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Thread: Integration after differentiation of exponential to exponential

  1. #1
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    Integration after differentiation of exponential to exponential

    I am trying to solve the following complex form. I will really appreciate the way to solve it. I have tried and failed myself. My results do not even qualify to be mention here.


    \begin{equation}
    P = e^{-ne^{- \sigma \lambda t}}
    \end{equation}


    \begin{equation}
    E(t) = \int_{0}^{+\infty} t \frac{\partial P}{\partial t}\ dt
    \end{equation}


    This is part of the paper I am trying to solve.
    The solution that the author has given is


    \begin{equation}
    E(t) = \frac{1}{\sigma \lambda} \sum^n_{k=1} (-1)^{k+1} {{n} \choose {k}} \frac{1}{k}
    \end{equation}
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  2. #2
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    Re: Integration after differentiation of exponential to exponential

    how can

    $\displaystyle \int_0^{+\infty}~t \dfrac{\partial P}{\partial t}~dt$

    be a function of $t$ ?
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  3. #3
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    Re: Integration after differentiation of exponential to exponential

    Why not? If P is kind of CDF, then the whole term is the Expectation of the pdf of the function... It seems all ok!
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  4. #4
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    Re: Integration after differentiation of exponential to exponential

    sure the integral exists.

    It's just not a function of t.

    Did you try integration by parts?
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  5. #5
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    Re: Integration after differentiation of exponential to exponential

    I don't get that they are exactly the same

    Integration after differentiation of exponential to exponential-clipboard01.jpg

    Mathematica doesn't recognize the difference of these two terms as 0 even after FullSimplify.

    Perhaps there is a typo somewhere?
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  6. #6
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    Re: Integration after differentiation of exponential to exponential

    Quote Originally Posted by sjaffry View Post
    Why not? If P is kind of CDF, then the whole term is the Expectation of the pdf of the function... It seems all ok!
    Please go back and check the full statement of the problem. If the integrand depends only on t and the integration is from 0 to infinity, then the result of the integral is a number, not a function of t. Further the fact that there is a partial derivative of P with respect to x implies that P is a function of at least on other variable.
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