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


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 $-$ which could be found freely available $-$ 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 withint 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!


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

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

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:



  • 1
    $\begingroup$ First of all thank you for your answer (+1). How did you encountered this identity? Is it a particular form of a more general relation $-$ as I conjectured $-$ or did you found it by pure chance while trying to evaluate the integral? $\endgroup$ – mrtaurho Jan 25 at 12:58
  • 1
    $\begingroup$ @mrtaurho The identity is just something I jerry-rigged together for the sole purpose of aiding the evaluation of a difficult integral, and so far as I know isn't a particular case of any well-known general functional relation. Knowing that the Appell $F_1$ function stands for an elementary integral, and knowing that most basic elementary functions can be expressed as a Gauss ${_2F_1}$ function, I was able to guess that some sort of identity could be constructed in this case. Note that the steps in my work rely HEAVILY on the fact that the hypergeometric parameters are half-integers. $\endgroup$ – David H Jan 25 at 13:34
  • $\begingroup$ I have to admit that I am even more impressed by your capability of eventually evaluating this integral knowing that it did not rely on a general relation. I take my hat off to you! $\endgroup$ – mrtaurho Jan 25 at 13:50

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