I'm attempting to solve the indefinite integral

$$S\left(v\right) = 2a\sqrt{\alpha E}\int\!v^{-1/2}\left(\frac{Q^2}{2E}-v^2\right)^{1/2}\left(v^3+\alpha^3\right)^{-1/2}\left(v^3+2\alpha^3\right)^{-1/6}\,\mathrm dv$$

Because this looks like it could be a Hypergeometric integral I have defined $x\equiv\frac{2E}{Q^2}v^2$, $z\equiv -\frac{Q^2}{4E\alpha^3}v$, $a = \frac 16$, $b = \frac 23$, and $c = \frac 94$ so that it can be rewritten in the form

$$S\left(x\right) \sim \int\!\left(\frac{x}{1+\beta x^{3/2}}\right)^{1/2}x^{b-1}\left(1-x\right)^{c-b-1}\left(1-zx\right)^{-a}\,\mathrm dx$$

Clearly the last three terms in this integral look like a hypergeometric integral, but I have a leading term which messes things up.

Note the integral form of the Hypergeometric function I find on Wikipedia is defined as

$$B(b,c-b)\,{}_2F_1(a,b;c;z) = \int_0^1 x^{b-1} (1-x)^{c-b-1}(1-zx)^{-a} \, dx \qquad Re(c) > Re(b) > 0$$

Also, my integral is not bounded. Does anyone have any suggestions for how to solve this (or even relevant transformations which could be useful)?

  • $\begingroup$ You can split the fraction up and then you have the standard integrand divided by a fractional pole. I am sure the standard contour integration can be used. Unfortunately I am not good enough yet to post; I am to likely to make a mistake. $\endgroup$ – rrogers Apr 28 '17 at 18:47
  • $\begingroup$ Contour integration can only be used for integrals with bounds. $\endgroup$ – William J. Cunningham Apr 28 '17 at 19:07
  • $\begingroup$ It did occur to me that you can take the denominator and expand it in an infinite series of x^#@% with the combinatorial negative binomial coefficients. This leads to an infinite series of hypergeometric terms. $\endgroup$ – rrogers Apr 28 '17 at 21:19
  • $\begingroup$ If the upper limit is inf then I don't see any better way than Mellin transforms; which look to be a bit messy but they directly deal with this your type of expression; multiplicative convolution over 0 inf. BTW: the intermediate terms do yield 2F1 as stepping stones. I have the Batemen book on transforms. It's available at: authors.library.caltech.edu/43489/1/Volume%201.pdf You have to go to Chapter VI . I do have a little knowledge about them and in the forward direction there is no real problem. Inverting the result might be iffy. I can't do it today though. $\endgroup$ – rrogers Apr 28 '17 at 21:21
  • $\begingroup$ The direct method seemed to run into problems, but if you have a good Mathematica/Maple system try doing the Mellin transform against $\alpha$ in your original equation. That should yield $I=\int v^{q+p\cdot s}\left(\frac{Q}{2E}-v^{2}\right)^{\frac{1}{2}}dv$ where s is the Mellin transform variable against $\alpha$ . I think that the integral is fairly easy (Maxima gave me an answer) but then the Mellin transform has to be reversed s->$alpha$ . Which looks doable but I am not sure. With Maple/Mathmatica these steps should be doable (and then checked). $\endgroup$ – rrogers Apr 29 '17 at 17:48

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