Some combinatorial identity. Let $a_1 \ge 0$ and $a_2 \ge 0$ be real numbers and let $n_1 \ge 0$ and $n_2 \ge 0$ be integers. Finally let $m\ge 1$ be another integer. By using the method of generating functions I have shown that the following identity holds:
\begin{eqnarray}
&&S = \sum_{i=0}^{m-1} \prod_{\xi=1}^2 \binom{i+a_\xi}{n_\xi} =\\ 
&&\sum_{j=0}^{n_2} (-1)^j \left(\binom{a_1+m+j}{n_1+1+j} \binom{a_2+m}{n_2-j} - \binom{a_1+j}{n_1+1+j} \binom{a_2}{n_2-j}\right)=\\
&&(-1)^{n_1+1}
 \sum\limits_{j=0}^{n_2} \left(\binom{n_1-a_1-m}{n_1+1+j} \binom{a_2+m}{n_2-j} - \binom{n_1-a_1}{n_1+1+j} \binom{a_2}{n_2-j}\right) \\
&&(-1)^{n_1}
\sum\limits_{j=0}^{n_1} \left(\binom{n_1-a_1-m}{j} \binom{a_2+m}{n_2+n_1+1-j} - \binom{n_1-a_1}{j} \binom{a_2}{n_2+n_1+1-j}\right)
\end{eqnarray}
Now, there are two questions.
Firstly, is there a generalization of the above result to higher dimensions, ie what happens if the number two in the upper limit of the product on the left hand side is replaced by some integer $d \ge  1$.
Secondly, what happens if the numbers $( n_\xi)_{\xi=1}^d$ are allowed to be real.
 A: Here, I will only answer the second part of the question, namely what happens when both the numbers $a_1$ and $a_2$ and  $n_1$ and $n_2$ are real. Let $i$ be a positive integer. Then  using the reflection formula for the Gamma function we can write the binomial factor as follows:
\begin{equation}
\binom{i+a_{\xi}}{n_{\xi}}=  B(1+i+a_\xi, -i-a_\xi+n_\xi) \cdot (-1)^i \cdot\frac{\sin(\pi(a_\xi - n_\xi))}{\pi} 
\end{equation}
where $B()$ is the beta function. Now, using the integral representation of the beta function and summing up the resulting geometric series under the sign of the integral we easily get:
\begin{eqnarray}
S = \left(\prod\limits_{\xi=1}^2 \frac{\sin\left(\pi (a_\xi - n_\xi)\right)}{\pi} \right) \cdot
\int\limits_{[0,1]^2} \frac{\prod\limits_{\xi=1}^2 t_\xi^{a_\xi} (1-t_\xi)^{-a_\xi + n_\xi} - \prod\limits_{\xi=1}^2 t_\xi^{a_\xi+m} (1-t_\xi)^{-a_\xi + n_\xi-m}}{1-t_1-t_2} d t_1 d t_2 
\end{eqnarray}
In order to evaluate the integral we write the denominator as $(1 - t_1-t_2) = (1-t_1)(1-t_2/(1-t_1))$, we absorb the first factor on the left hand side into the integrand and we expand the second factor in an infinite series. Then we integrate the series term by term and finally we apply the reflection formula back again to get rid of the sine prefactors. Having done this we easily arrive at the formula:
\begin{eqnarray}
&&S = \sum\limits_{j=0}^\infty \left(\binom{a_1+m+j}{n_1+1+j} \binom{a_2+m}{n_2-j} - \binom{a_1+j}{n_1+1+j} \binom{a_2}{n_2-j}\right) (-1)^j \\
&&\frac{\sin(\pi (n_1-a_1))}{\sin(\pi a_1)} 
\sum\limits_{j=0}^\infty \left( \binom{n_1-a_1-m}{n_1+1+j} \binom{a_2+m}{n_2-j} - \binom{n_1-a_1}{n_1+1+j} \binom{a_2}{n_2-j} \right)
\end{eqnarray}
Now, if $n_1$ and $n_2$ are integers the sum above is truncated at $j=n_2$ and we retrieve the formula given in the formulation of the question. 
In order to analyze convergence of the series above is useful to rewrite our sum in terms of factorials as follows:
\begin{eqnarray}
&&S = -\frac{ \sin(\pi n_2) }{\pi} \cdot \\
&& \sum\limits_{j=0}^\infty  \left({\mathcal C}^{a_2+m}_{a_1+m}(n_1) \cdot \frac{(a_1+m+j)!(-1-n_2+j)!}{(n_1+1+j)! (a_2+m-n_2+j)!} -  {\mathcal C}^{a_2}_{a_1}(n_1)\cdot \frac{(a_1+j)!(-1-n_2+j)!}{(n_1+1+j)! (a_2-n_2+j)!} \right)
\end{eqnarray}
where 
\begin{equation}
{\mathcal C}^{a_2}_{a_1}(n_1) := \frac{(a_2)!}{(a_1-n_1-1)!}
\end{equation}
Using the Stirling approximation to factorials and comparing the series to the harmonic series we deduce that series converge whenever $a_1-a_2 < n_1+1$. 
Ultimately the sum in question can be written in terms of the hypergeometric function as follows:
\begin{eqnarray}
&&S =  -\frac{ \sin(\pi n_2) }{\pi} \cdot \left\{\right.\\
&& \mbox{$\frac{(a_2+m)!}{(a_1+m-n_1-1)!} \frac{(-n_2-1)!(a_1+m)!}{(1+n_1)!(-n_2+a_2+m)!}
F_{3,2}\left[\begin{array}{rrr} -n_2 && a_1+m+1 & 1 \\ 2+n_1 & -n_2+a_2+m+1 & \end{array}; 1\right]$} - \\
&&\mbox{$\frac{(a_2)!}{(a_1-n_1-1)!} \frac{(-n_2-1)!(a_1)!}{(1+n_1)!(-n_2+a_2)!}
F_{3,2}\left[\begin{array}{rrr} -n_2 && a_1+1 & 1 \\ 2+n_1 & -n_2+a_2+1 & \end{array}; 1\right]$}
\left.\right\}
\end{eqnarray}
