The usual Binomial Theorem states that $$\sum_{m=0}^n \binom{n}{m}x^my^{n-m}=(x+y)^n, $$ and it may even be extended to the case where $n$ is not necessarily a positive integer. However, I now find myself trying to find a closed form for $$\sum_{m=0}^{n} \frac{\Gamma(1-an)}{\Gamma(1-am)\Gamma(1-a(n-m))}x^m,$$ where $a\in(\frac{1}{2},1)$ is irrational. I've looked everywhere to no avail. Something somewhat related is the Pochhammer $k$-symbol, but I couldn't find a way to use the few results available for it. Is there some sort of simple closed form for it? I found an equally complex representation:
$$\sum_{m=0}^{n}x^{m}\exp\left(\sum_{k=1}^{\infty}\frac{a^{2}m\left(n-m\right)}{\left(k-a\left(n-m\right)\right)\left(k-a m\right)}\right),$$
where the sum may be expressed in terms of the digamma function (but is still pretty useless for me). In fact, what I am really trying to find out is the value of the integral (or whether it is even finite): $$\int_{0}^{1}u^{-a\left(n-1\right)}-\sum_{m=0}^{n-1}\frac{a\Gamma\left(1-a n\right)}{\Gamma\left(1-a\left(n-m\right)\right)\Gamma\left(1-a m\right)}u^{-a\left(n-m-1\right)}\left(1-u\right)^{-a m}du,$$ which should have a simple form as a function of $a$ if I'm not wrong...
Thanks in advance for your help.