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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.

AmbretteOrrisey
  • 981
  • 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
  • @AmbretteOrrisey The integral in that question is actually the $a$-derivative of your integral now that I look at it (with suitable parameter redefinition), assuming the bounds of your integral are $(r^,\infty)$, where $r^$ is the turning point. – eyeballfrog Nov 14 '18 at 22:22
  • ¶ The question of mine that this arises from is the normalisation of Rutherford differential cross-section (not verbatim name) problem that I posted a few days ago. ¶ One of the keys on my virtual keypad acts as a post button: I wish it wouldn't, as I keep accidentally sending fragments of comments! – AmbretteOrrisey Nov 14 '18 at 22:22
  • @That's the one! the relation a (impact parameter) =(b/2)cot(φ/2) is obtained by substituing r=∞ & r=b into the non -'Yukawa-ised' form of this, which is tractable. – AmbretteOrrisey Nov 14 '18 at 22:30
  • I made another typo - a much more important one ... I've fixed it now! – AmbretteOrrisey Nov 14 '18 at 22:37
  • @AmbretteOrrisey OK now it is equivalent to the one inside the cosine in the other question (substitute r->a/u). Admittedly not super helpful since neither seems to have an elementary antiderivative. – eyeballfrog Nov 14 '18 at 22:39
  • Is it!? Right so the matter of the variable being reciprocated in the argument of the ewponential to the other is not a big obstacle, and they teanspire to be the same! Thankyou for that observation. That's good then, because it mezns – AmbretteOrrisey Nov 14 '18 at 22:42
  • or tends_to mean that this integral has some degree of _ubiquiosity in Yukawa-pitential-type setups, & therefore ought at least to be somewhat studied. You might even define __a new function___ as being the solution of it. That's effectively all that's meant, often, when you say that an integral _is tractable. It just means you've accepted the integral of it amongst the canon of functions. That's 'kind of' what it is with elliptic functions ... but admittedly they have so very vast scope of interconnections with other functions, we cannot but, really, call them bona fide func – AmbretteOrrisey Nov 14 '18 at 22:49

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