6
$\begingroup$

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?

$\endgroup$
2

1 Answer 1

6
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ 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. $\endgroup$ Jan 21, 2010 at 21:51
  • $\begingroup$ Kevin: a good suggestion, but wouldn't it be even better if Zev did the work for me? $\endgroup$ Jan 21, 2010 at 22:06
  • $\begingroup$ A partial fraction decomposition means there is an antiderivative in terms of the integral of (sin x)/x. $\endgroup$ Jan 21, 2010 at 22:59
  • $\begingroup$ 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? $\endgroup$ Jan 22, 2010 at 0:10
  • 2
    $\begingroup$ FC, perhaps you should post your comments as a separate answer $\endgroup$ Jan 22, 2010 at 18:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.