A $\Gamma$- function identity: $\sum_{j=0}^m \frac{\Gamma(x-j)\Gamma(m-x+j)}{\Gamma(j+1)\Gamma(m-j+1)} = 0$ Mathematica told me the following identity,
$$x\notin \mathbb{Z}, \quad m\in \mathbb{Z}^+,\quad  \sum_{j=0}^m \frac{\Gamma(x-j)\Gamma(m-x+j)}{\Gamma(j+1)\Gamma(m-j+1)} = 0 .$$
I tried to prove it but not successful. Could anyone please take a look at the above identity?
I rearranged it to the form,
$$ \frac{\pi}{\sin\pi x} \sum_{j=0}^m (-1)^{j} \frac{\prod_{i=1}^{m-1}(i+j-x)}{j!(m-j)!}$$
but doesn't seem to help.
 A: Suppose  we seek  to  prove that  for $x$  not  an integer  and $m$  a
positive integer
$$\sum_{q=0}^m 
\frac{\Gamma(x-q)\Gamma(m-x+q)}{\Gamma(q+1)\Gamma(m-q+1)} = 0.$$
We have
$$\Gamma(x-q) \prod_{p=1}^{q} (x-p) = \Gamma(x)$$
and
$$\Gamma(m-x+q) = \Gamma(1-x) \prod_{p=1}^{m+q-1} (p-x).$$
This yields for the sum
$$\frac{1}{m!} \frac{\pi}{\sin(\pi x)}
\sum_{q=0}^m {m\choose q}
\frac{\prod_{p=1}^{m+q-1} (p-x)}{\prod_{p=1}^{q} (x-p)}
\\ = \frac{1}{m!} \frac{\pi}{\sin(\pi x)}
\sum_{q=0}^m {m\choose q} (-1)^q
\frac{\prod_{p=1}^{m+q-1} (p-x)}{\prod_{p=1}^{q} (p-x)}
\\ = \frac{1}{m!} \frac{\pi}{\sin(\pi x)}
\sum_{q=0}^m {m\choose q} (-1)^q
\prod_{p=q+1}^{m+q-1} (p-x)
\\ = \frac{1}{m} \frac{\pi}{\sin(\pi x)}
\sum_{q=0}^m {m\choose q} (-1)^q
{m+q-1-x\choose m-1}.$$
Introduce
$${m+q-1-x\choose m-1} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m}}
(1+z)^{m+q-1-x}
\; dz.$$
We get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m}}
(1+z)^{m-1-x}
\sum_{q=0}^m {m\choose q} (-1)^q (1+z)^q
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m}}
(1+z)^{m-1-x}
(-z)^m
\; dz
\\ = \frac{(-1)^m}{2\pi i}
\int_{|z|=\epsilon}
(1+z)^{m-1-x}
\; dz = 0.$$
This is the claim.
Remark.   To  ensure   the  integrand   being  analytic   in  a
neighborhood of zero  (recall that $x$ is not an  integer) we may take
e.g.  $(1+z)^\alpha = \exp(\alpha\log(1+z))$ with the principal branch
of the logarithm that has argument from $-\pi$ to $\pi$ (branch cut on
the  negative real  axis). We  can  multiply two  powers $\alpha$  and
$\beta$ of  $1+z$ as long as we  use the same branch  of the logarithm
throughout.
Even better we can use  the formal power series definition to prove
that   $$(1+z)^\alpha  (1+z)^\beta   =   (1+z)^{\alpha+\beta}$$  where
$\alpha$ and $\beta$ are not integers. We get
$$[z^n] (1+z)^\alpha (1+z)^\beta =
\sum_{p=0}^n {\alpha \choose p} {\beta \choose n-p}
\\ = \frac{1}{n!} 
\sum_{p=0}^n {n\choose p} \alpha^{\underline{p}}
\beta^{\underline{n-p}}.$$
The coefficient here on $\alpha^{p_1} \beta^{p_2}$ is
$$\frac{1}{n!} \sum_{p=0}^n {n\choose p}
(-1)^{p_1+p} \left[p\atop p_1\right]
(-1)^{p_2+n-p} \left[n-p\atop p_2\right]
\\ = \frac{(-1)^{n+p_1+p_2}}{n!} \sum_{p=0}^n {n\choose p}
\left[p\atop p_1\right]
\left[n-p\atop p_2\right].$$
On the other hand we have
$$[\alpha^{p_1} \beta^{p_2}] {\alpha+\beta\choose n}
= [\alpha^{p_1} \beta^{p_2}] \frac{1}{n!} \sum_{p=0}^n 
(\alpha+\beta)^p (-1)^{n+p} \left[n\atop p\right]
\\ = \frac{1}{n!} {p_1+p_2\choose p_1}
(-1)^{n+p_1+p_2} \left[n\atop p_1+p_2\right].$$
The quantity other  than the sign and the  factorial counts the number
of ways we  can partition $[n]$ into $p_1$ red  and $p_2$ blue cycles,
using everything. The sum says we first pick the values that will form
the  red  cycles  and  partition  them into  $p_1$  cycles.   Then  we
partition the remaining values, which will be blue, into $p_2$ cycles.
The closed formula  says that we first partition  $[n]$ into $p_1+p_2$
cycles and  then choose the $p_1$  red cycles, leaving  the rest blue.
This concludes the  argument where we have not  used complex variables
and/or branches of the logarithm.
