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Can somebody help me in proving the following equation?

\begin{align*}&\textstyle \sum _{d=0} ^{n} \frac{1}{d!(n-d)!} \frac{\Gamma (b+d) \Gamma (b+n-d) \Gamma (c-n+d) \Gamma (c-b+1-n + 2d) \Gamma (-c-d) \Gamma (-c-b+1 +n -2d)}{\Gamma (b) \Gamma (b) \Gamma (c-n+2d) \Gamma (c-b+1-n+d) \Gamma (-c+n-2d) \Gamma (-c-b+1-d)} \notag \\ &= \frac{1}{n!} \frac{\Gamma(2b+n)}{\Gamma(2b)}, \end{align*} where $n$ is an integer, $b$ and $c$ are complex numbers. In particular, the dependence on $c$ vanishes.

I explicitly checked this up to $n=7$ order by mathematica, so I believe this equation indeed holds. But I do not know how to prove this for now. Are there special identities that allow this equation to hold?

If I manipulate the equation a little bit, I get the following equation.

\begin{align*} 1 = \textstyle \sum _{d=0} ^{n} \frac{n!}{d!(n-d)!} \frac{B (b+d, b+n-d) B (c-n+d, -c-b+1+n-2d) B (c-b-n+2d+1, -c-d)}{B (b,b) B (c-n+2d,-c-b-d+1) B (c-b-n+d+1, -c+n-2d)} \end{align*}

Now both $b$ and $c$ dependencies vanish.

The proof by induction will not be useful because I want to do a similar manipulations to more complicated combinations of gamma functions.

Thanks!

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  • $\begingroup$ Could you supply the background in which you have to show such an identity. I hope it's not random playing with Mathematica :) $\endgroup$
    – Fan Zheng
    Apr 19, 2015 at 5:49
  • $\begingroup$ I want to reduce a specific combination of pochhammer symbols summed over two nonnegetive integers d1 and d2 into a single hypergeometric function. For this I first changed the sum into n and d, going from 0 to inf and the other going from 0 to n respectively. The sum over d is the expression that I wrote down. $\endgroup$
    – genideal
    Apr 19, 2015 at 16:40

1 Answer 1

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(too long or too complicated as a comment).

There are at least two relevant properties in your problem.

The part of the expression inside your initial sum that does not depend on $c$ is already equal to the result.

\begin{align*}\sum _{d=0} ^{n} \frac{1}{d!(n-d)!} \frac{\Gamma (b+d) \Gamma (b+n-d)}{\Gamma (b) \Gamma (b)} = \frac{1}{n!} \frac{\Gamma(2b+n)}{\Gamma(2b)} = \frac{1}{n!}\prod_{i=0}^{n-1}(2b+i), \end{align*}

a binomial sum (or a finite product if you like) you can manipulate in many ways.

You succeeded in finding a one-arbitrary-complex-parameter weight expression that leaves that sum invariant. It is a very interesting property that can be exploited if one sees this sum as a discrete probability distribution. The sum without the $c$ fractions is probably a limit when $c$ approaches an integer.

Another related point is that the original summand is invariant by $d\rightarrow n-d$ so you would always obtain a symmetric distribution.

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