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Thread: 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.

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

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

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

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

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

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

    So rewrite it as $\displaystyle \displaystyle \lim_{t \to \infty}\left[\frac{t - 1}{e^{-t}}\right]$, which is now of the form $\displaystyle \displaystyle \frac{\infty}{\infty}$ and L'Hospital's Rule can be used.
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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, $\displaystyle \displaystyle \lim_{t \to -\infty}[e^t(t - 1)]$ is of the indeterminate form $\displaystyle \displaystyle 0 \times \infty$.

    So rewrite it as $\displaystyle \displaystyle \lim_{t \to \infty}\left[\frac{t - 1}{e^{-t}}\right]$, which is now of the form $\displaystyle \displaystyle \frac{\infty}{\infty}$ 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 $\displaystyle -\infty$ to $\displaystyle \infty$ or is that a typo?

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

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

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