# Higher-order asymptotics of generalized hypergeometric function

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?

• See references mentioned here: mathoverflow.net/a/184829/454 – Gerald Edgar Nov 7 '20 at 22:06
• @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. – Predrag Punosevac Nov 8 '20 at 3:50

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.

• 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. – Predrag Punosevac Nov 8 '20 at 3:46