Gamma function versions of combinatorial identites? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T04:00:17Z http://mathoverflow.net/feeds/question/12079 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/12079/gamma-function-versions-of-combinatorial-identites Gamma function versions of combinatorial identites? Zev Chonoles 2010-01-17T07:10:50Z 2010-01-21T22:06:57Z <p>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</p> <p>$\sum_{k=0}^n \binom{n}{k} = 2^n$ and $\sum_{k=0}^n \binom{n}{k}^2 = \binom{2n}{n}$</p> <p>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?</p> http://mathoverflow.net/questions/12079/gamma-function-versions-of-combinatorial-identites/12573#12573 Answer by David Speyer for Gamma function versions of combinatorial identites? David Speyer 2010-01-21T21:46:13Z 2010-01-21T22:06:57Z <p>Chapter 5.5 of <em>Concrete Mathematics</em> discusses generalizing binomial coefficient identities to the Gamma function. It doesn't discuss the two integrals you mention, though.</p> <p>Doing a bit of thinking on my own, if $n$ is a positive integer then $$\int_{z=0}^n \binom{n}{z} dz = \int_{z=0}^n \frac{n! dz}{\Gamma(1+z) \Gamma(n+1-z)}$$ $$\int_{z=0}^{n} \frac{n! dz}{(n-z)(n-1-z) \cdots (1-z) \Gamma(1-z) \Gamma(1+z)}.$$</p> <p>We have $\Gamma(1+z) \Gamma(1-z) = \pi z/\sin (\pi z)$, if I haven't made any dumb errors, so this is $$\int_{0}^n \frac{ n! \sin (\pi z) \ dz}{\pi z (n-z)(n-1-z) \cdots (1-z)}.$$</p> <p>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.</p>