Simplifying a hypergeometric function

Question

Why $$(1-z)^{1+a}~_2F_1(1+(m+1)a, 2+a; 2+(m+1)a;z)$$ can be simplified to $$_2F_1(1, ma, 2+(m+1)a;z)$$ ?

The above step comes from: https://arxiv.org/pdf/cond-mat/0004434.pdf

Thanks.

Answer 1

It's the Eluler transformation for the hypergeometric function: $${}_{2}F_{1}(\alpha,\beta;\gamma;z)=(1-z)^{\gamma-\alpha-\beta}\,{}_{2}F_{1}(\gamma-\alpha,\gamma-\beta;\gamma;z)$$

The proof consists of writing the Euler integral for the hypergeometric function and changing the variable unders the integral according to:

$$t=\frac{1-y}{1-zy}$$

The Euler integral is:

$${_2 F_1} (\alpha,\beta;\gamma;z)=\frac{1}{B(\beta,\gamma-\beta)} \int_0^1 t^{\beta-1} (1-t)^{\gamma-\beta-1} (1- zt)^{-\alpha} dt$$

By direct substitution with:

$$dt=-\frac{1-z}{(1-zy)^2}dy$$

$$1-t=\frac{(1-z)y}{1-zy}$$

$$1-zt=\frac{1-z}{1-zy}$$

We can confirm the transformation.