Deriving the partial fraction decomposition of a hypergeometric function

Question

I am studying a research paper and I don't understand how to derive a partial fraction representation of algebraic expressions whose image I am posting here.

Démonstration

On écrit $$R_n(t)(t+j)^a =F(t)^3\times G(t)^3 \times H(t)^{a-6} \times I(t)$$ ou $I(t)=t+\frac{n}{2}$ et $$ F(t)=\frac{(t-n)_n}{(t)_{n+1}}(t+j), \quad G(t)=\frac{(t+n+1)_n}{(t)_{n+1}}(t+j), \quad H(t)=\frac{n!}{(t)_{n+1}}(t+j). $$ Décomposons $F(t), G(t)$, et $H(t)$ en fractions partielles :$$ F(t)=1+\sum_{p=0\\p\neq j}^n\frac{j-p}{t+p}f_p, \quad G(t)=1+\sum_{p=0\\p\neq j}^n\frac{j-p}{t+p}g_p, \quad H(t)=\sum_{p=0\\p\neq j}^n\frac{j-p}{t+p}h_p, \quad $$

Can someone please derive one of the $F(t)$ , $G(t)$ and $H(t)$ and tell a reference for studying more about how to obtain such representation. I will be really thankful for the help.

Edit 1

Answer 1

Here we derive the partial fraction decomposition of $F(t)$. The other functions can be transformed similarly.

We have \begin{align*} F(t)&=\frac{(t-n)_n}{(t)_{n+1}}(t+j)\\ &=\frac{(t-1)(t-2)\cdots(t-n)}{t(t+1)\cdots(t+n)}(t+j)\\ &=\color{blue}{\frac{\prod_{k=1}^n(t-k)}{\prod_{l=0}^n(t+l)}(t+j)=\sum_{{q=0}\atop{q\ne j}}^n\frac{\alpha_q}{t+q}}\tag{1} \end{align*}

We see the left-hand side of (1) is a rational function in $t$ with denominator $\prod_{{l=0}\atop{t\ne j}}^n(t+l)$ which is a polynomial of degree $n$ with simple zeros at integral values $0\leq p\leq n$, $p\ne j$.

We write the left-hand side in (1) as partial fraction decomposition with unknown constants $\alpha_q$, $0\leq q\leq n$, $q\ne j$.

We determine the unknown constants $\alpha_q$ as follows. We consider $0\leq p\leq n$, $p\ne j$ and multiply (1) with $t+p$. We obtain \begin{align*} \frac{\prod_{k=1}^n(t-k)}{\prod_{{l=0}\atop{t\ne p}}^n(t+l)}(t+j)=\sum_{{q=0}\atop{q\ne j, q\ne p}}^n\frac{\alpha_q}{t+q}(t+p)+\alpha_p\tag{2} \end{align*}

Note the summand $\alpha_p$ is separated from the sum of the right-hand side, since this is the term where $t+p$ cancels.

We evaluate (2) at $t=-p$ and obtain \begin{align*} \frac{\prod_{k=1}^n(-p-k)}{\prod_{{l=0}\atop{l\ne p}}^n(-p+l)}(-p+j)=\alpha_p\tag{3} \end{align*}

Since the numerator at the left-hand side of (3) is \begin{align*} \prod_{k=1}^n(-p-k)=(-p-1)(-p-2)\cdots(-p-n)=(-p-n)_{n}\tag{4} \end{align*} we obtain from (1), (3) and (4) \begin{align*} \frac{\prod_{k=1}^n(t-k)}{\prod_{l=0}^n(t+l)}(t+j)&=\sum_{{p=0}\atop{p\ne j}}^n\frac{\alpha_p}{t+p}\\ &=\sum_{{p=0}\atop{p\ne j}}^n\frac{j-p}{t+p}\cdot\underbrace{\frac{(-p-n)_{n}}{\prod_{{l=0}\atop{l\ne p}}^n(-p+l)}}_{f_p} \end{align*} and the claim follows.