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.
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.