Rewriting Appell's Hypergeometric Function $F_1$ in terms of Gauss' Hypergeometric Function $_2F_1$

Question

While going through David H. answer on What is $\int_0^1 \frac{\log \left(1-x^2\right) \sin ^{-1}(x)^2}{x^2} \, dx$? I have encountered a step in between I do not really understand. Within the second half of Part $3$ he has rewritten an Appellian Hypergeometric Function $F_1(a;b,b';c;x,y)$ in terms of two Gaussian Hypergeometric Functions $_2F_1(a,b;c;x)$ in the following way

$$\small F_1\left(1;1,\frac12;2;x,xz\right)~=~\frac2{x\sqrt{1-zx}}~_2F_1\left(1,\frac12;\frac32;\frac{1-z}{1-xz}\right)-\frac2x~_2F_1\left(1,\frac12;\frac32;1-z\right)$$

Relying on the general idea of Appell's Hypergeometric Function as an extension of Gauss' Hypergeometric Function I conjectured that the given identity is a particular case of a more general formula. Searching through various websites aswell as articles I was not able to find something helpful. It might be the case that I have overlooked something.

However, while searching I became more familiar with the machiney of Hypergeometric Functions in general. Thus, I have some guesses regarding the identity

1. First of all I concerning the factor $2$: quite often a regularisation using Beta Functions is done within the integral representations of Hypergeometric Functions. Therefore I guess the $2$ can be represented through $B\left(\frac12,1\right)=B\left(\frac12,\frac32-\frac12\right)=2$. The arguments of this Beta Function correspond to the values of the $_2F_1$ functions.
2. Secondly, it seems to me that the arguments of the latter $_2F_1$ functions correspond to the two arguments of the $F_1$ function as $$F_1(\dots;x,y)=~_2F_1\left(\dots;\frac{1-\frac yx}{1-y}\right)-~_2F_1\left(\dots;1-\frac yx\right)$$
3. Last but not least, the factors infront of the $_2F_1$ functions which depend on $x,y$ might be given by $$F_1(a;b,b';c;x,y)=~x^{-b}y^{-b'}~_2F_1(\dots)-~x^{-b}~_2F_1(\dots)$$

I am still not really sure how to deduce the general formula from the given particular case since on could combine the arguments of the $F_1$ function in many ways in order to get the one from the $_2F_1$ functions. I tried to utilize the integral representations of both invoked functions but it did not really lead anywhere. The usage of the series representations seems to be pointless her but I could be proved wrong. I had no luck with the particular case either.

Could someone provide a proof for the given identity? Moreover I would be interested in the general formula which was used here and in a proof of it aswell.

Thanks in advance!

Answer 1

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Given real parameters $\left(\alpha,\beta,\gamma,\delta\right)\in\mathbb{R}^{4}$ such that $0<\alpha<\delta$ and real arguments $\left(x,y\right)\in\left(-\infty,1\right)^{2}$, we can express the Appell $F_{1}$ function via the integral representation

$$\begin{align} F_{1}{\left(\alpha;\beta,\gamma;\delta;x,y\right)} &=\frac{1}{\operatorname{B}{\left(\alpha,\delta-\alpha\right)}}\int_{0}^{1}\mathrm{d}t\,\frac{t^{\alpha-1}\left(1-t\right)^{\delta-\alpha-1}}{\left(1-xt\right)^{\beta}\left(1-yt\right)^{\gamma}}.\\ \end{align}$$

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Starting from the integral representation of the $F_{1}$ function for the particular set of parameters that we're interested in, we obtain an integral of a simple algebraic function with elementary antiderivative: for any fixed but arbitrary $\left(x,y\right)\in\left(-\infty,1\right)^{2}$,

$$\begin{align} F_{1}{\left(1;1,\frac12;2;x,y\right)} &=\frac{1}{\operatorname{B}{\left(1,1\right)}}\int_{0}^{1}\mathrm{d}t\,\frac{1}{\left(1-xt\right)\sqrt{1-yt}}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{1}{\left(1-xt\right)\sqrt{1-yt}}\\ &=\int_{1}^{0}\mathrm{d}u\,\frac{\left(-1\right)\left(1-x\right)}{\left(1-xu\right)^{2}}\cdot\frac{\left(1-xu\right)\sqrt{1-xu}}{\left(1-x\right)\sqrt{\left(1-y\right)-\left(x-y\right)u}};~~~\small{\left[t=\frac{1-u}{1-xu}\right]}\\ &=\frac{1}{\sqrt{1-y}}\int_{0}^{1}\mathrm{d}u\,\frac{1}{\sqrt{\left(1-xu\right)\left[1-\left(\frac{x-y}{1-y}\right)u\right]}}.\\ \end{align}$$

Suppose $a\in\left(0,1\right)$ and $x<1\land x\neq0$. Setting $y=ax$, we then find

$$\begin{align} F_{1}{\left(1;1,\frac12;2;x,ax\right)} &=\frac{1}{\sqrt{1-ax}}\int_{0}^{1}\mathrm{d}u\,\frac{1}{\sqrt{\left(1-xu\right)\left[1-\left(\frac{x-ax}{1-ax}\right)u\right]}}\\ &=\frac{1}{\sqrt{1-ax}}\int_{0}^{1}\mathrm{d}u\,\frac{1}{\sqrt{\left(1-xu\right)\left[1-\left(\frac{1-a}{1-ax}\right)xu\right]}}\\ &=\frac{1}{x\sqrt{1-ax}}\int_{0}^{x}\mathrm{d}v\,\frac{1}{\sqrt{\left(1-v\right)\left[1-\left(\frac{1-a}{1-ax}\right)v\right]}};~~~\small{\left[u=\frac{v}{x}\right]}\\ &=\frac{1}{x\sqrt{1-ax}}\int_{1-x}^{1}\mathrm{d}w\,\frac{1}{\sqrt{w\left[1-\left(\frac{1-a}{1-ax}\right)\left(1-w\right)\right]}};~~~\small{\left[v=1-w\right]}\\ &=\frac{1}{x}\int_{1-x}^{1}\mathrm{d}w\,\frac{1}{\sqrt{w\left[\left(1-ax\right)-\left(1-a\right)\left(1-w\right)\right]}}\\ &=\frac{1}{x}\int_{1-x}^{1}\mathrm{d}w\,\frac{1}{\sqrt{w\left[a\left(1-x\right)+\left(1-a\right)w\right]}}\\ &=\frac{1}{x}\int_{0}^{1}\mathrm{d}w\,\frac{1}{\sqrt{w\left[a\left(1-x\right)+\left(1-a\right)w\right]}}\\ &~~~~~-\frac{1}{x}\int_{0}^{1-x}\mathrm{d}w\,\frac{1}{\sqrt{w\left[a\left(1-x\right)+\left(1-a\right)w\right]}}\\ &=\frac{1}{x}\int_{0}^{1}\mathrm{d}t\,\frac{1}{\sqrt{\left(1-t\right)\left[\left(1-ax\right)-\left(1-a\right)t\right]}};~~~\small{\left[w=1-t\right]}\\ &~~~~~-\frac{1}{x}\int_{0}^{1}\mathrm{d}t\,\frac{1}{\sqrt{t\left[a+\left(1-a\right)t\right]}};~~~\small{\left[w=\left(1-x\right)t\right]}\\ &=\frac{1}{x\sqrt{1-ax}}\int_{0}^{1}\mathrm{d}t\,\frac{1}{\sqrt{\left(1-t\right)\left[1-\left(\frac{1-a}{1-ax}\right)t\right]}}\\ &~~~~~-\frac{1}{x}\int_{0}^{1}\mathrm{d}u\,\frac{1}{\sqrt{\left(1-u\right)\left[1-\left(1-a\right)u\right]}};~~~\small{\left[t=1-u\right]}\\ &=\frac{2}{x\sqrt{1-ax}}\,{_2F_1}{\left(\frac12,1;\frac32;\frac{1-a}{1-ax}\right)}-\frac{2}{x}\,{_2F_1}{\left(\frac12,1;\frac32;1-a\right)},\\ \end{align}$$

where in the last line above we've used the Euler integral representation formula to express the remaining integrals in terms of the ${_2F_1}$ function:

$$\int_{0}^{1}\mathrm{d}t\,\frac{t^{\beta-1}\left(1-t\right)^{\gamma-\beta-1}}{\left(1-zt\right)^{\alpha}}=\operatorname{B}{\left(\beta,\gamma-\beta\right)}\,{_2F_1}{\left(\alpha,\beta;\gamma;z\right)};~~~\small{z<1\land0<\beta<\gamma}.$$

$$\tag*{$\blacksquare$}$$