5
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    Thanks
    Last edited by Mourits; Jun 10th 2010 at 10:17 PM.
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