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Definite integral: exponential with squared exponent

  1. Jan 28, 2009 #1

    I'm trying to solve the following:
    [tex]f(x) = \int^\infty_{-\infty}ce^{yx-y^2/2} dy[/tex]
    where c is a constant
    My only idea thus far was that since it is an even function, the expression can be simplified to:
    [tex]= 2c\int^\infty_0 e^{y(x-{1/2}y)} dy[/tex]

    but I'm stuck here.

    Anyone know how to do this?
  2. jcsd
  3. Jan 28, 2009 #2


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    This is a Gaussian integral (which is ubiquitous in theoretical physics, so if you have any aspirations in that direction pay extra attention :smile:)

    The trick is to complete the square: write*
    [tex]y^2 - 2 y x = (y - a)^2 - b[/tex]
    where a and b are independent of y, so you get
    [tex]f(x) = c e^{-b/2} \int_{-\infty}^{\infty} e^{- (y - a)^2 / 2} \, dy.[/tex]
    Then you can do a variable shift and use the standard result
    [tex]I = \int_{-\infty}^\infty e^{-x^2 / 2} = \sqrt{2 \pi}[/tex]
    which you can easily prove (if you've never done it, try it: consider [itex]I^2[/itex] and switch to polar coordinates).

    * From this line onwards, I take no responsibility for sign errors and wrong factors of 1/2 - please check yourself :)
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