# Asymptotic expansion of hypergeometric function near $z=1$

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 $$z>1$$.

• I forgot to mention that c=b+a. – user136032 Feb 20 '19 at 0:49

The Mathematica 11.3 command

Series[Hypergeometric2F1[a, b, a + b, z], {z, 1, 1}, Assumptions -> z > 1]


produces $$-\frac{\Gamma (a+b) (\psi ^{(0)}(a)+\psi ^{(0)}(b)+\log (z-1)+i \pi +2 \gamma )}{\Gamma (a) \Gamma (b)}+\frac{(z-1) \Gamma (a+b) (a b \log (z-1)-2 a b+i \pi a b+2 \gamma a b+a b \psi ^{(0)}(a+1)+a b \psi ^{(0)}(b+1))}{\Gamma (a) \Gamma (b)}+O\left((z-1)^2\right)$$ The Maple 2018 command

series(hypergeom([a, b], [a+b], z), z = 1, 2) assuming z>1


performs $$-{\frac {\Gamma \left( a+b \right) \left( \ln \left( z-1 \right) +i \pi+2\,\gamma+\Psi \left( a \right) +\Psi \left( b \right) \right) }{ \Gamma \left( a \right) \Gamma \left( b \right) }}+{\frac {\Gamma \left( a+b \right) ab \left( \ln \left( z-1 \right) +i\pi-2+2\, \gamma+\Psi \left( a+1 \right) +\Psi \left( b+1 \right) \right) \left( z-1 \right) }{\Gamma \left( a \right) \Gamma \left( b \right) }}+O \left( \left( z-1 \right) ^{2} \right) .$$ 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: $$\left(\frac{\pi \csc ((-a-b+c) \pi ) \Gamma (c)}{\Gamma (a+b-c+1) \Gamma (c-a) \Gamma (c-b)}-\frac{a b \pi \csc ((-a-b+c) \pi ) \Gamma (c) (z-1)}{\Gamma (a+b-c+2) \Gamma (c-a) \Gamma (c-b)}+\frac{a (a+1) b (b+1) \pi \csc ((-a-b+c) \pi ) \Gamma (c) (z-1)^2}{2 \Gamma (a+b-c+3) \Gamma (c-a) \Gamma (c-b)}+O\left((z-1)^3\right)\right)+(z-1)^{-a-b+c} e^{2 i \pi (-a-b+c) \left\lfloor -\frac{\arg (z-1)}{2 \pi }\right\rfloor } \left(\frac{(-1)^{-a-b+c+1} \pi \csc ((-a-b+c) \pi ) \Gamma (c)}{\Gamma (a) \Gamma (b) \Gamma (-a-b+c+1)}+\frac{(-1)^{-a-b+c+1} (a-c) (c-b) \pi \csc ((-a-b+c) \pi ) \Gamma (c) (z-1)}{\Gamma (a) \Gamma (b) \Gamma (-a-b+c+2)}-\frac{\left((-1)^{-a-b+c} (a-c-1) (a-c) (c-b) (-b+c+1) \pi \csc ((-a-b+c) \pi ) \Gamma (c)\right) (z-1)^2}{2 (\Gamma (a) \Gamma (b) \Gamma (-a-b+c+3))}+O\left((z-1)^3\right)\right)$$

• I know that result But I think is wrong. Thanks anyway. – user136032 Feb 20 '19 at 0:50
• @user136032 If you know it, you should say so in the question. If you think it is wrong, you should explain why. – Igor Rivin Feb 20 '19 at 0:52