# Asymptotic expansion at infinity given a branch cut

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.

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

• Your title is confusing: rho[] does not seem to be an interpolating function. – Michael E2 Feb 25 at 14:16
• I am sorry for the confusion. The inverse of rho[ ] is an interpolating function – user583893 Feb 25 at 14:17
• 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. – user583893 Feb 25 at 14:22
• 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? – Dominic Feb 25 at 15:18

## 1 Answer

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

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

• @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. – user64494 Feb 25 at 16:12
• $b_{0}=1$ 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 $q$? – user583893 Feb 25 at 16:29
• @user583883: The assumption Assumptions->q<1 includes negative values of q. Don't hesitate to ask for further explanation in need. – user64494 Feb 25 at 16:39
• Is there a way to invert your result, what I need is $r(\rho)$ – user583893 Feb 25 at 16:44
• 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 ∞?" – Michael E2 Feb 25 at 17:18