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  1. Dec 14, 2007 #1
    Some time ago I had to do a calculation concerning a physical problem. In this calculation some integrals were needed to be solved and the method on how to do this was briefly described. I was able to solve the problem completely and found the method so beautiful that I would like to share it with all of you. Consider the following integrals:

    [tex]\int \frac{1}{1+\epsilon \cdot cos \theta}d\theta[/tex]

    [tex]\int \frac{1}{\left(1+\epsilon \cdot cos \theta\right)^2}d\theta[/tex]

    [tex]\int \frac{1}{\left(1+\epsilon \cdot cos \theta\right)^3}d\theta[/tex]

    [tex]\int \frac{cos \theta}{1+\epsilon \cdot cos \theta}d\theta[/tex]

    [tex]\int \frac{cos \theta}{\left(1+\epsilon \cdot cos \theta\right)^2}d\theta[/tex]

    [tex]\int \frac{cos \theta}{\left(1+\epsilon \cdot cos \theta\right)^3}d\theta[/tex]

    [tex]\int \frac{sin \theta}{1+\epsilon \cdot cos \theta}d\theta[/tex]

    [tex]\int \frac{sin \theta}{\left(1+\epsilon \cdot cos \theta\right)^2}d\theta[/tex]

    [tex]\int \frac{sin \theta}{\left(1+\epsilon \cdot cos \theta\right)^3}d\theta[/tex]

    Some of these are straightforward, but some are not. In order to solve the non trivial ones in a systematic way, one can use the following substitution:

    [tex]1+\epsilon \cdot cos \theta = \frac{1-\epsilon^2}{1-\epsilon \cdot cos \gamma}[/tex]


    [tex]0\leq \theta \leq 2\pi[/tex]

    [tex]0\leq \gamma \leq 2\pi[/tex]

    The following relations can be obtained:

    [tex]cos \theta = \frac{cos \gamma -\epsilon}{1-\epsilon \cdot cos \gamma}[/tex]

    [tex]sin \theta = \frac{\sqrt{1-\epsilon^2} sin \gamma}{1-\epsilon \cdot cos \gamma}[/tex]

    [tex]cos \gamma = \frac{\epsilon +cos \theta}{1+\epsilon \cdot cos \theta}[/tex]

    [tex]sin \gamma = \frac{\sqrt{1-\epsilon^2} sin \theta}{1+\epsilon \cdot cos \theta}[/tex]

    [tex]d \theta = \frac{\sqrt{1-\epsilon^2}}{1-\epsilon \cdot cos \gamma}d \gamma[/tex]

    [tex]d \gamma = \frac{\sqrt{1-\epsilon^2}}{1+\epsilon \cdot cos \theta}d \theta[/tex]

    After rewriting some of the integrals into smaller ones, substituting this and rearranging it is possible to solve them in a fairly easy way. As an example, let's take the one before the last, it was:

    [tex]\int \frac{sin \theta}{\left(1+\epsilon \cdot cos \theta\right)^2}d\theta[/tex]

    Using the substitution we get:

    [tex]\int \frac{sin \gamma}{1-\epsilon^2}d\gamma=\frac{-cos \gamma}{1-\epsilon^2}+C[/tex]

    The solution is now obtained by substituting either

    [tex]\gamma = arccos \left( \frac{\epsilon +cos \theta}{1+\epsilon \cdot cos \theta} \right)[/tex]


    [tex]\gamma = arcsin \left( \frac{\sqrt{1-\epsilon^2} sin \theta}{1+\epsilon \cdot cos \theta} \right)[/tex]

    This substitution is called the "Sommerfeld substitution" after the "inventor" and is one of the nicest substitutions I've ever encountered for solving integrals in the real with "classical" functions. I hope this is helpful in solving other types of integrals you might be working on.

    best regards,

  2. jcsd
  3. Feb 10, 2008 #2
    your information was very useful thanks
  4. Mar 14, 2008 #3
    very nice
    and very useful
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