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Math Help - Difficult integration

  1. #1
    Member roshanhero's Avatar
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    Difficult integration

    \int_{-\infty }^{\infty }xe^\frac{-x^2}{a^2}dx
    I tried to solve it by integration by parts, but all messed up.
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  2. #2
    Moo
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    Hello,

    Note that the function f(x)=xe^{-x^2/a^2} is an odd function. And since it's integrated over an interval that is symmetric to 0, the value of the integral is just 0 !
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  3. #3
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    Hello, roshanhero!

    This integration is straight-forward . . .


    \int_{-\infty }^{\infty }xe^{-\frac{x^2}{a^2}}dx

    We have: . \int e^{-\frac{x^2}{a^2}}\left(x\,dx\right)

    \text{Let: }\:u \,=\,-\frac{x^2}{a^2} \quad\Rightarrow\quad du \,=\,-\frac{2x}{a^2}\,dx \quad\Rightarrow\quad x\,dx \,=\,-\frac{a^2}{2}\,du

    \text{Substitute: }\;\int e^u\left(-\frac{a^2}{2}\,du\right) \;=\;-\frac{a^2}{2}\int e^u\,du

    Got it?

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  4. #4
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    Quote Originally Posted by Moo View Post
    Hello,

    Note that the function f(x)=xe^{-x^2/a^2} is an odd function. And since it's integrated over an interval that is symmetric to 0, the value of the integral is just 0 !
    No. The reason this integral is zero is not because the function is odd. If that were the case, then the Cauchy distribution would have a mean equal to zero ....

    Of related interest: http://www.mathhelpforum.com/math-he...ancel-out.html
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  5. #5
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    in order to say it's zero, we need to prove convergence.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by roshanhero View Post
    \int_{-\infty }^{\infty }xe^\frac{-x^2}{a^2}dx
    I tried to solve it by integration by parts, but all messed up.
    xe^{-\frac{x^2}{a^2}}= - \frac{a^2}{2}\frac{d}{dx}e^{-\frac{x^2}{a^2}}

    Then the fundamental theorem of calculus will allow you to show that

    \lim_{a,b \to \infty}\int_{-a }^{b }xe^\frac{-x^2}{a^2}dx=0

    Alternatively recognise that you are being asked to find a multiple of of the mean of a zero mean normally distributed RV, which is therefore 0.

    CB
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