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Math Help - Divergent Improper Integral

  1. #1
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    Divergent Improper Integral

    I could use some help with an integral. The problem asks to determine whether a given integral converges or diverges. I know that the answer is "diverges" but I'm not sure why.

    \int_{-\infty}^0 xe^{-2x} dx

    I used integration by parts to get: [xe^{-2x} - \int e^{-2x} dx] = xe^{-2x}-e^{-2x}

    Then I evaluated \lim_{b\to-\infty}[xe^{-2x}-e^{-2x}]_{b}^{0} = (-1)-(be^{-2b}-e^{-2b})

    I understood \lim_{b\to-\infty}be^{-2b} to require the use of L' Hopital's rule which I applied as follows: \lim_{b\to-\infty}\frac{b}{e^{2b}}=\frac{1}{2e^2b}=\frac{1}{-\infty}=0 If this were true, the improper integral would evaluate to -1 - 0 = -1 and would converge. Can anyone tell me where I went wrong?
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  2. #2
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    \lim_{b\to -\infty}b\;=\;-\infty

    \lim_{b\to -\infty}e^{2b}\;=\;0

    Perhaps your usage of L' Hopital was inappropriate.

    Pardon the notation, but...

    \lim_{b\to -\infty}b\cdot e^{-2b}\;=\;-\infty \cdot \infty\;=\;Unbounded

    Why do you need any fancy theorem?
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  3. #3
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by kaiser0792 View Post
    I could use some help with an integral. The problem asks to determine whether a given integral converges or diverges. I know that the answer is "diverges" but I'm not sure why.

    \int_{-\infty}^0 xe^{-2x} dx

    I used integration by parts to get: [xe^{-2x} - \int e^{-2x} dx] = xe^{-2x}-e^{-2x}

    Then I evaluated \lim_{b\to-\infty}[xe^{-2x}-e^{-2x}]_{b}^{0} = (-1)-(be^{-2b}-e^{-2b})

    I understood \lim_{b\to-\infty}be^{-2b} to require the use of L' Hopital's rule which I applied as follows: \lim_{b\to-\infty}\frac{b}{e^{2b}}=\frac{1}{2e^2b}=\frac{1}{-\infty}=0 If this were true, the improper integral would evaluate to -1 - 0 = -1 and would converge. Can anyone tell me where I went wrong?
    Well,

    [xe^{-2x} - \int e^{-2x} dx] = xe^{-2x}-e^{-2x}

    Is not true.

    [xe^{-2x} - \int e^{-2x} dx] = xe^{-2x} + \frac {e^{-2x}} {2}

    But, just noticing and going back to your integration by parts, the above is not true as well (you did the by parts incorrectly).

     \int_{ - \infty }^0 xe^{-2x} dx = \frac { - xe^{-2x} } {2} -  \frac{-1}{2} \int e^{-2x} =  \frac { - xe^{-2x} } {2} - \frac{ e^{-2x}}{4}


    .........I did the limits on this before on your by parts, but since TKHunny already did this and I don't feel like doing it again with the correct equation I will leave it be. His answer satisfies why.
    Last edited by AllanCuz; April 18th 2010 at 07:16 PM.
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  4. #4
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    I realized after I posted that, my careless integration errors, i.e., missing the factor of 1/2.

    But my real error was interpreting 1/(e^-2b) as b approaches negative infinity as resulting in 1/e^infinity rather than e^infinity. Thank you for clearing things up all, I'm okay now, I think!!
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