Given the hypergeometric function $_2F_1[a,b,c,z]$ in the interval $z\in(1,\infty)$. What is the proper asymptotic expansion of the aforesaid function near $z=1$, when one is approaching from $z>1$.
2 Answers
The Mathematica 11.3 command
Series[Hypergeometric2F1[a, b, a + b, z], {z, 1, 1}, Assumptions > z > 1]
produces $$\frac{\Gamma (a+b) (\psi ^{(0)}(a)+\psi ^{(0)}(b)+\log (z1)+i \pi +2 \gamma )}{\Gamma (a) \Gamma (b)}+\frac{(z1) \Gamma (a+b) (a b \log (z1)2 a b+i \pi a b+2 \gamma a b+a b \psi ^{(0)}(a+1)+a b \psi ^{(0)}(b+1))}{\Gamma (a) \Gamma (b)}+O\left((z1)^2\right) $$ The Maple 2018 command
series(hypergeom([a, b], [a+b], z), z = 1, 2) assuming z>1
performs $${\frac {\Gamma \left( a+b \right) \left( \ln \left( z1 \right) +i \pi+2\,\gamma+\Psi \left( a \right) +\Psi \left( b \right) \right) }{ \Gamma \left( a \right) \Gamma \left( b \right) }}+{\frac {\Gamma \left( a+b \right) ab \left( \ln \left( z1 \right) +i\pi2+2\, \gamma+\Psi \left( a+1 \right) +\Psi \left( b+1 \right) \right) \left( z1 \right) }{\Gamma \left( a \right) \Gamma \left( b \right) }}+O \left( \left( z1 \right) ^{2} \right) . $$ Numerical calculations (on demand) for $a=1,b=1/2,z=1.01$ confirm both the Mathematica's result and the Maple's result.
Mathematica says: $$\left(\frac{\pi \csc ((ab+c) \pi ) \Gamma (c)}{\Gamma (a+bc+1) \Gamma (ca) \Gamma (cb)}\frac{a b \pi \csc ((ab+c) \pi ) \Gamma (c) (z1)}{\Gamma (a+bc+2) \Gamma (ca) \Gamma (cb)}+\frac{a (a+1) b (b+1) \pi \csc ((ab+c) \pi ) \Gamma (c) (z1)^2}{2 \Gamma (a+bc+3) \Gamma (ca) \Gamma (cb)}+O\left((z1)^3\right)\right)+(z1)^{ab+c} e^{2 i \pi (ab+c) \left\lfloor \frac{\arg (z1)}{2 \pi }\right\rfloor } \left(\frac{(1)^{ab+c+1} \pi \csc ((ab+c) \pi ) \Gamma (c)}{\Gamma (a) \Gamma (b) \Gamma (ab+c+1)}+\frac{(1)^{ab+c+1} (ac) (cb) \pi \csc ((ab+c) \pi ) \Gamma (c) (z1)}{\Gamma (a) \Gamma (b) \Gamma (ab+c+2)}\frac{\left((1)^{ab+c} (ac1) (ac) (cb) (b+c+1) \pi \csc ((ab+c) \pi ) \Gamma (c)\right) (z1)^2}{2 (\Gamma (a) \Gamma (b) \Gamma (ab+c+3))}+O\left((z1)^3\right)\right)$$

$\begingroup$ I know that result But I think is wrong. Thanks anyway. $\endgroup$ Commented Feb 20, 2019 at 0:50

1$\begingroup$ @user136032 If you know it, you should say so in the question. If you think it is wrong, you should explain why. $\endgroup$ Commented Feb 20, 2019 at 0:52