In " Special functions: an introduction to classical functions of mathematical physics" by Nico M. Temme, at page 113 is reported this formula:
$$_2F_1(a,b;c;z)=\frac{\Gamma(c)\Gamma(b-a)}{\Gamma(b)\Gamma(c-a)}\,(-z)^{-a}\,_2F_1(a,1-c+a\ ;\,1-b+a\ ;\frac{1}{z})+ \\+\,\frac{\Gamma(c)\Gamma(a-b)}{\Gamma(a)\Gamma(c-b)}\,(-z)^{-b}\,_2F_1(b,1-c+b\ ;1-a+b\ ;\,\frac{1}{z})$$
Which, from what I understand, is the extension of the solution of hypergeometric differential equation for $\vert{z}\vert>1$, and therefore the convergent series representation of the $_2F_1(a,b;c;z)$ for $\vert{z}\vert>1$.
How is to possible to demonstrate this formula, for $z\in\mathbb{R}$, starting from this integral formulation of Gauss Hypergeoemtric series?
$$_2F_1(a,b;c;z)=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_0^1t^{b-1}\,(1-t)^{c-b-1}\,(1-tz)^{-a}dt $$
I have searched in various texts but unfortunately no one explicitly reports the demonstration. In particular "Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables" by M. Abramowitz and I. Stegun at page 559,suggest to get this result by Mellin-Barnes integral,but i would like to know if there is a (perhaps simpler) proof that does not include the calculation of the residuals.
Best Answer
Let's actualy assume $z\in\mathbb{C}$, ${\rm Im} z >0$, as that will allow ust to use a simple contour without worrying about the singularity at $z^{-1}$.
By choosing a countour made of the real line (avoiding singularities at $0$ an $1$ from above and closing it with the upper semicircle, it can be shown that (for $a,b,c$ from aprropriate range) $$ 0 = \Big(\int_{-\infty}^0 + \int_0^1 + \int_1^\infty \Big) t^{b-1}(1-t)^{c-b-1}(1-tz)^{-a} dt$$
We have \begin{align} & \int_{-\infty}^0 t^{b-1}(1-t)^{c-b-1}(1-tz)^{-a} dt = \\ & \qquad = - e^{i\pi b} \int_0^{\infty} s^{b-1}(1+s)^{c-b-1}(1+sz)^{-a} ds =^{s=\frac{u}{1-u}} \\ & \qquad= - e^{i\pi b} \int_0^1 u^{b-1}(1-u)^{a-c} (1-u(1-z))^{-a}du = \\ & \qquad= - e^{i\pi b} \frac{\Gamma(b)\Gamma(1+a-c)}{\Gamma(1+a+b-c)}\, _2F_1(a,b;1+a+b-c;1-z) \\ & \int_0^1 t^{b-1}(1-t)^{c-b-1}(1-tz)^{-a} dt = \\ &\qquad = \frac{\Gamma(b)\Gamma(c-b)}{\Gamma(c)}\, _2F_1(a,b;c;z) \\ & \int_1^\infty t^{b-1}(1-t)^{c-b-1}(1-tz)^{-a} dt = \\ & \qquad = -e^{i\pi(b-c)}\int_1^\infty t^{b-1}(t-1)^{c-b-1}(1-tz)^{-a} dt = ^{t=\frac{1}{u}} \\ &\qquad = -e^{i\pi(b-c)} \int_0^1 u^{a-c} (1-u)^{c-b-1}(u-z)^{-a}du = \\ &\qquad = -e^{i\pi(b-c)} (-z)^{-a} \int_0^1 u^{a-c} (1-u)^{c-b-1}(1-uz^{-1})^{-a}du = \\ &= -e^{i\pi(b-c)}\frac{\Gamma(1+a-c)\Gamma(c-b)}{\Gamma(1+a-b)} (-z)^{-a}\,_2F_1(a,1+a-c;1+a-b;z^{-1}) \end{align} In total we have $$ 0 = - e^{i\pi b} \frac{\Gamma(b)\Gamma(1+a-c)}{\Gamma(1+a+b-c)}\, _2F_1(a,b;1+a+b-c;1-z) + \frac{\Gamma(b)\Gamma(c-b)}{\Gamma(c)}\, _2F_1(a,b;c;z) -e^{i\pi(b-c)}\frac{\Gamma(1+a-c)\Gamma(c-b)}{\Gamma(1+a-b)} (-z)^{-a}\,_2F_1(a,1+a-c;1+a-b;z^{-1}) $$ By exchanging $a\leftrightarrow b$ and using a symmetry of a hypergeometric function, we get another identity $$ 0 = - e^{i\pi a} \frac{\Gamma(a)\Gamma(1+b-c)}{\Gamma(1+a+b-c)}\, _2F_1(a,b;1+a+b-c;1-z) + \frac{\Gamma(a)\Gamma(c-a)}{\Gamma(c)}\, _2F_1(a,b;c;z) -e^{i\pi(a-c)}\frac{\Gamma(1+b-c)\Gamma(c-a)}{\Gamma(1-a+b)} (-z)^{-b}\,_2F_1(b,1+b-c;1-a+b;z^{-1}) $$ By eliminating $_2F_1(a,b;1+a+b-c;1-z)$ from these equations, we get \begin{align}& \Big(e^{i\pi (c-b)}\frac{\Gamma(c-b)}{\Gamma(c)\Gamma(1+a-c)} - e^{i\pi (c-a)}\frac{\Gamma(c-a)}{\Gamma(c)\Gamma(1+b-c)}\Big) \,_2F_1(a,b;c;z) = \\ &\qquad = \frac{\Gamma(c-b)}{\Gamma(b)\Gamma(1+a-b)} (-z)^{-a}\,_2F_1(a,1+a-c;1+a-b;z^{-1}) + \\ &\qquad\quad - \frac{\Gamma(c-a)}{\Gamma(a)\Gamma(1-a+b)} (-z)^{-b}\,_2F_1(b,1+b-c;1-a+b;z^{-1}) \end{align} Now, by using the identity $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$ it can be shown that $$ e^{i\pi (c-b)}\frac{\Gamma(c-b)}{\Gamma(c)\Gamma(1+a-c)} - e^{i\pi (c-a)}\frac{\Gamma(c-a)}{\Gamma(c)\Gamma(1+b-c)} = \frac{\sin(\pi(b-a))}{\pi}\frac{\Gamma(c-a)\Gamma(c-b)}{ \Gamma(c)} $$ $$ \frac{\Gamma(c-b)}{\Gamma(b)\Gamma(1+a-b)} = \frac{\sin(\pi(b-a))}{\pi} \frac{\Gamma(c-b)\Gamma(b-a)}{\Gamma(b)} $$ $$ \frac{\Gamma(c-a)}{\Gamma(a)\Gamma(1-a+b)} = -\frac{\sin(\pi(b-a))}{\pi} \frac{\Gamma(c-a)\Gamma(a-b)}{\Gamma(a)} $$ So after dividing the previous identity by a constant we get \begin{align}& \,_2F_1(a,b;c;z) = \\ &\qquad = \frac{\Gamma(c)\Gamma(b-a)}{\Gamma(b)\Gamma(c-a)} (-z)^{-a}\,_2F_1(a,1+a-c;1+a-b;z^{-1}) + \\ &\qquad\quad + \frac{\Gamma(c)\Gamma(a-b)}{\Gamma(a)\Gamma(c-b)} (-z)^{-b}\,_2F_1(b,1+b-c;1-a+b;z^{-1}) \end{align}