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

## Homework Statement

Integrals of this type:

$$\int \frac{1}{\sqrt{2E-2(\frac{1}{r}-\frac{1}{2})+e^{-r^2-z^2}}} dz$$

does anyone know where I can find it?

## The Attempt at a Solution

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CompuChip
Homework Helper

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

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

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

$$\int_{-a}^a \frac{1}{\sqrt{c + e^{-r^2} e^{-z^2}}} \, dz = \int_{-a}^{-\delta} \frac{1}{\sqrt{c + e^{-r^2} e^{-z^2}}} \, dz + \int_{-\delta}^{\delta} \frac{1}{\sqrt{c + e^{-r^2} e^{-z^2}}} \, dz + \int_{\delta}^a \frac{1}{\sqrt{c + e^{-r^2} e^{-z^2}}} \, dz$$
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
$$\int_{-a}^a \frac{1}{\sqrt{c + e^{-r^2} e^{-z^2}}} \, dz \approx \frac{1}{\sqrt{c}} \cdot (a - \delta) + \int_{-\delta}^{\delta} \left( c_0 + c_2 z^2 + c_4 z^4 + \cdots \right) \, dz + \frac{1}{\sqrt{c}} \cdot (a - \delta)$$
where the c2i 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
$$\frac{2a}{\sqrt{c}}$$