About the confluent versions of Appell Hypergeometric Function and Lauricella Functions

I know there are two important properties about Appell Hypergeometric Function and Lauricella Functions:

$F_1(a,b_1,b_2,c;x,y)=\dfrac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}\int_0^1t^{a-1}(1-t)^{c-a-1}(1-xt)^{-b_1}(1-yt)^{-b_2}~dt$ , where $\text{Re}(c)>\text{Re}(a)>0$

$F_D^{(n)}(a,b_1,\ldots,b_n,c;x_1,\ldots,x_n)=\dfrac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}\int_0^1t^{a-1}(1-t)^{c-a-1}(1-x_1t)^{-b_1}\cdots(1-x_nt)^{-b_n}~dt$ , where $\text{Re}(c)>\text{Re}(a)>0$

Now I want to simplfy $\int_0^1t^{a-1}(1-t)^{c-a-1}(1-xt)^{-b}e^{yt}~dt$ and $\int_0^1t^{a-1}(1-t)^{c-a-1}(1-x_1t)^{-b_1}\cdots(1-x_nt)^{-b_n}e^{yt}~dt$ , where $\text{Re}(c)>\text{Re}(a)>0$ .

I know I can borrow the concpet from Confluent Hypergeometric Function so that I can express the two integrals as:

$\int_0^1t^{a-1}(1-t)^{c-a-1}(1-xt)^{-b}e^{yt}~dt=\dfrac{\Gamma(a)\Gamma(c-a)}{\Gamma(c)}\lim\limits_{k\to\infty}F_1\left(a,b,k,c;x,\dfrac{y}{k}\right)$

$\int_0^1t^{a-1}(1-t)^{c-a-1}(1-x_1t)^{-b_1}\cdots(1-x_nt)^{-b_n}e^{yt}~dt=\dfrac{\Gamma(a)\Gamma(c-a)}{\Gamma(c)}\lim\limits_{k\to\infty}F_D^{(n+1)}\left(a,b_1,\ldots,b_n,k,c;x_1,\ldots,x_n,\dfrac{y}{k}\right)$

But I don't like to leave the unsimplfied limits as the final result. How can I simplfy further?

Solution 1:

Consider the integral \begin{align} I_{n} = \int_{0}^{1} e^{y t} \ t^{a-1} (1-t)^{c-a-1} (1-x_{1} t)^{-b_{1}} \cdots (1-x_{n} t)^{-b_{n}} \ dt. \end{align} Expanding the exponential into series form leads to \begin{align} I_{n} &= \sum_{n=0}^{\infty} \frac{y^{n}}{n!} \ \int_{0}^{1} t^{a+n-1} (1-t)^{c-a-1} (1-x_{1} t)^{-b_{1}} \cdots (1-x_{n} t)^{-b_{n}} \ dt \\ &= B(a,c-a) \ \sum_{n=0}^{\infty} \frac{(a)_{n} y^{n}}{n! (c)_{n}} F_{D}^{(n)}(a+n, b_{j}; c+n; x_{j}) \end{align} where $\{b_{j}\}_{1}^{n}$ and $\{x_{j}\}_{1}^{n}$ represent all the elements in the respective sets. The Lauricella function $F_{D}^{(n)}$ can be represented in series form by use of \begin{align} & F_{D}^{(n)}(a, b_{1}, \cdots, b_{n}; c; x_{1}, \cdots, x_{n}) \\ &= \sum_{m_{1},\cdots, m_{n}=0}^{\infty} \frac{(a)_{m_{1}+\cdots+m_{n}} (b_{1})_{m_{1}} \cdots (b_{n})_{m_{n}} }{ (c)_{m_{1}+\cdots+m_{n}} } \frac{x_{1}^{m_{1}}}{m_{1}!} \cdots \frac{x_{n}^{m_{n}}} {m_{n}!} \end{align} and leads the integral into the form \begin{align} I_{n} &= B(a, c-a) \ \sum_{k,m_{1},\cdots, m_{n}=0}^{\infty} \frac{(a)_{k+m_{1}+\cdots+m_{n}} (b_{1})_{m_{1}} \cdots (b_{n})_{m_{n}} }{ (c)_{k+m_{1}+\cdots+m_{n}} } \frac{x_{1}^{m_{1}}}{m_{1}!} \cdots \frac{x_{n}^{m_{n}}}{m_{n}!} \frac{y^{k}}{k!} \\ &= B(a, c-a) \ F_{D}^{(n+1)}(a, b_{1}, \cdots, b_{n} ; c; x_{1}, \cdots, x_{n}, y) \end{align}

Now, when $n=1$ in the integral the result becomes, when use is made of $F_{D}^{(2)} = F_{1}$, \begin{align} I_{1} &= \int_{0}^{1} e^{y t} \ t^{a-1} (1-t)^{c-a-1} (1-x t)^{-b} \ dt \\ &= B(a, c-a) F_{1}(a, -, b; c; x, y) = B(a, c-a) \Phi_{1}(a,b,c; x,y) \end{align} where $F_{1}$ is the Appell function and $\Phi_{1}$ is a Humbert series.

Hence \begin{align} & \int_{0}^{1} e^{y t} \ t^{a-1} (1-t)^{c-a-1} (1-x_{1} t)^{-b_{1}} \cdots (1-x_{n} t)^{-b_{n}} \ dt \\ &= B(a, c-a) \ F_{D}^{(n+1)}(a, b_{1}, \cdots, b_{n} ; c; x_{1}, \cdots, x_{n}, y) \end{align} and \begin{align} & \int_{0}^{1} e^{y t} \ t^{a-1} (1-t)^{c-a-1} (1-x t)^{-b} \ dt \\ &= B(a, c-a) F_{1}(a, -, b; c; x, y) = B(a, c-a) \Phi_{1}(a,b,c; x,y). \end{align}