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I tried to perform a series expansion in the variable $r$ of the following expression (the following expression is a solution of a differential equation) at $r \rightarrow \infty$.

FullSimplify[{Series[
1/r (-1)^-β (-r^2)^-β (r^2)^((1 + β)/
 2) (1 + r^2)^(α/
  2) (C[2] Hypergeometric2F1[
     1/2 (1 - Sqrt[1 + m^2] + α - β), 
     1/2 (1 + Sqrt[1 + m^2] + α - β), 
     1 - β, -r^2] + (-1)^β (-r^2)^β C[
     1] Hypergeometric2F1[
     1/2 (1 - Sqrt[1 + m^2] + α + β), 
     1/2 (1 + Sqrt[1 + m^2] + α + β), 
     1 + β, -r^2]), {r, Infinity, 4}]} /. m^2 -> 3]

However I am getting the following errors

FullSimplify::infd: "Expression Gamma[1/2\ (-1+α-β)+1/2\ (-3-α+β)] simplified to ComplexInfinity."
FullSimplify::infd: "Expression C[2]\ Gamma[1-β]\ Gamma[1/2\ (-1+α-β)+1/2\ (-3-α+β)] simplified to ComplexInfinity."
FullSimplify::infd: "Expression ((-1)^-β\C[2]\Gamma[1-β]\Gamma[1/2(-1+\α-β)+1/2(-3-α+β)])/(Gamma[1/2(-1+α-\β)]\Gamma[1-β+1/2(-3-α+β)]) simplified to \ComplexInfinity. "

and the further output is stopped. So I have two intertwined questions related to this

  1. How do I get an expansion without any errors in terms of ....
  2. .... in terms of powers of $r$; $\ln(r^2)$ as well as some special functions like the Pochhammer symbols[1] and the digamma function[2] for the above expression. Is it possible to expand the series in such a way that I can get such an expansion, something close to the form as written below?

    $\qquad r^{l-1} + ...... + r^{-(l+1)}A(\alpha, \beta, r)[ln(r^2) + B(\alpha, \beta, r)] + ... $

    where dots indicate lower powers of $r$, and $A(\alpha, \beta, r)$ is a function of the Pochhammer symbol, and $B(\alpha, \beta, r)$ is a function of the digamma function.

[1] - https://reference.wolfram.com/language/ref/Pochhammer.html

[2] - https://reference.wolfram.com/language/ref/PolyGamma.html

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To answer the first question, try making the replacement m^2 -> 3 before carrying out the series expansion:

inputExpr = 1/r (-1)^-β (-r^2)^-β (r^2)^((1 + β)/2) (1 + r^2)^(α/2) *
  (C[2] Hypergeometric2F1[1/2 (1-Sqrt[1+m^2]+α- β), 1/2 (1+Sqrt[1+m^2]+α-β), 1-β, -r^2] + 
  (-1)^β (-r^2)^β C[1] Hypergeometric2F1[1/2 (1-Sqrt[1+m^2]+α+β), 1/2 (1+Sqrt[1+m^2]+α+β), 1+β, -r^2]);

FullSimplify[Series[inputExpr /. m^2 -> 3, {r, Infinity, 4}]]

returns a result without errors:

enter image description here


For the second question, it is generally difficult to get a result in exactly the desired form automatically, especially because the correct result may cruicially depend on properties of parameters like α and β.

For example, if r, α and β are real-valued, then it is safe to approximate

$$(1+r^2)^{\alpha/2} \sim (r^2)^{\alpha /2} \Big(1+\frac{\alpha }{2r^2}+\frac{(\frac{\alpha }{2}-1) \alpha }{4 r^4} + \ldots\Big)$$

Try making that replacement and a simpler expansion will result (Mathematica 11.1):

FullSimplify[
  Series[
    PowerExpand[inputExpr /. 
      {m^2 -> 3, (1 + r^2)^(α/2) -> (r^2)^(α/2) Series[1 + 1/r^2, {r, Infinity, 6}]^(α/2)}], 
  {r, Infinity, 4}]
]

enter image description here

Be sure to check the result is correct, for example by numerically plotting the full and approximate forms of inputExpr.

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  • $\begingroup$ thanks for the answer. :) what do you suggest for the second part of the question, i.e, the expansion in terms of the special functions? $\endgroup$ – Bruce Lee Jun 6 '17 at 13:48
  • $\begingroup$ @BruceLee Is it safe to say $\frac{(-1)^{-\beta } \left(r^2+1\right)^{\alpha /2} \left(-r^2\right)^{-\beta } \left(r^2\right)^{\frac{\beta +1}{2}}}{r}$ is equal to $(-1)^{-2 \beta } \left(r^2+1\right)^{\alpha /2} r^{-\beta }$? $\endgroup$ – QuantumDot Jun 6 '17 at 13:54
  • $\begingroup$ $r^2$ in this case is real, so ya it is safe to say that. $\endgroup$ – Bruce Lee Jun 6 '17 at 14:44
  • $\begingroup$ @BruceLee I have updated my answer to address the second part of your question. $\endgroup$ – QuantumDot Jun 6 '17 at 16:14

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