# Prove an transformation formula for Gauss hypergeometric function $_2F_1(a,b;c;z)$

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.

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}
• The integrated function is holomorphic in the upper half of the complex plane (${\rm Im} t > 0$), so integral over any closed curve contained within this region is equal to $0$. In this case, the curve that we're using is an interval $[-R,R]$ closed with a semi-circle. For the right parameters $a,b,c$ the function vanishes quickly enough when $t\rightarrow\infty$, and the integral over the semicircle vanishes when we go with $R\rightarrow\infty$. That means that the integral over the real line is $0$. – Adam Latosiński May 1 at 20:14
• One more issue that we need to be careful about is whether we can approach points $t=0$ and $t=1$ (which may be singularities) with no problems, but again, for $a,b,c$ in an appropriate range, it is possible. – Adam Latosiński May 1 at 20:14