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## Gamma function versions of combinatorial identites?

We can extend the binomial coefficient $\binom{n}{k}$ to $\mathbb{R}$ or $\mathbb{C}$ by defining $\binom{x}{y}=\frac{\Gamma(x+1)}{\Gamma(y+1)\Gamma(x-y+1)}$. Do any the standard binomial coefficient identities have generalizations to this setting? Just as two simple examples, we have

$\sum_{k=0}^n \binom{n}{k} = 2^n$ and $\sum_{k=0}^n \binom{n}{k}^2 = \binom{2n}{n}$

What are $\int_0^x \binom{x}{y} dy$ and $\int_0^x \binom{x}{y}^2 dy$, and are the answers analogous to the discrete case? Is there any combinatorial significance we can give to these integrals? Has this already been tried?

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See here: math.niu.edu/~rusin/known-math/99/hypergeom_func – Steve Huntsman Jan 17 2010 at 8:50
Ah, thanks for the reference - though it doesn't seem like they were able to prove the relationship, only tested some numerical examples. – Zev Chonoles Jan 17 2010 at 17:34

Chapter 5.5 of Concrete Mathematics discusses generalizing binomial coefficient identities to the Gamma function. It doesn't discuss the two integrals you mention, though.

Doing a bit of thinking on my own, if $n$ is a positive integer then $$\int_{z=0}^n \binom{n}{z} dz = \int_{z=0}^n \frac{n! dz}{\Gamma(1+z) \Gamma(n+1-z)}$$ $$\int_{z=0}^{n} \frac{n! dz}{(n-z)(n-1-z) \cdots (1-z) \Gamma(1-z) \Gamma(1+z)}.$$

We have $\Gamma(1+z) \Gamma(1-z) = \pi z/\sin (\pi z)$, if I haven't made any dumb errors, so this is $$\int_{0}^n \frac{ n! \sin (\pi z) \ dz}{\pi z (n-z)(n-1-z) \cdots (1-z)}.$$

I suspect this integrand does not have an elementary anti-derivative, because it reminds me of $\int \sin t \ dt/t$. But there might be some special trick which would let you compute the integral between these specific bounds.

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David: if you want to push this further, then compute the integral numerically to high accuracy for a few small values of n and look up the results in Plouffe pi.lacim.uqam.ca . If they're not there then you're perhaps right to be pessimistic. – Kevin Buzzard Jan 21 2010 at 21:51
You dropped a factor of &pi; in the denominator. – Michael Lugo Jan 21 2010 at 22:02
For any real number s, Int^{-\infty}^{\infty} \binom{s,x} dx = 2^s, where the binomial expression is interpreted in terms of the Gamma function. This is an elementary exercise using contour integration. – Lavender Honey Jan 21 2010 at 22:17
Another argument: one can use hypergeometric functions to prove that \sum_{n = -\infty}^{\infty} \binom{s,z+n} = 2^s and \sum_{n = -\infty}^{\infty} \binom{s,z+n}^2 = \binom{2s,s}. One gets the appropriate infinite integrals by integrating from 0 to 1, and recovers the classical sums by letting z = 0. – Lavender Honey Jan 22 2010 at 0:52
FC, perhaps you should post your comments as a separate answer – Zev Chonoles Jan 22 2010 at 18:45