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Math Help - Integration

  1. #1
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    Integration

    Integrate x.e^(-x^2) from - infinity to + infinity
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  2. #2
    Junior Member
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    Zagreb, Croatia
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     \int_{-\infty}^{\infty} x\cdot e^{-x^2}dx ?
    If that is so, then the integral is 0, becoase integral of odd function from -infinity to +infinity is 0.
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  3. #3
    MHF Contributor
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    Here is one way.

    INT.(-inf.-->+inf.)[x e^(-x^2)]dx

    Let u = e^(-x^2)
    So du = e^(-x^2)*(-2x dx)

    INT.(-inf.-->+inf.)[x e^(-x^2)]dx
    = INT.(-inf.-->+inf.)[e^(-x^2)](x dx)
    = INT.(-inf.-->+inf.)[e^(-x^2)](-2x dx)*(-1/2)
    = (-1/2)INT.(-inf.-->+inf.)[e^(-x^2)](-2x dx)

    That is (-1/2)INT.[du]
    which is (-1/2)(u) +C
    so,

    = (-1/2)[e^(-x^2)](-inf.-->+inf.)
    = (-1/2)[e^(-(+inf.)^2) -e^(-(-inf.)^2)]
    = (-1/2)[e^(-inf.) -e^(-inf)]
    = 0 ---------answer.
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