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A Beautiful Looking Gaussian Integral - Please HELP!

  1. Jul 21, 2008 #1
    I have the following Gaussian Integral:

    [tex] \int_{0}^{\infty}2\pi r\left |{\int_{-l/2}^{l/2}\frac{e^\frac{-r^2}{bH}}{(1+ix)(k^{''} - ik^{'}x)H}}dx\right |^2dr [/tex]


    [tex]H = \frac{1+x^2}{k^{''} - ik^{'}x} - i\frac{x - \zeta}{k^{'}}[/tex]

    Assume any characters not defined are constants.

    I agree it does look fantastic, and I'm currently trying to numerically solve it (duh..). I'm using MATLAB and I run into the problem:
    "Warning: System is inconsistent. Solution does not exist."
    Yet I know a solution exists because I'm looking at an article that solved it numerically.

    Any Hints/Advice?
  2. jcsd
  3. Jul 21, 2008 #2
    whats the article? it doesn't say how they solved it? what notation are you taking for granted as known? for example is the zeta the riemann zeta fn ( i know not likely since it doesn't have arguments ) but yea if you tell em for what values of the constants you're trying to numerically integrate i'll put it into mathematica and tell you what i get.
  4. Jul 21, 2008 #3
    Hey ice109,
    I've attached the article. You'll see the equation on page 1 and 2.
    No it doesn't say how they solved it, they just quote the results.
    For the notation:
    zeta, k'', k' and everything else are just real numbers (constants). Well, for my current purpose they are.

    I'm trying to (in a way) reproduce the results in this paper with my own parameters.
    Ok, I havn't got all the parameters with me now, but tomorrow I'll grab them and post it here.
  5. Jul 21, 2008 #4
    Here are my numbers:

    l = 7e-3
    b = 14.516e-6
    k'' = 11.466e6
    k' = 4.387e6
    zeta is actually given by:
    where z is a vector that varies from 0 to 10e-3, f is 2.51e-3. But you can try solving it for one particular value of z.

    I am curious as too how you will solve this since I need to vary several parameters and obtain different results.
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