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Math Help - Improper Integral Evaluation

  1. #1
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    Improper Integral Evaluation

    Does anyone know how to find the integral of abs(x)e^((-x)^2)dx from negative infinity to infinity? I can't quite figure out how to do this...
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  2. #2
    MHF Contributor Calculus26's Avatar
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    See attachment
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  3. #3
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    hmmm... strange it's not indicative of anything we have been working on...
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  4. #4
    Super Member Matt Westwood's Avatar
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    No you're all right, that's not the same problem at all, which is:

    \int_{-\infty}^{\infty} |x| e^{-x^2} dx

    (sorry but I don't believe e^{(-x)^2} because that won't converge.)

    You can split it into two bits: one for x positive, one for x negative. Let's do

    \int_0^{\infty} x e^{-x^2} dx

    Put z = x^2 and that gives you an easy integral in e^z.

    Now do the same for

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

    and add them together.

    If I'm wrong about it NOT being e^{(-x)^2} then it could be the answer is supposed to be zero because one half equals minus the other half ...?
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  5. #5
    MHF Contributor Calculus26's Avatar
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    e^(-x^2) is always positive

    Int e^(-x^2) from -inf to pos inf is sqrt(pi)-- this is one of those well known results - the Gaussian Integral See Gaussian integral - Wikipedia, the free encyclopedia

    however i read the original pblm here incorrectly --didn't see the abs|x| out front so for your problem it works out to 1
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