simplifying scattering integral, goldstein derivation 3.4

[JuliusS]

Hi everyone, could anyone give me a hint on Goldstein derivation 3.4? Starting from

\theta = \pi - 2 \int_{r_{m}}^{\infty} \frac{s / r^{2} dr}{\sqrt{1 - V(r)/E - s^{2}/r^{2}}}


they do a change of variables to get

\theta = \pi - 4 s \int_{0}^{1} \frac{\rho d\rho}{\sqrt{r_{m}^{2} (1 - V(r)/E)^{2} - s^{2} (1-\rho^{2})}}


where

1 - V(r_{m})/E - s^{2}/r^{2} = 0


Naturally I want the mystery function \rho(r) . I have gotten to the expression


\theta = \pi - 2 \int_{0}^{1} \frac{s du}{\sqrt{r_{m}^{2}(1 - V(u)/E) - s^{2}u^{2}}}


by making the transformation u = r_{m} / r, but no further. I haven't been able to find this transformed integral in the literature either. Note that this is from the third edition, 6th printing of Goldstein; earlier versions had an error where a square exponent was omitted.

Thanks!

[JuliusS]

Many thanks to Ravinder Abrol for solving this problem. There is a typo in the most current printing of Goldstein, the correct equation should be

\theta = \pi - 4 s \int_{0}^{1} \frac{\rho d\rho}{\sqrt{r_{m}^{2} (1 - V(r)/E) - s^{2} (1-\rho^{2})^{2}}}


with the mystery function


1 - \rho(r)^{2} = r_{m}/r


Interestingly, the online errata for the earlier printings

http://astro.physics.sc.edu/goldstein/4-5To6.html

stated that the (1 - ...) term should be squared, but in the most recent printing they squared the wrong term.

I have confirmed with Mathematica that the transformation does remove the singularity, and that the integral value is equal.