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Integrating exp(x^2)

  1. Sep 18, 2010 #1
    Hello everyone,

    This is my first post on the forum; I'm pretty sure my question fits in this section.

    I am having a lot of difficulty finding the definite integral
    [tex]\int^{+\infty}_{-\infty}e^{-2ax^{2}}[/tex] dx
    where a is positive and real.
    I know the answer is [tex]\\sqrt{\frac{\pi}{2a}}, but I have no idea at all how to get there.
     
  2. jcsd
  3. Sep 18, 2010 #2
    hint: write another integral of that form, but with y as the variable. multiply them together. convert to polar coordinates, and integrate. then take the square root.
     
  4. Sep 18, 2010 #3
    I will try this, but why would this work? As far as I understand it, the product of the integrals is not necessarily equal to the integral of the products.

    Is this what you are suggesting I try?

    [tex]\int{\inf}_{-\inf}e{-ax{2}}e{-ay{2}}dxdy
     
  5. Sep 18, 2010 #4
    [tex] \int \int f(x) g(y) dx dy = \int f(x) dx \int g(y) dy [/tex]

    (most of the time)

    and yes, i'm suggesting you look at:

    [tex] \sqrt{\int \int e^{-\alpha x^2} e^{-\alpha y^2} dxdy [/tex]
     
  6. Sep 18, 2010 #5
    As a bonus, once you convince yourself that
    [tex] \int_{-\infty}^{\infty} e^{-\alpha x^2} dx = \sqrt{{\frac{\pi}{\alpha}} [/tex],
    you can easily calculate integrals of the form
    [tex] \int_{-\infty}^{\infty}x^{2n} e^{-\alpha x^2} dx [/tex]
    by differentiating the first expression (under the integral sign, on the left) with respect to alpha n times. This is a very useful fact.
     
  7. Sep 18, 2010 #6
    To justify the interchange you can use "[URL [Broken] Theorem[/URL].
     
    Last edited by a moderator: May 4, 2017
  8. Sep 19, 2010 #7
    Thank you for all the help. I'm well on my way, but I am having trouble with the limits of integration. Since x and y go from -inf to inf, what are the new limits?

    Since r = sqrt(x^2+y^2), it seems that r goes from inf to inf, which would result in 0. I know, logically, r should be 0 to infinity, but how do I calculate that?
     
  9. Sep 19, 2010 #8

    HallsofIvy

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    This is "Fubini's Theorem"
     
  10. Sep 19, 2010 #9

    HallsofIvy

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    1) [tex]\int_{-\infty}^\infty e^{-2ax^2}dx= 2\int_0^\infty e^{-2ax^2}dx [/itex]
    by symmetry.

    2) If you let [tex]I= \int_{-\infty}^\infty} e^{-2ax^2}dx[/tex]
    then [tex]\frac{I}{2}= \int_0^\infty e^{-2ax^2}dx[/tex]
    and it is also true that
    [tex]\frac{I}{2}= \int_0^\infty e^{-2ay^2}dy[/tex]
    since that is the same integral with a different "dummy" variable.

    so
    3) [tex]I^2/4= \left(\int_0^\infty e^{-2ax^2}dx\right)\left(\int_0^\infty e^{-2ay^2}dy\right)[/tex]
    [tex]= \int_0^\infty \int_0^\infty e^{-2a(x^2+ y^2)} dy dx[/tex]
    by Fubini's theorem.

    You can interpret that last as a double integral over the first quadrant and change to polar coordinates. Remember that the "differential of area" in polar coordinates is [itex]r drd\theta[/itex] and that [itex]\theta[/itex] goes from 0 to [itex]\pi/2[/itex] in the first quadrant.
     
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