# An Awkward Integral Arising in Scattering of $\alpha$-Particle by Nucleus

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.

• is that $=$ sign a typo? – clathratus Nov 14 '18 at 22:09
• Oh yes! Certainly is! Thankyou. – AmbretteOrrisey Nov 14 '18 at 22:10
• Is this related to this question? – eyeballfrog Nov 14 '18 at 22:13
• I can see that that one 8 – AmbretteOrrisey Nov 14 '18 at 22:14
• 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. – AmbretteOrrisey Nov 14 '18 at 22:17