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Math Help - Integration by Parts

  1. #1
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    Integration by Parts

    Hi, I am trying to integrate this function, but do not know how to go past


     \int_{\theta}^{\infty}xe^{-(x-\theta)}dx

    Let  u = x, dv=e^{-(x-\theta)}, du = dx, v=-e^{-(x-\theta)}

    Therefore,

     -xe^{-(x-\theta)}\bigg|_{\theta}^{\infty}-\int_{\theta}^{\infty}-e^{-(x-\theta)}
     [\underbrace{-xe^{-\infty}}_{0}+xe^{-(\theta-\theta)}]-[e^{-x-\theta}\bigg|_{\theta}^{\infty}]

    The X will not cancel, and I know that the answer, through Maple, is  1+\theta .

    The command I used in maple is
    Code:
    int(x*exp(-(x-theta)),x=theta..infinity);
    . Perhaps I have done this incorrect.

    Thank you!
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  2. #2
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    Re: Integration by Parts

    Quote Originally Posted by SC313 View Post
    Hi, I am trying to integrate this function, but do not know how to go past


     \int_{\theta}^{\infty}xe^{-(x-\theta)}dx

    ...
    Re-write the integral to:

     \int_{\theta}^{\infty}xe^{-(x-\theta)}dx = e^{\theta} \cdot \int_{\theta}^{\infty}xe^{-x}dx

    and then proceed.
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  3. #3
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    Re: Integration by Parts

    Quote Originally Posted by SC313 View Post
    ...
    Therefore,

     -xe^{-(x-\theta)}\bigg|_{\theta}^{\infty}-\int_{\theta}^{\infty}-e^{-(x-\theta)}
     [\underbrace{-xe^{-\infty}}_{0}+xe^{-(\theta-\theta)}]-[e^{-x-\theta}\bigg|_{\theta}^{\infty}]

    The X will not cancel, and I know that the answer, through Maple, is  1+\theta .

    The command I used in maple is
    Code:
    int(x*exp(-(x-theta)),x=theta..infinity);
    . Perhaps I have done this incorrect.

    Thank you!
    You stopped exactly one step befor the finish:


     -xe^{-(x-\theta)}\bigg|_{\theta}^{\infty}-\int_{\theta}^{\infty}-e^{-(x-\theta)}
     [\underbrace{-xe^{-\infty}}_{0}+\underbrace{xe^{-(\theta-\theta)}}_{=x}]\underbrace{-[e^{-x-\theta}\bigg|_{\theta}^{\infty}]}_{-(0-1)}

    Expand the last bracket and you'll get x + 1 as the final result.
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  4. #4
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    Re: Integration by Parts

    The OP seems to have forgotten that when evaluating at the lower limit of \displaystyle \begin{align*} \theta \end{align*} that ALL values of x are replaced with \displaystyle \begin{align*} \theta \end{align*}.

    Also for the upper limit of \displaystyle \begin{align*} \infty \end{align*} it should really be

    \displaystyle \begin{align*} \lim_{\epsilon \to \infty} -\epsilon \, e^{-\left( \epsilon - \theta \right) } &= \lim_{\epsilon \to \infty} -\frac{\epsilon}{e^{\epsilon - \theta}} \\ &= \lim_{\epsilon \to \infty} -\frac{1}{e^{\epsilon - \theta}} \textrm{ by L'Hospital's Rule} \\ &= 0 \end{align*}
    Thanks from earboth
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  5. #5
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    Re: Integration by Parts

    Quote Originally Posted by Prove It View Post
    The OP seems to have forgotten that when evaluating at the lower limit of \displaystyle \begin{align*} \theta \end{align*} that ALL values of x are replaced with \displaystyle \begin{align*} \theta \end{align*}.

    ...
    Not only the OP, the same happened to me too
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  6. #6
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    Re: Integration by Parts

    The final answer is zero for this improper integral.
    Minoas
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