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

[Nusc]

Homework Statement

Integrals of this type:

<br /> \int \frac{1}{\sqrt{2E-2(\frac{1}{r}-\frac{1}{2})+e^{-r^2-z^2}}} dz<br />

does anyone know where I can find it?

Homework Equations

The Attempt at a Solution

 

[CompuChip]

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

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

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 1/\sqrt{c}, 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

\int_{-a}^a \frac{1}{\sqrt{c + e^{-r^2} e^{-z^2}}} \, dz<br /> = <br /> \int_{-a}^{-\delta} \frac{1}{\sqrt{c + e^{-r^2} e^{-z^2}}} \, dz<br /> + <br /> \int_{-\delta}^{\delta} \frac{1}{\sqrt{c + e^{-r^2} e^{-z^2}}} \, dz<br /> +<br /> \int_{\delta}^a \frac{1}{\sqrt{c + e^{-r^2} e^{-z^2}}} \, dz<br />
where you choose \delta \approx 2 (for sufficiently large a >> delta) conveniently so that the integrand is as good as constant outside the interval [-\delta, \delta].
Then
\int_{-a}^a \frac{1}{\sqrt{c + e^{-r^2} e^{-z^2}}} \, dz<br /> \approx <br /> \frac{1}{\sqrt{c}} \cdot (a - \delta)<br /> + <br /> \int_{-\delta}^{\delta} \left( c_0 + c_2 z^2 + c_4 z^4 + \cdots \right) \, dz <br /> +<br /> \frac{1}{\sqrt{c}} \cdot (a - \delta)<br />
where the c_2i are the Taylor series coefficients, e.g.
c_0 = \frac{1}{\sqrt{c + e^{-r^2}}};
c_2 = \frac{e^{-r^2}}{2 \left(c+e^{-r^2}\right)^{3/2}};
c_4 = \frac{1-2 c e^{r^2}}{8 \sqrt{c+e^{-r^2}} \left(c e^{r^2}+1\right)^2}
etc. (you can get it as accurate as you want by including more terms).

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