# Lauricella integral transformation?

The Lauricella hypergeometric function $F_D$ has two different integral representations I know of. First one can be found on wikipedia:

$$\scriptsize F_D^{(n)} (a,b_1,\dots,b_n,c;x_1,\dots,x_n)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}\int_0^1 t^{a-1}(1-t)^{c-a-1}(1-x_1t)^{-b_1}\cdots(1-x_nt)^{-b_n}~dt,\qquad \operatorname*{Re}(c)>\operatorname*{Re}(a)>0$$

And second one can be found in eq. $(1.9)$ of this paper:

$$\small \frac{\Gamma(\beta_1)\cdots \Gamma(\beta_n)\Gamma(\gamma-\beta_1-\cdots-\beta_n)}{\Gamma(\gamma)}F_D(\alpha,\beta_1,\cdots,\beta_n,\gamma,x,\cdots,x_n)$$ $$\small =\mathop{\int\cdots\int}_{z_1,\cdots, z_n\geq 0\\\\ 1-z_1-\cdots-z_n\geq 0} z_1^{\beta_1-1}\cdots z_n^{\beta_n-1}(1-z_1-\cdots -z_n)^{\gamma-\beta_1-\cdots-\beta_n-1}\times (1-x_1 z_1 -\cdots -x_n z_n)^{-\alpha} dz_1\cdots dz_n$$

Since both definitions should be equivalent, there must be a transformation one could apply to the first integral to get the second and vice versa. Unfortunately, I could not guess what transformation that is. Does anyone knowledgeable see it and can tell me how to transform one integral into the other? Thanks for any suggestion!

• I have been looking at this recently and think I know the way into it if your willing to put up with an old man's backtracking over and over :) Basically Lucy Slater's "Generalized HyperGeometric Functions" section 8.2 for Appell F_1() provides what I consider to be templates for progress. The first formula is reasonably easy (it's Slater's second). The second formula is more interesting. Slater does start a process. But it's algebraic and would be recursive. I will give a personal view, slightly different, in the next comment. – rrogers May 4 '18 at 14:14
• Since the second formula is basically over a n-dimensional tetrahedron, I would think that recursively defining (n-1) dimensional slicing would provide an evaluation of x,y coefficients. If the coefficients, one by one, are equal then you have a implicit transform. To me a good alternative is to treat both as conformal maps on the complex plane; they "look" like that. That would be neat since conformal maps are invertable; but my grasp is of how is very low. Interesting though. – rrogers May 4 '18 at 14:21
• @rrogers please see my answer I just posted below. – Kagaratsch May 4 '18 at 20:13
• Your last equation sprawls into a sidebar on my computer. Could you, or would you mind if I inserted a CRLF? – rrogers May 4 '18 at 20:38
• @rrogers Thanks for pointing this out! I split the last line. – Kagaratsch May 4 '18 at 20:58

We want to demonstrate the equivalence of the two integral representations (3.24) and (3.25) of the hypergeometric function of type $(n+1,m+1)$ in Aomoto and Kita (of which the Lauricella function is a special case (2,m+1)).

We start with the definition of the hypergeometric series of type $(n+1,m+1)$ as given in eq. (3.5) of Aomoto and Kita: \begin{align} \label{Fdef}{\tag{1}} F(\alpha_i,\beta_j,\gamma,x)=\sum_{\nu_{ij}=0}^\infty\frac{\prod_{i=1}^n(\alpha_i)_{\sum_{j=1}^{m-n-1}\nu_{ij}}\prod_{j=1}^{m-n-1}(\beta_j)_{\sum_{i=1}^n\nu_{ij}}}{(\gamma)_{\sum_{i=1}^n\sum_{j=1}^{m-n-1}\nu_{ij}}\prod_{i=1}^n\prod_{j=1}^{m-n-1}\nu_{ij}!}\prod_{i=1}^n\prod_{j=1}^{m-n-1}x_{ij}^{\nu_{ij}}, \end{align} where the infinite sum is over all dummy indices $\nu_{ij}$ with $i=1,2,...,n$ and $j=1,2,...,m-n-1$. Here $(x)_y$ is the Pochhammer symbol.

We will also require an integral identity given in eq. (3.17-3.19) of Aomoto and Kita: \begin{align} \label{id1}\tag{2} \frac{\prod_{i=1}^n(\alpha_i)_{\sum_{j=1}^{m-n-1}\nu_{ij}}}{(\gamma)_{\sum_{i=1}^n\sum_{j=1}^{m-n-1}\nu_{ij}}}&=c_1\int_{{u_1>0,..,u_n>0}\atop{1-\sum_{i=1}^nu_i>0}}\prod_{i=1}^ndu_i\,u_i^{\sum_{j=1}^{m-n-1}\nu_{ij}+\alpha_i-1}\left(1-\sum_{i=1}^nu_i\right)^{\gamma-\sum_{i=1}^n\alpha_{i}-1},\\ c_1&=\frac{\Gamma(\gamma)}{\Gamma(\gamma-\sum_{i=1}^n\alpha_i)\prod_{i=1}^n\Gamma(\alpha_i)}, \end{align} or alternatively, simply replacing labels and names of dummy indices, we can also write the same integral identity as: \begin{align} \label{id2}\tag{3} \frac{\prod_{j=1}^{m-n-1}(\beta_j)_{\sum_{i=1}^n\nu_{ij}}}{(\gamma)_{\sum_{i=1}^n\sum_{j=1}^{m-n-1}\nu_{ij}}}&=c_2\int_{{u_1>0,..,u_n>0}\atop{1-\sum_{j=1}^{m-n-1}u_j>0}}\prod_{j=1}^{m-n-1}du_j\,u_j^{\sum_{i=1}^{n}\nu_{ij}+\beta_j-1}\left(1-\sum_{j=1}^{m-n-1}u_i\right)^{\gamma-\sum_{j=1}^{m-n-1}\beta_j-1},\\ c_2&=\frac{\Gamma(\gamma)}{\Gamma(\gamma-\sum_{j=1}^{m-n-1}\beta_j)\prod_{j=1}^{m-n-1}\Gamma(\beta_j)}, \end{align} Now we can plug $(\ref{id1})$ into $(\ref{Fdef})$ and observe: \begin{align} F(\alpha_i,\beta_j,\gamma,x)=&c_1\int_{{u_1>0,..,u_n>0}\atop{1-\sum_{i=1}^nu_i>0}}\prod_{i=1}^ndu_i\,u_i^{\alpha_i-1}\left(1-\sum_{i=1}^nu_i\right)^{\gamma-\sum_{i=1}^n\alpha_{i}-1}\\ &\prod_{j=1}^{m-n-1}\sum_{\nu_{ij}=0}^\infty(\beta_j)_{\sum_{i=1}^n\nu_{ij}}\prod_{i=1}^n\frac{(u_ix_{ij})^{\nu_{ij}}}{\nu_{ij}!},\\ =&c_1\int_{{u_1>0,..,u_n>0}\atop{1-\sum_{i=1}^nu_i>0}}\prod_{i=1}^ndu_i\,u_i^{\alpha_i-1}\left(1-\sum_{i=1}^nu_i\right)^{\gamma-\sum_{i=1}^n\alpha_{i}-1}\prod_{j=1}^{m-n-1}\left(1-\sum_{i=1}^nu_ix_{ij}\right)^{-\beta_j},\notag \end{align} where in the last step we have recognized multinomial expansions and resummed them. This gives us the first integral representation (3.24) in Aomoto and Kita.

Similarly, we can instead plug (\ref{id2}) into (\ref{Fdef}) and analogously observe: \begin{align} F(\alpha_i,\beta_j,\gamma,x)=&c_2\int_{{u_1>0,..,u_n>0}\atop{1-\sum_{j=1}^{m-n-1}u_j>0}}\prod_{j=1}^{m-n-1}du_j\,u_j^{\sum_{i=1}^{n}\nu_{ij}+\beta_j-1}\left(1-\sum_{j=1}^{m-n-1}u_i\right)^{\gamma-\sum_{j=1}^{m-n-1}\beta_j-1}\\ &\prod_{i=1}^{n}\sum_{\nu_{ij}=0}^\infty(\alpha_i)_{\sum_{j=1}^{m-n-1}\nu_{ij}}\prod_{j=1}^{m-n-1}\frac{(u_jx_{ij})^{\nu_{ij}}}{\nu_{ij}!},\\ =&c_2\int_{{u_1>0,..,u_n>0}\atop{1-\sum_{j=1}^{m-n-1}u_j>0}}\prod_{j=1}^{m-n-1}du_j\,u_j^{\sum_{i=1}^{n}\nu_{ij}+\beta_j-1}\left(1-\sum_{j=1}^{m-n-1}u_i\right)^{\gamma-\sum_{j=1}^{m-n-1}\beta_j-1}\\ &\prod_{i=1}^{n}\left(1-\sum_{j=1}^{m-n-1}u_ix_{ij}\right)^{-\alpha_i}.\notag \end{align} Here we have again recognized and resummed multinomial expansions. This gives us the dual integral representation (3.25) in Aomoto and Kita.

With this the equivalence of the two dual representations is demonstrated.

• @rrogers For instance in (1) the very first summation $\sum_{\nu_{i,j}=0}^\infty$ is just a shorthand notation for $\sum_{\nu_{1,1}=0}^\infty...\sum_{\nu_{n,1}=0}^\infty...\sum_{\nu_{1,m-n-1}=0}^\infty...\sum_{\nu_{n,m-n-1}=0}^\infty$. Basically, all sums over $\nu_{i,j}$ where either $i$ or $j$ or both were not defined before implies summation over labels for all $i,j$. In cases where it was defined before, it is unambiguous. – Kagaratsch May 5 '18 at 14:26
• Sorry I erased my comment. Let me repeat part: is $$\nu^{ij}$$, a function or a set? i.e. is it an array of numbers or "an array of dummy indices"? I don't have Aomoto and Kita but am trying to read your result. I still don't see the mapping between the line (one-dimensional) integral and the tetrahedron/solid integral. Aside from the fact they yield the same power series in x,y after being integrated. I thought that was what you asked for. – rrogers May 5 '18 at 14:34
• BTW: do you have a reference for the negative multinomial expansion; similiar to the binomial one? I looked around and, while it's very reasonable, I couldn't actually find a reference. – rrogers May 5 '18 at 14:45
• All the $\nu_{ij}$ are literally just labels for summation, we could have called all of them just $a,b,c,d,...$ etc. For multinomial expansion, see here en.wikipedia.org/wiki/Multinomial_theorem . One can write pochhammer symbols in terms of gamma functions and see that here we have a direct generalization of the integer case discussed in the above link. The sum representation is the defining representation of a hypergeometric function. By showing that two integral representations reduce to the same sum representation we prove that they are equivalent. – Kagaratsch May 5 '18 at 15:56
• The $\nu^{i,j}$ in $x_{ij}^{\nu_{i,j}}$ looks like a scalar power? I was looking for validation for the expansion $1/(x_{1]+x_{2}...x_{n})^\alpha$ but I can work it out for myself. – rrogers May 5 '18 at 16:39