Identity involving double sum with binomials (asked previously in MSE here)
In the course of a calculation, I have met the following complicated identity. Let $A$ and $a$ be positive integers. Then I believe that
$$ \sum_{B\ge A,b\ge a} (-1)^{a+b}{b\choose a} {B-1\choose A-1}\frac{(B-1)!b!}{(B+b)(x-b)^{(B+b)}}=\frac{(A-1)!a!}{(A+a)(x-A+1)^{(A+a)}},$$
where $x$ is some variable and $(x)^{(a)}=x(x+1)\cdots(x+a-1)$ is the rising factorial. This is what I would like to prove.
Notice how the left hand side in principle has poles at all integer values of $x$, while the right hand side only has poles at integers smaller than $A-1$.

Rewriting the l.h.s. of the conjectured identity and using the properties of beta function, we have:
\begin{split}
\text{l.h.s.} &= \sum_{B, b} (-1)^{a+b}{b\choose a} {B-1\choose A-1}\frac{1}{(B+b)\binom{B+b-1}b (x+B-1) \binom{x+B-2}{B+b-1} }\\
&= \sum_{B, b} (-1)^{a+b}{b\choose a} {B-1\choose A-1} \int_0^1 \int_0^1 (1-y)^{x-b-1} y^{B+b-1} (1-z)^bz^{B-1}\,{\rm d}y\,{\rm d}z\\
&= \int_0^1 \int_0^1 (1-y)^{x} y^{-2} (z(1-z))^{-1} \sum_{B, b} (-1)^{a+b}{b\choose a} {B-1\choose A-1}  (y(1-z)/(1-y))^{b+1} (yz)^{B}\,{\rm d}y\,{\rm d}z\\
&= -\int_0^1 \int_0^1 (1-y)^{x} y^{-2}(z(1-z))^{-1} \left( \frac{y(1-z)}{1-yz} \right)^{a+1} \left(\frac{yz}{1-yz}\right)^{A}\,{\rm d}y\,{\rm d}z\\
&= -\int_0^1 \int_0^1 (1-y)^{x} (yz)^{A-1} (y-yz)^a (1-yz)^{-(A+a+1)}\,{\rm d}y\,{\rm d}z.
\end{split}
Substituting $(u,v)=(\frac{1-y}{1-yz},yz)$, or $(y,z) = (1-u(1-v),\frac{v}{1-u(1-y)})$ we further get
\begin{split}
... &= \int_0^1 \int_0^1 u^x (1-u)^a v^{A-1} (1-v)^{x-A} \frac1{1-u(1-v)}\,{\rm d}u\,{\rm d}v\\
&= \sum_{i\geq 0}\int_0^1 \int_0^1 u^{x+i} (1-u)^a v^{A-1} (1-v)^{x-A+i}\,{\rm d}u\,{\rm d}v \\
&= \sum_{i\geq 0} \frac{1}{(a+1)A\binom{x+i+a+1}{a+1}\binom{x+i}{A}}  = \sum_{i\geq 0} \frac{1}{(a+1)A\binom{A+a+1}A \binom{x+i+a+1}{A+a+1}},\\
&= \frac{x+a+1}{(a+1)A\binom{A+a+1}A (A+a) \binom{x+a+1}{A+a+1}} = \frac{1}{(A+a)^2\binom{A+a-1}{a}\binom{x+a}{A+a}},
\end{split}
which matches the r.h.s.
QED
