Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Transforming limits of integration to a bounded region

  1. Mar 26, 2010 #1

    I've been working on a problem which requires the numerical evaluation of an improper integral. I would like to transform the limits of integration on [tex][0,\infty)[/tex] to the bounded region [tex][a,b][/tex] by replacing the variable [tex]\omega[/tex] with another variable. Here is the integral:

    u(t,\tau)=\frac{1}{\pi}\int_{0}^{\infty}\! G(\omega)\, d\omega

    G(\omega)=4\sqrt{\pi}\frac{\omega^{2}}{\omega_{0}^ {3}}\mbox{exp}\left(-\frac{\omega^{2}}{\omega_{0}^{2}}\right)\mbox{cos \left(\omega t-\left(\frac{\omega}{\omega_{0}}\right)^{-\gamma}\omega\tau\right)\mbox{exp}\left(-\frac{1}{2Q}\left(\frac{\omega}{\omega_{0}}\right) ^{-\gamma}\omega t\right)}

    How should I proceed? Is there a relevant reference which could point me in the proper direction?
  2. jcsd
  3. Mar 26, 2010 #2
    Perhaps contour integration could provide a technique to evaluate real definite integrals? Is there a good textbook/monograph available on contour integration?
  4. Mar 27, 2010 #3
    I've found a monograph with some recommended substitutions [1], but I can't immediately see how this would transform the limits of integration.

    Here are three recommended substitutions found in the monograph to transform the limits of integration from [tex][0,\infty)[/tex] to [tex][0,1)[/tex], using the variable [tex]y[/tex]:

    \omega = -\alpha \mbox{log}(1 - y), \alpha > 0

    \omega = \frac{y}{1-y}

    \omega = \left ( \frac{y}{1 - y} \right) ^ 2

    But what is particularly confusing is that as [tex]y \rightarrow 1[/tex], [tex]\omega \rightarrow |\infty|[/tex].

    Is there anything that can be done to transform [tex][0,\infty)[/tex] to [tex][0,1][/tex] or similar?

    [1] A.R. Krommer and C.W. Ueberhuber, Computational Integration, Philadelphia: Society for Industrial and Applied Mathematics, 1998.
  5. Mar 28, 2010 #4
    Is there a way to numerically integrate this integral? I've tried to perform the integration on a truncated interval such as [tex][0,1000][/tex] or [tex][0,10000][/tex] instead of [tex][0,\infty)[/tex] but I've found that the truncated integral cannot adequately approximate the improper integral.

    There's got to be a way to properly do this.
  6. Mar 28, 2010 #5
    Oh, and I've also learned that as [tex]\omega \rightarrow 0[/tex], [tex]G(\omega) \rightarrow \infty[/tex], so the interval over which the integration is performed would have to be [tex](0,\infty)[/tex], or the transformed interval would have to be [tex](0,1)[/tex].
  7. Mar 28, 2010 #6
    So perhaps the best way to proceed would be to use a numerical integration procedure which does not use the endpoints of the interval. One of these methods is the "open" Newton-Coates integration method applied to the transformed interval [tex](0, 1)[/tex].
  8. Mar 29, 2010 #7
    Okay, well - the following method worked for me. An algorithm to perform adaptive quadrature can be found in the paper by Shampine [1]. This method has been implemented in Matlab as "quadgk." Running the quadgk function on the interval [tex][0,\infty)[/tex] worked well for [tex]u(t, \tau)[/tex] above.

    [1] L. Shampine, “Vectorized adaptive quadrature in MATLAB,” Journal of Computational and Applied Mathematics, vol. 211, Feb. 2008, pp. 131-140.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook