In an answer to T. Amdeberhan's [recent question](https://mathoverflow.net/questions/254881/binomial-again-and-again/254959#254959), Noam D. Elkies gives, as a by-product, an elementary computation of the classical integral
$$\int\_{-\infty}^{+\infty}{\sin x\over x}dx=\pi,$$
and suggests *a trigonometric definition* of binomial coefficients ${n\choose x}$ for non-negative integer $n$ and real or complex $x$, as:
$${n\choose x}:=\frac{n!}{\pi} \, \frac{\sin \pi x}{(n-x)(n-1-x) \cdots (1-x) x}\ ,
$$
which is consistent with the definition via Gamma function, $\frac{n!}{\Gamma(1+x)\Gamma(1+n-x)}
$, using the functional equation and the symmetry relation $\Gamma(x) \Gamma(1-x)=\pi/\sin\pi x$. (We may even hide the sine function and $\pi$, using the infinite product for $\sin(\pi x)$, which gives a plain infinite product
$${n\choose x}=\prod\_{k=1}^n \Big(1+{x\over k}\Big)\prod\_{k>n} \Big(1-{x^2\over k^2}\Big),$$
but this formula seems somehow less handy to work with). 


I'd like to elaborate this idea, mainly for didactic purposes, but not only, and see what can be deduced from the trigonometric definition without resorting to the Gamma function.


* It is clear that it actually reduces to the classical definition for $x\in\mathbb{Z}$, taking a limit as $x\to k$.
* The symmetry property ${\big({n\atop x}\big)}={n\choose n-x \ }$ and the recursive relations ${\big({n\atop x}\big)}={n\over x}{n-1\choose x-1 \ }$ and ${n-1\choose x}+{n-1\choose x-1\ }={n\choose x}$, just follow from elementary identities.
* Many identities seem to have a continuous counterpart, like Amdeberhan's
$$\int\_{\mathbb{R}}{n\choose x}a^xb^{n-x}dx=\sum\_{k\in\mathbb{Z}} {n\choose k}a^kb^{n-k}=(a+b)^n.$$


While these properties motivate and justify as quite natural the choice of the above definition for the binomials ${n\choose x \ }$ for real $x$, it remains the want of a necessity for this choice, that is, a uniqueness result analogous to the Artin-Bohr-Mollerup characterization of the Gamma function.


So here is my question: 



> 
> Is there a Gamma-free characterization of the extended binomials $\textstyle{n\choose x}$? In other
>  words, what property, together with one or more of the above relations,
>  would lead to the above trigonometric definition?
> 
> 
> 



Gamma-free definition of binomial coefficients
Yes, there is a Bohr-Mollerup (or Artin-Bohr-Mollerup?)
criterion for $n \choose x$. It's probably known,
but seems easier to write up anew than to find in the literature.


**Proposition**: 
*The function $n\choose x$ of an integer $n \geq 0$ and a real $x$
is characterized by the initial value ${0 \choose 0} = 1$
and the recurrences
$$
(x+1) {n \choose x+1} = (n-x) {n \choose x},
\quad\
{n \choose x} + {n \choose x+1} = {n+1 \choose x+1}
$$
(which imply that $n \choose x$ agrees with the usual values when
$x \in \bf Z$), together with the condition that
$n \choose x$ is a logarithmically convex function of $x \in (0,n)$
for each $n$.*


(We do not need to impose the symmetry condition 
${n \choose x} = {n \choose n-x}$, 
but it follows automatically from uniqueness.)


*Proof sketch*:
We can show that the trigonometric $n \choose x$ satisfies this 
convexity condition using the product formula for the sine:
each factor $(x+k)(x-n-k)$ (with $k=0,1,2,\ldots$) is logarithmically 
convex (downwards) on $0 \leq x \leq n$, so the same is true of their product.


The converse can be proved a la Bohr-Mollerup.
Any function of $n$ and $x$ satisfying the recurrences 
is of the form $f(x) {n \choose x}$ for some $1$-periodic function $f$.
But for large $n$ and $x$ near $n/2$ the logarithm of $n \choose x$, 
though convex, is nearly flat; e.g.
$$
\log {2m \choose m-1} - 2 \log {2m \choose m} + \log {2m \choose m+1}
= O(1/m).
$$
Thus if $f$ were nonconstant then for large $n$ the fluctuations would
make $f(x) {n \choose x}$ nonconvex. Hence $f$ is constant, and then
the initial value ${0 \choose 0} = 1$ makes $f(x)=1$ identically. $\Box$


