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Math Help - limit of exponential function

  1. #1
    Bud
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    limit of exponential function

    Hi there,

    I found this line in a book which I would like to understand:

    K \int^{+\infty}_{-\infty}e^{-ax^{2}}dx = K \left(\frac{\pi}{a}\right)^{\frac{1}{2}}

    I tried to follow this step, but I did not succeed. I tried to calculated the limits after I determined the integral using l'hopital's rule:

    K \int^{+\infty}_{-\infty}e^{-ax^{2}}dx = \left[ -\frac{1}{2ax}e^{-ax^{2}} \right]^{+\inf}_{-\inf}

    Does somebody know how to handle this?
    Thanks in advance

    Cheers
    Bud
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  2. #2
    Flow Master
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    Quote Originally Posted by Bud View Post
    Hi there,

    I found this line in a book which I would like to understand:

    K \int^{+\infty}_{-\infty}e^{-ax^{2}}dx = K \left(\frac{\pi}{a}\right)^{\frac{1}{2}}

    I tried to follow this step, but I did not succeed. I tried to calculated the limits after I determined the integral using l'hopital's rule:

    K \int^{+\infty}_{-\infty}e^{-ax^{2}}dx = \left[ -\frac{1}{2ax}e^{-ax^{2}} \right]^{+\inf}_{-\inf} Mr F says: This is wrong. If you differentiate the anti-derivative you found, it is clear that you do not get the integrand. In fact, no anti-derivative exists in closed form using elementary functions.

    Does somebody know how to handle this?
    Thanks in advance

    Cheers
    Bud
    Read this: Gaussian Integral -- from Wolfram MathWorld
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  3. #3
    MHF Contributor Amer's Avatar
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    this can't be done like that suppose

    \displaystyle \int_{-\infty}^{\infty}  e^{-ax^2} dx = I

    then

    \displaystyle I^2 = \int_{\infty}^{\infty} e^{-ax^2} dx \int_{-\infty}^{\infty} e^{-ay^2} dy

    rite now

    \displaystyle I^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-a(x^2+y^2)} dx dy

    transfer to polar

    r^2 = x^2 +y^2

    and the Jacobean is r

    \displaystyle I^2 = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-ar^2} r dr d\theta

    can u continue
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  4. #4
    Bud
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    Thanks you two. You proposed the same solution, i.e. this trick of combining two one-dimensional Gaussians and then changing to polar coordinates.

    From there it was simple. Thanks!
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