1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gaussian function

  1. Aug 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Screenshot2012-08-19at122009AM.png
    Screenshot2012-08-19at122012AM.png
    2. Relevant equations
    3. The attempt at a solution

    Looking at equations 17 and 18, I don't see how that follows. If you substitute infinity for x you're going to get infinity divided by some real number which is infinity
     
  2. jcsd
  3. Aug 18, 2012 #2

    chiro

    User Avatar
    Science Advisor

    Hey g.lemaitre.

    The first result is not intuitive and it is based on what is calle the error function or erf(x). The function does converge because e^(-x) when x gets really large goes quickly to 0.

    http://en.wikipedia.org/wiki/Error_function

    For the second one, you need to use a substitution of u = x^2. If you have done a year of calculus, this should be straight-forward.
     
  4. Aug 19, 2012 #3

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    [itex]e^{- \infty}=0[/itex]

    There is no need for the error function when evaluating the first integral, just use the fact that [itex]\int_{ - \infty }^{ \infty } f(x)dx = \int_{ - \infty }^{ \infty } f(y)dy [/itex] to calculate the square of the integral, by switching to polar coordinates.
     
  5. Aug 19, 2012 #4

    chiro

    User Avatar
    Science Advisor

    Do you mean ydx and xdy instead of f(x)dx f(y)dy? Sorry to nitpick but changing x to y is change a dummy variable change rather than a variable description change.
     
  6. Aug 19, 2012 #5

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    The whole point is to exploit a change of the "dummy" integration variable, to transform the problem from one of finding a one-dimensional integral, to one of finding a two-dimensional integral, as the latter turns out to be easier:

    [tex] \left( \int_{-\infty}^{\infty} f(x)dx \right)^2 = \left( \int_{-\infty}^{\infty} f(x)dx \right)\left( \int_{-\infty}^{\infty} f(y)dy \right) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x)f(y)dxdy[/tex]

    To see why it's easier, just switch to polar coordinates.
     
  7. Aug 19, 2012 #6

    chiro

    User Avatar
    Science Advisor

    That makes it a lot clearer. Thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Gaussian function
  1. Gaussian Function (Replies: 1)

Loading...