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I wonder whether anyone knows how to do the integral

$$\int\frac{dr}{r\sqrt{r^2 -br\exp(-kr)-a^2}} .$$

It arises when the Rutherford scattering problem is instisted upon being treated classically instead of by treating the incoming $\alpha$-Particle as a wave, as is done in the Born approximation. The wave-mechanical treatment does yield a solution in elementary functions, whence it might seem that this integral is tractable, and that one could perhaps reverse-engineer the answer to this question from that ... but then not necessarily, as that would only yield the value of the integral at certain special end-points, and not the actual content of it as a function of $r$.

Also the $b$ in this is not the same $b$ as in the non-Yukawa-ised form of this this problem & simply brought over from it: rather $b$ will now be given by a lambertw function ... but that's not so very bad atall. Infact, it'll just be

$$\frac{w(kb_0)}{k} ,$$

with $b_0$ being the $b$ in the non-Yukawa-ised form. The closest approach of the α-particle being well within the shielding corresponds to $kb_0$ being a small fraction of 1, & therefore $b$ being not much less than $b_0$ ... $b≈b_0(1-kb_0(1-\frac{3}{2}kb_0(1-\frac{16}{9}kb_0)))$, infact.

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  • $\begingroup$ is that $=$ sign a typo? $\endgroup$ – clathratus Nov 14 '18 at 22:09
  • $\begingroup$ Oh yes! Certainly is! Thankyou. $\endgroup$ – AmbretteOrrisey Nov 14 '18 at 22:10
  • $\begingroup$ Is this related to this question? $\endgroup$ – eyeballfrog Nov 14 '18 at 22:13
  • $\begingroup$ I can see that that one 8 $\endgroup$ – AmbretteOrrisey Nov 14 '18 at 22:14
  • $\begingroup$ is similar, & also arises in a Yukawa potential -type problem. That question is not mine, however; and the integral is somewhat different in detail, having u in the denominator in the argument of the exponential, and no u in the denominator outside the square-root. $\endgroup$ – AmbretteOrrisey Nov 14 '18 at 22:17

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