Is this Lauricella $\text F_\text D$ to hypergeometric R, from DLMF, conversion formula correct?

Here is another post about uncommon standard special functions in the form of Elliptic Integrals which will be special cases of the Lauricella D function. Most elliptic integrals can be put in terms of a double hypergeometric functions like the Jacobi functions and others like the Arithmetic Geometric Mean, Elliptic Logarithm, and Legendre Ellipctic functions can all be put into closed form with a single or double hypergeometric series, but the Carlson Elliptic functions and Bulirsch’s Elliptic Integrals. Let’s start with some definitions which will allow integral and sum representations for the more complex elliptic integrals of a triple hypergeometric function with the Pochhammer Symbol. The Lauricella D function is also in Wolfram functions

$$\text F_\text D^{(n)}(a,b_1,…,b_n,c;x_1,…,x_n)=\sum_{m_1\ge0}\cdots\sum_{m_n\ge0}\frac{(a)_{\sum_{j=1}^n m_j}}{(c)_{\sum_{j=1}^n m_j}}\prod_{j=1}^n \frac{(b_j)_{m_j}x_j^{m_j}}{m_j!}=\frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}\int_0^1t^{a-1}(1-t)^{c-a-1}\prod_{j=1}^n (1-x_j t)^{-b_j}dt$$

When the Hyperelliptic hypergeometrc function from DLMF can give any elliptic integral as a special case:

$$\text R_a(b_1,..,b_n;z_1,…,z_n)=\frac 1{\text B(-a,(-a)’)}\int_0^\infty t^{(-a)’-1}\prod_{j=1}^n(t+z_j)^{-b_j}dt,(-a)’=a+\sum_{j=1}^n b_j$$

The problem here is that the integral bounds for the Lauricella function, so some substitutions help fix the bounds to put the Hyperelliptic function into Lauricella form: $$\frac{\Gamma(a)\Gamma(c-a)}{\Gamma(c)}\text F^{(n)}_\text D(a,b_1,…,b_n,c;x_1,…,x_n)=\int_0^1t^{a-1}(1-t)^{c-a-1}\prod_{j=1}^n (1-x_j t)^{-b_j}dt \mathop=^{t\to\frac1t}-\int_\infty ^1t^{-a+1}(t-1)^{c-a-1}t^{-(c-a-1)}\prod_{j=1}^n (t-x_j )^{-b_j}t^{b_j}t^{-2}dt =\int_1^\infty t^{-c} (t-1)^{c-a+1}t^{b_1}\cdots t^{b_n}\prod_{j=1}^n (t-x_j)^{-b_j} dt =\int_1^\infty t^{-c+\sum\limits_{j=1}^n b_j}(t-1)^{c-a-1}\prod_{j=1}^n(t-x_j)^{-b_j}\mathop=^{t\to t+1}_{t-1\to t} \int_0^\infty (t+1)^{-c+\sum\limits_{j=1}^n b_j}t^{c-a-1}\prod_{j=1}^n(t+1-x_j)^{-b_j} $$

Now assume the two hypergemoetric functions are equal:

$$\frac{\Gamma(a)\Gamma(c-a)}{\Gamma(c)}\text F^{(n)}_\text D(a,b_1,…,b_n,c;x_1,…,x_n) = \int_0^\infty (t+1)^{-c+\sum\limits_{j=1}^n b_j}t^{c-a-1}\prod_{j=1}^n(t+1-x_j)^{-b_j}= \text B(-a,(-a)’) \text R_a(b_1,..,b_n;z_1,…,z_n)=\int_0^\infty t^{(-a)’-1}\prod_{j=1}^n(t+z_j)^{-b_j}dt $$

Therefore the following is implied. Note that we had the same $a$ variable in both hypergeometric functions, so let’s represent one variable for the hyperelliptic function as $a=\alpha$:

$$t^{\alpha+\sum\limits_{j=1}^n b_j-1}\prod_{j=1}^n(t+z_j)^{-b_j} = (t+1)^{-c+\sum\limits_{j=1}^n b_j}t^{c-a-1}\prod_{j=1}^n(t+1-x_j)^{-b_j} \implies z_j=1-x_j,c=\sum_{j=1}^n b_j=(-a)’-a,c-a-1= \alpha+\sum\limits_{j=1}^n b_j-1 =\alpha +c-1\implies -a\to\alpha,a=-\alpha$$

and:

$$\implies \frac{\Gamma(-\alpha)\Gamma\left(\sum\limits_{j=1}^n b_j+\alpha\right)}{\Gamma\left(\sum\limits_{j=1}^n b_j\right)}\text F^{(n)}_\text D\left(-\alpha,b_1,…,b_n, \sum\limits_{j=1}^n b_j;1-x_1,…,1-x_n \right)= \text B\left(-\alpha,\alpha+\sum_{j=1}^n b_j\right) \text R_\alpha(b_1,..,b_n;z_1,…,z_n)\implies \text R_\alpha(b_1,..,b_n;z_1,…,z_n) = \frac{\Gamma(-\alpha)\Gamma\left(\sum\limits_{j=1}^n b_j+\alpha\right)}{\Gamma\left(\sum\limits_{j=1}^n b_j\right) \text B\left(-\alpha,\alpha+\sum\limits_{j=1}^n b_j\right)} \text F^{(n)}_\text D\left(-\alpha,b_1,…,b_n, \sum\limits_{j=1}^n b_j;1-x_1,…,1-x_n \right) $$

The coefficients cancel. I have derived this simple looking formula before, and got the same result, but am not sure if it is correct: $$\boxed{\text F^{(n)}_\text D\left(-a,b_1,…,b_n, \sum\limits_{j=1}^n b_j;1-x_1,…,1-x_n \right)=\text R_a(b_1,…,b_n;x_1,…,x_n) =\frac{\int_0^\infty t^{a-1}\prod\limits_{j=1}^n(1+tx_j)^{-b_j} dt}{\text B\left(-a,a+\sum\limits_{j=1}^nb_j\right)}= \sum_{m_1\ge0}\cdots\sum_{m_n\ge0}\frac{(-a)_{\sum_{j=1}^n m_j}}{\left(\sum\limits_{j=1}^nb_j\right)_{\sum_{j=1}^n m_j}}\prod_{j=1}^n\frac{(b_j)_{m_j}(1-x_j)^{m_j}}{m_j!} }$$

This is the basic expression of the conversion formula, but is it correct? We can now put any elliptic integral into a Lauricella hypergeometric form and have many new decomposition formula for the Lauricalla D function which is a standard function.

Maybe there will be some examples of special cases of the formula if it is correct. Specifically, the Carlson and Bulirsch’s elliptic functions, which also appear on Wolfram, many of which can only be expressed as a special case of the Lauricella D function. Please correct me and give me feedback!

This identity seems to be correct. Another derivation is presented below. By changing $t=(1-v)/v$ in the integral representation of the multivariate hypergeometric function (here abreviated $\operatorname R_a$), \begin{equation} \operatorname R_a(b_1,\ldots,b_n;z_1,\ldots,z_n)=\frac 1{\operatorname B(-a,(-a)')}\int_0^\infty t^{(-a)'-1}\prod_{j=1}^n(t+z_j)^{-b_j}\,dt \end{equation} we have \begin{align} dt&=-\frac{1}{v^2}\\ t+z_j&=\frac{1-v(1-z_j)}{v} \end{align} and thus \begin{align} \operatorname R_a&=\frac 1{\operatorname B(-a,(-a)')}\int_0^1 (1-v)^{(-a)'-1}v^{-1-(-a)'+\sum_{j=1}^n b_j}\prod_{j=1}^n\left(1-v(1-z_j)\right)^{-b_j}\,dv\\ &=\frac{\Gamma(-a+(-a)')}{\Gamma(-a)\Gamma((-a)')}\int_0^1 v^{-1-a}(1-v)^{(-a)'-1}\prod_{j=1}^n\left(1-v(1-z_j)\right)^{-b_j}\,dv \end{align} By comparison with the Lauricella function \begin{equation} \operatorname F_\text D^{(n)}(\alpha,\beta_1,\ldots,\beta_n,c;x_1,\ldots,x_n)=\frac{\Gamma(c)}{\Gamma(\alpha)\Gamma(c-\alpha)}\int_0^1t^{\alpha-1}(1-t)^{c-\alpha-1}\prod_{j=1}^n (1-x_j t)^{-\beta_j}\,dt \end{equation} with $\alpha=-a,\beta_j=b_j,c=-a+(-a)'=\sum_{j=1}^n b_j$ and $x_j=1-z_j$, the proposed identity follows.