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How one should do the series expansion of Beta function $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ around an arbitrary negative integer let's say $x=-k$ and $y=-l$? I want a symbolic expression of the coefficients of the expansion in terms of $k$ and $l$.

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  • $\begingroup$ may be relevant: math.stackexchange.com/questions/3195847/… $\endgroup$
    – ydd
    Commented Aug 8, 2023 at 22:20
  • $\begingroup$ @ydd That is relevant. But I want to avoid knowing all those details. I thought one should be able to directly expand the function around arbitrary points in mathematica. $\endgroup$
    – Hkw
    Commented Aug 9, 2023 at 4:18

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You can try directly expanding the function. For example, the first three terms:

Series[Beta[w, v], {w, -k, 3}, {v, -l, 3}]

That produces a large expression about -k and -l in terms of Gamma and PolyGamma functions.

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  • $\begingroup$ Plugging in integer values produces Indeterminate coefficient values though: Series[Beta[w, v], {w, -k, 1}, {v, -l, 1}] /. {l -> 1, k -> 1} $\left(\text{Indeterminate}+\text{Indeterminate} (v+1)+O\left((v+1)^2\right)\right)+(w+1) \left(\text{Indeterminate}+\text{Indeterminate} (v+1)+O\left((v+1)^2\right)\right)+O\left((w+1)^2\right)$ $\endgroup$
    – ydd
    Commented Aug 8, 2023 at 22:16
  • $\begingroup$ @Lexington1776 Doing the series expansion doesn't really work. Even the leading term is not being reproduced by the way you are suggesting. $\endgroup$
    – Hkw
    Commented Aug 9, 2023 at 4:12
  • $\begingroup$ @ydd The beta function is singular at those negative integers. Try doing the expansion directly around -1. Not using the l->1. You will get some coefficients* 1/x+k.. $\endgroup$
    – Hkw
    Commented Aug 9, 2023 at 4:15

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