Integrating exp(-x^2)

Discussion in 'Undergraduate Math' started by Stan Brown, Dec 24, 2010.

  1. Stan Brown

    Stan Brown Guest

    This came up at work yesterday, and neither of us could remember how
    to do it. I pulled out my copy of Thomas and couldn't find it,
    though I did find a statement at the beginning of the "Methods of
    Integration" chapter that it would be dealt with later and would
    involve infinite series.

    Can someone remind me, or point me toward an online reference?
    Thanks!

    P.S. We actually want the definite integral from x=0 to infinity.

    And again, it's not the answer but the method that we're looking for.
     
    Stan Brown, Dec 24, 2010
    #1
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  2. Stan Brown

    Paul Sperry Guest

    <http://www.wolframalpha.com/input/?i=integrate+e^-x^2> gives an answer
    but not much of a method.

    <http://mathforum.org/library/drmath/view/69832.html> gives a
    derivation (which I have not read) and an answer.

    The two answers do not agree - I think (but am not sure) that Wolfram
    is correct.
     
    Paul Sperry, Dec 24, 2010
    #2
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  3. Turn it into a double integral and convert to polar coordinates.
     
    Pubkeybreaker, Dec 24, 2010
    #3
  4. On Fri, 24 Dec 2010 01:47:26 -0500, Stan Brown
    What you want is half of the integral over R; call its value
    I. Then I^2 = int[exp(-x^2) dx] * int[exp(-y^2) dy] =
    int int[exp(-(x^2 + y^2)) dy dx], where all integrals are
    from -inf to inf, so that you're integrating over the whole
    plane. Now convert everything to polar coordinates.

    Brian
     
    Brian M. Scott, Dec 24, 2010
    #4
  5. Stan Brown

    Virgil Guest

    If I = Integral_0^oo e^(-x^2) dx

    Then I^2 = Integral_0^oo Integral_0^oo e^(-x^2-y^2 ) dx dy

    = Integral_0^(pi/2) Integral_0^oo e^(-r^2) r dr dtheta

    Which is fairly simple to evaluate.

    To show that the two infinite double integrals actually converge to the same value is
    also not difficult.

    Consider a quarter circle region centered at origin contained in a minimal square
    contained in a larger but also minimal quarter circle with center at origin.

    Clearly the integral over the square region has a value between the integrals over the
    two quarter circle regions.

    Then show that as the radii of the quarter circle goes towards oo, the integral over the
    region between the quarter circles goes to zero, thus the difference in the 3 integrals
    over the finite regions goes to zero as the regions of integration become infinite.
     
    Virgil, Dec 24, 2010
    #5
  6. Stan Brown

    Stan Brown Guest

    Wow! Thanks, Brian!
     
    Stan Brown, Dec 24, 2010
    #6
  7. Stan Brown

    Stan Brown Guest

    Thanks, Paul, for the links.

    When I read through Dr. Math's method from your link, it seems to
    match the method that Brian Scott posted, Dr. Math integrates from -
    infinity to +infinity and gets sqrt(pi) = 1.772... Wolfram is
    integrating from 0 to +infinity and gets 0.886... which is half of
    sqrt(pi), so I think the two actually do agree.
     
    Stan Brown, Dec 24, 2010
    #7
  8. Stan Brown

    Stan Brown Guest

    Thanks for responding.
     
    Stan Brown, Dec 24, 2010
    #8
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