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Unintegrateable fn.[s]

  1. Feb 26, 2008 #1


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    [tex]\int e^{-x^2} dx[/tex]

    if that can't be expressed in terms of elementary functions, how did they compute

    [tex] \int ^{\infty} _{- \infty} e^{-x^2} dx =\sqrt{\pi}[/tex]

    (I think I have the limits wrong, but I know it has [itex]\infty[/itex] as the upper or lower limit)
  2. jcsd
  3. Feb 26, 2008 #2
    well, one way of doing so is i guess expanding [tex] e^{-x^{2}}[/tex] as a power series, using taylor series. But i also think one can compute it using double integrals. I have just heard about this though, since i have no idea how to deal with double integrals yet!
  4. Feb 26, 2008 #3
    [tex] I = \int_{-\infty}^{\infty}e^{-x^2}dx[/tex]
    [tex] I^2 = \int_{-\infty}^{\infty}e^{-x^2} dx\times \int_{-\infty}^{\infty} e^{-y^2}dy[/tex]

    [tex] I^2 = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}dxdy[/tex]
    [tex] I^2 = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-(r^2)}rdrd\theta[/tex]
    [tex] I^2 = \pi - \pi e^{-\infty}[/tex]
    Last edited: Feb 26, 2008
  5. Feb 26, 2008 #4
    THis looks cute, although i do not understand a damn thing what u did! I mean i haven't yet dealt with double integrals!
  6. Feb 26, 2008 #5

    D H

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    This is the Gaussian integral. The wikipedia article (http://en.wikipedia.org/wiki/Gaussian_integral" [Broken]) on this integral goes over the derivation of this integral and does a rigorous job (check out the "careful" proof of the identity).
    Last edited by a moderator: May 3, 2017
  7. Feb 26, 2008 #6
    EDIT: Note: Read the article D H posted, and not the gibberish below.

    Honestly, I can't believe I remembered how to do it. I've seen it a couple of times in class. To me it is, a trick.

    But here's the jist of it. That x^2 looks like a beast, and integrating from -infinity to infinity seems like a problem.

    We know that if we multiply two exponentials e^u*e^y we get e^(u+y). So when we multiply the two integrals together and get x^2+y^2 this should be screaming, convert me into polar coordinates.

    So we multiply the two integrals together, and convert the x^2+y^2 into r^2.

    First though, why is that even possible? Well remember that when you integrate with "numbers" you get a number. What I mean by this is the following.

    If we integrate [tex] \int_0^1 x dx [/tex] we get a number right? What about when we integrate [tex] \int_0^u x dx [/tex]? Well the second case returns a function dependent on [itex] u [/itex].

    So in the first case, [tex] \int_0^1 x dx [/tex], why not just call this a number, how about [itex] I [/itex]. So this makes sense to be able to multiply two numbers together, eg. [itex] I\timesI = \int_0^1 x dx \times \int_0^1 x dx [/tex]. Think about why we can "push" them together.

    I think the most interesting part about it, was changing to polar coordinates. The part where we change from sweeping out -infinity to infinity in the x and y direction in rectangular coordinates to sweeping out all values by rotating from 0 to 2pi and extending the "arm" from 0 to infinity.
    Last edited: Feb 26, 2008
  8. Feb 26, 2008 #7
    There are other ways to solve the above as well. You will learn these if you take a course in complex variables.
  9. Feb 27, 2008 #8
    You will also need Fubini's theorem in order to justify that you may in this case convert the product of two integrals into a double integral.
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