Here's another proof of the key integral
$$
\int_{-\infty}^\infty {n \choose x} \, dx = 2^n
$$
for $n=0,1,2,\ldots$, which is elementary modulo the classical definite integral
$$
\int_{-\infty}^\infty \sin t \, \frac{dt}{t} = \pi.
$$
To my surprise, with a bit more work we also get what might be
a new derivation of the latter formula as well.

Start from **Pietro Majer**'s rewriting of $n \choose x$ as
$$
\frac{n!}{\pi} \, \frac{\sin \pi x}{(n-x)(n-1-x) \cdots (1-x) x}
$$
(which we can take as the *definition* of $n \choose x$ for real $x$,
thus replacing the Gamma function with a more elementary trigonometric
function; note that the last factor in the denominator is $x$, not $0-x$).
Now expand in partial fractions:
$$
\frac1{(n-x)(n-1-x) \cdots (1-x) x} = \sum_{i=0}^n \frac{c_i}{x-i},
$$
to get
$$
\int_{-\infty}^\infty {n \choose x} \, dx
= \frac{n!}{\pi} \sum_{i=0}^n
c_i \int_{-\infty}^\infty \sin \pi x \, \frac{dx}{x-i}.
$$
The integral is
$(-1)^i \int_{-\infty}^\infty \sin \pi x \, \frac{dx}{x} = (-1)^i \pi$, so
$$
\int_{-\infty}^\infty {n \choose x} \, dx = n! \sum_{i=0}^n (-1)^i c_i.
$$
But $c_i$ can be computed by letting $x \to i$ in the
partial fraction expansion; we find that $(1)^i n! c_i = {n \choose i}$, so
finally
$$
\int_{-\infty}^\infty {n \choose x} \, dx = \sum_{i=0}^n {n \choose i} = 2^n,
$$
as claimed.

Now suppose we didn't know the value, call it $I$, of
$\int_{-\infty}^\infty \sin t \, \frac{dt}{t}$.
Then our analysis still gives
$$
\int_{-\infty}^\infty {n \choose x} \, dx = \frac{I}{\pi} 2^n.
$$
But I claim that for large even $n$ the integral is asymptotic to $2^n$,
whence $I=\pi$. The idea is:

i) the Riemann sum $\sum_{x=-\infty}^\infty {n \choose x}$ is $2^n$ exactly;

ii) the integral of ${n \choose x} \, dx$ over $x$ outside the interval
$[0,n]$ is asymptotically small compared with $2^n$; and

iii) the integral from $0$ to $n$ is within $n \choose n/2$ of
the Riemann sum $\sum_{i=0}^n {n \choose i}$, and is thus asymptotic to $2^n$.

Now (i) is just the binomial expansion of $(1+1)^n$, which we've used already,
while (ii) was already proved by **Pietro Majer** (in fact he obtained a
much stronger bound of $1$, while we need only $o(2^n)$ which is easier to prove).
It remains to show (iii). But this requires only that $n \choose x$
is increasing on $0 < x < n/2$ and decreasing on $n/2 < x < n$
(we can then compare the integral with the lower and upper Riemann sums).
But we readily see from the product formula for $\sin \pi x$ that
$\log{n \choose x}$ is concave downwards on $0 < x < n$;
since this function is symmetric about $x=n/2$, we conclude that
it is increasing on $x<n/2$ and decreasing on $x>n/2$, **QED**.