# Pfaff transformations of the hypergeometric function proof

$${}_2F_1 \left(a,b;c;z\right)=(1-z)^{-a} {}_2F_1 \left(a,c-b;c;\frac{z}{z-1}\right)$$

and

$${}_2F_1 \left(a,b;c;z\right)=(1-z)^{-b} {}_2F_1 \left(c-a,b;c;\frac{z}{z-1}\right)$$

$$\textit{proof}$$

Start by the integral representation

$${}_2F_1 \left(a,b;c;z\right)=\frac{1}{\beta(b,c-b)}\int_0^1 \frac{t^{b-1}(1-t)^{c-b-1}}{(1-tz)^a}\, dt$$

By the the transformation $t\to 1-t$

$$\int_0^1 \frac{(1-t)^{b-1}t^{c-b-1}}{(1-(1-t)z)^a}\, dt=\int_0^1 \frac{(1-t)^{b-1}t^{c-b-1}}{(1-z+tz)^a}\, dt$$

Which can be written as

$$\frac{(1-z)^{-a}}{\beta(b,c-b)}\int_0^1t^{c-b-1}(1-t)^{b-1} \left( 1-t\frac{z}{z-1} \right)^{-a} dt$$

Note this is the integral representation of

$$(1-z)^{-a} {}_2F_1 \left(a,c-b;c;\frac{z}{z-1}\right)\,\,$$

Also using that

$${}_2F_1 \left(a,b;c;z\right)= {}_2F_1 \left(b,a;c;z\right)$$

We deduce that

$${}_2F_1 \left(a,b;c;z\right)=(1-z)^{-b} {}_2F_1 \left(c-a,b;c;\frac{z}{z-1}\right)$$

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