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I have a question about higher-order asymptotics of generalized hypergeometric functions. According to https://dlmf.nist.gov/15.4 the following is well known: $$ _2F_1(a,b;a+b;z)\sim -\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\log(1-z),\ \ z\rightarrow1^{-}. $$ My collaborator was able to coax Wolfram Mathematica into giving a higher-order estimate $$ _2F_1(a,b;a+b;z)\sim -\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\left[\log(1-z)+\psi(a)+\psi(b)+2\gamma\right]+\mathcal{O}((1-z)\log(1-z)) $$ as $z\rightarrow 1^{-}$ for $a$ and $b$ real and positive and where $\psi(z)$ is the digamma function and $\gamma$ is the Euler–Mascheroni constant.

Can anybody provide me with any reference or a hint of why the above Mathematica estimate would be correct?

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    $\begingroup$ See references mentioned here: mathoverflow.net/a/184829/454 $\endgroup$ – Gerald Edgar Nov 7 '20 at 22:06
  • $\begingroup$ @GeraldEdgar the link you gave me was super useful! I am going over Evans and Stanton's paper regardless of the answer I accepted which is sufficient for the result we are trying to push through. $\endgroup$ – Predrag Punosevac Nov 8 '20 at 3:50
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In Abramowitz and Stegun, Formula 15.3.11, the equation reads for $m=0,$ $${}_2F_1(a,b,a+b) = -\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)} \sum_{n=0}^\infty \frac{(a)_n (b)_n}{n!^2}(1-z)^n \Big( $$ $$\log(1-z) -2\psi(n+1) + \psi(a+n) + \psi(b+n) \Big)$$ Your asymptotic approximation is the $n=0$ term.

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  • $\begingroup$ It is actually formula 15.3.10 I needed but that is a cosmetic error. You gave me exactly what I needed. I didn't want to edit your answer. Maybe I have a different edition of the Abramowitz and Stegun. $\endgroup$ – Predrag Punosevac Nov 8 '20 at 3:46

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