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Math Help - Improper Integrals - Convergence/Divergence

  1. #1
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    Improper Integrals - Convergence/Divergence

    Evaluate the following problem and determine if it converges or diverges. If it converges, calculate what it converges to.

    \displaystyle\int^0_{-\infty} xe^x\,dx

    \displaystyle\int^0_{-\infty} xe^x\,dx=\lim_{t\to-\infty}\int^0_t xe^x\,dx=\lim_{t\to-\infty}(xe^x-e^x)\bigg|_{0}^{t}

    =[e^0(0-1)]-[e^{-\infty}(-\infty-1)]

    =[-1]-[0(-\infty)]{}

    I know the answer is that it conveges to -1, but wouldn't it diverge since I end up with -1-(0*\infty) ?

    Thanks!
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  2. #2
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    First of all, I don't know why you switched the order of the terminals...

    Second, \displaystyle \lim_{t \to -\infty}[e^t(t - 1)] is of the indeterminate form \displaystyle 0 \times \infty.

    So rewrite it as \displaystyle \lim_{t \to \infty}\left[\frac{t - 1}{e^{-t}}\right], which is now of the form \displaystyle \frac{\infty}{\infty} and L'Hospital's Rule can be used.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    First of all, I don't know why you switched the order of the terminals...

    Second, \displaystyle \lim_{t \to -\infty}[e^t(t - 1)] is of the indeterminate form \displaystyle 0 \times \infty.

    So rewrite it as \displaystyle \lim_{t \to \infty}\left[\frac{t - 1}{e^{-t}}\right], which is now of the form \displaystyle \frac{\infty}{\infty} and L'Hospital's Rule can be used.
    That was a typo, sorry....I certainly did not mean to switch the terminals. On the third line of your response, did you purposely switch the limit from -\infty to \infty or is that a typo?

    If it is a typo, here is what I came up with...

    \displaystyle\int^0_{-\infty} xe^x\,dx=\lim_{t\to-\infty}\int^0_t xe^x\,dx=\lim_{t\to-\infty}(e^x(x-1))\bigg|^0_t=-1-\lim_{t\to-\infty}[e^t(t-1)]

    \displaystyle =-1-\lim_{t\to-\infty}\frac{t-1}{e^{-t}}=^{LH}-1-\lim_{t\to-\infty}\frac{1}{-e^{-t}}=-1-0=converges\to -1
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