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Basically, I have obtained the function $\rho (r)$ below as a result of integrating

$$\rho(r)=\int_{b_0}^{r}\frac{dx}{\sqrt{1-(b_{0}/x)^{1-q}}}$$ which results to

 b0 = 1;
 rho[r_] := (2 b0)/(1 - q) Sqrt[1 - (b0/r)^(1 - q)] Hypergeometric2F1[1/2, (q - 2)/(q - 1), 3/2, 1 - (b0/r)^(1 - q)];

where $b$ is just some positive constant while $-\infty<q<1$.

What I need is $r(\rho)$ which is the inverse of the function above (which I can implement with numerics). I am also interested in the series expansion of $r(\rho)$ at $\infty$ but unknowingly, the hypergeometric function above has a branch cut at $1-\left(\frac br\right)^{1-q}=1$ (i.e. $r\rightarrow \infty$). My question would be, am I allowed to expand $r(\rho)$ at $\infty$ given that there is a branch cut at infinity? What are the possible complications? What can I possibly do to safely obtain an asymptotic expansion of $r(\rho)$ at $\infty$? I am also thinking of converting the integral above into a differential equation for $r(\rho)$ and expand it at infinity to bypass the hypergeometric function branch cut problem but I am not really sure of this. Thanks for the help

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  • $\begingroup$ Your title is confusing: rho[] does not seem to be an interpolating function. $\endgroup$ – Michael E2 Feb 25 at 14:16
  • $\begingroup$ I am sorry for the confusion. The inverse of rho[ ] is an interpolating function $\endgroup$ – user583893 Feb 25 at 14:17
  • $\begingroup$ Given the rho[ ], I can numerically obtain its inverse. What I want to find is the asymptotic expansion of the inverse at infinity but I am not sure whether I am allowed to do that since there is a branch cut in the hypergeometric function. $\endgroup$ – user583893 Feb 25 at 14:22
  • $\begingroup$ Might you just try letting $b_0=1$ and $q=1/2$ and try to get it to work for that one particular example first? $\endgroup$ – Dominic Feb 25 at 15:18
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The Mathematica 12.2 code

b0 = 1; AsymptoticIntegrate[1/Sqrt[1-(b0/x)^(1 - q)],{x, b0, r},{r,Infinity,1}, Assumptions->q<1]

performs

$$\frac{r^q}{2 q}+\frac{3 (q-1) r^{2 q-1}}{8 q (2 q-1)}+\frac{3 r^{2 q-1}}{8 q (2 q-1)}-\frac{\sqrt{\pi } \Gamma \left(\frac{q}{q-1}\right)}{\Gamma \left(\frac{1}{2}+\frac{1}{q-1}\right)}+r $$

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  • $\begingroup$ @user583883: The case of an arbitray positive $b_0$ is reduced to the case $b_0=1$ by the change $x=b_0 t$. I leave the rest on your own. Don't hesitate to ask for further explanation in need. $\endgroup$ – user64494 Feb 25 at 16:12
  • $\begingroup$ $b_{0}=1$ is fine. I guess my problem now is to find a way to invert what you got. Can you also extend your assumption to include negative values of $q$? $\endgroup$ – user583893 Feb 25 at 16:29
  • $\begingroup$ @user583883: The assumption Assumptions->q<1 includes negative values of q. Don't hesitate to ask for further explanation in need. $\endgroup$ – user64494 Feb 25 at 16:39
  • $\begingroup$ Is there a way to invert your result, what I need is $r(\rho)$ $\endgroup$ – user583893 Feb 25 at 16:44
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    $\begingroup$ Actually, it seems to be the questtion @user583893 asks above: "My question would be, am I allowed to expand 𝑟(𝜌) at ∞...?... What can I possibly do to safely obtain an asymptotic expansion of 𝑟(𝜌) at ∞?" $\endgroup$ – Michael E2 Feb 25 at 17:18

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