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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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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 '10 at 17:34
up vote 5 down vote accepted

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 . If they're not there then you're perhaps right to be pessimistic. – Kevin Buzzard Jan 21 '10 at 21:51
Kevin: a good suggestion, but wouldn't it be even better if Zev did the work for me? – David Speyer Jan 21 '10 at 22:06
A partial fraction decomposition means there is an antiderivative in terms of the integral of (sin x)/x. – Douglas Zare Jan 21 '10 at 22:59
Thanks for the answer, David, and thanks for pointing that out, FC! Is there a good explanation of why the limits change? (I haven't learned complex analysis yet, unfortunately). Does this approach work on other similar integrals using the gamma function? – Zev Chonoles Jan 22 '10 at 0:10
FC, perhaps you should post your comments as a separate answer – Zev Chonoles Jan 22 '10 at 18:45

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