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Homework Help: How to solve the following integral? Does it even have a solution?

  1. Apr 22, 2010 #1
    1. The problem statement, all variables and given/known data


    Integrals of this type:

    [tex]
    \int \frac{1}{\sqrt{2E-2(\frac{1}{r}-\frac{1}{2})+e^{-r^2-z^2}}} dz
    [/tex]

    does anyone know where I can find it?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 22, 2010 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    Re: Integrals

    I don't think that there is an exact solution.

    For notational convenience, let me write
    [tex]
    \int \frac{1}{\sqrt{2E-2(\frac{1}{r}-\frac{1}{2})+e^{-r^2-z^2}}} dz
    =
    \int \frac{1}{\sqrt{c + e^{-r^2} e^{-z^2}}} \, dz
    [/tex]

    If you plot the graph, you will see that it gets close to its asymptote y = 1/sqrt(c) rather quickly. So you could approximate it by [itex]1/\sqrt{c}[/itex], and expand around z = 0 in a series of which you only include the first several terms (they are polynomials in z with complicated coefficients, depending just on c and r though). I.e. something like

    [tex]\int_{-a}^a \frac{1}{\sqrt{c + e^{-r^2} e^{-z^2}}} \, dz
    =
    \int_{-a}^{-\delta} \frac{1}{\sqrt{c + e^{-r^2} e^{-z^2}}} \, dz
    +
    \int_{-\delta}^{\delta} \frac{1}{\sqrt{c + e^{-r^2} e^{-z^2}}} \, dz
    +
    \int_{\delta}^a \frac{1}{\sqrt{c + e^{-r^2} e^{-z^2}}} \, dz
    [/tex]
    where you choose [itex]\delta \approx 2[/itex] (for sufficiently large a >> delta) conveniently so that the integrand is as good as constant outside the interval [itex][-\delta, \delta][/itex].
    Then
    [tex]\int_{-a}^a \frac{1}{\sqrt{c + e^{-r^2} e^{-z^2}}} \, dz
    \approx
    \frac{1}{\sqrt{c}} \cdot (a - \delta)
    +
    \int_{-\delta}^{\delta} \left( c_0 + c_2 z^2 + c_4 z^4 + \cdots \right) \, dz
    +
    \frac{1}{\sqrt{c}} \cdot (a - \delta)
    [/tex]
    where the c2i are the Taylor series coefficients, e.g.
    [tex]c_0 = \frac{1}{\sqrt{c + e^{-r^2}}}[/tex];
    [tex]c_2 = \frac{e^{-r^2}}{2 \left(c+e^{-r^2}\right)^{3/2}}[/tex];
    [tex]c_4 = \frac{1-2 c e^{r^2}}{8 \sqrt{c+e^{-r^2}} \left(c e^{r^2}+1\right)^2}[/tex]
    etc. (you can get it as accurate as you want by including more terms).

    Note that for [itex]a \to \infty[/itex] the integral will diverge, as the leading contribution is something like
    [tex]\frac{2a}{\sqrt{c}}[/tex]
     
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