Integral representation of Gauss Hypergeometric function proof

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

$$\textit{proof}$$

Start by the RHS

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

Using the expansion of \( (1-tz)^{-a} \) we have

$$\int_0^1 t^{b-1}(1-t)^{c-b-1} \sum_{k=0}^\infty\frac{(a)_k}{k!}\, (tz)^k $$

Interchanging the integral with the series

$$ \sum_{k=0}^\infty\frac{(a)_k}{k!}\, z^k \, \int_0^1 t^{k+b-1}(1-t)^{c-b-1}\, dt $$

Recalling the beta function we have

$$ \sum_{n=0}^\infty\frac{(a)_k \Gamma(k+b) \Gamma(c-b)}{ \Gamma(k+c)}\, \frac{z^k}{k!} $$

Using the identity that

$$\beta(c-b,b) = \frac{\Gamma(b)\Gamma(c-b)}{\Gamma(c)}$$

and

$$\frac{\Gamma(z+k)}{\Gamma(z)} = (z)_k$$

We deduce that

$$ \beta(c-b,b)\sum_{k=0}^\infty\frac{(a)_k \Gamma(k+b) \Gamma(c)}{\Gamma(b) \Gamma(k+c)}\, \frac{z^k}{k!}= \beta(c-b, b) \sum_{k=0}^\infty \frac{(a)_k (b)_k}{(c)_k} \frac{z^k}{k!} \,\,$$

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