Continuous generalization of $\sum_{k=0}^n {n \choose k} = 2^n$? We know that 
$$\sum_{k=0}^n {n \choose k} = 2^n.$$
A continuous generalization of the formula would be
$$\int_0^{n+1} \frac{\Gamma(n+1)}{\Gamma(n-x+1) \Gamma(x+1)} dx = 2^n?,$$
but this is incorrect numerically, though close, as I've checked it for $n=4$ and $n=5$. It is even closer when the bounds of integration is changed to $[-1, n+1]$ (e.g. $n=3$ gives $8.036$, $n=4$ gives $16.0274$, $n=8$ gives $256.013$).
Is there a formula for the difference of the integral and $2^n$ with either set of bounds? Can the order of the error be determined as dependent on $n$? It should be trivial to show that the expression grows as $2^n$ by doing a Riemann sum comparison and using the original formula, right? Is there a better generalization?

Here is the Wolfram alpha plot of the integrand for $n=4$:

 A: The continuous generalization is given by the central limit theorem of de Moivre-Laplace, and in particular, the formula
$$\int_0^\infty \exp\left(-x^2\right) = \frac{\sqrt{\pi}}2.$$
The integrand is a good approximation to the binomial coefficient, modulo scaling (which is, in a way, the content of the central limit theorem; see the linked-to page).
A: Integrate over the entire real line and it will work exactly,
even without assuming that $n$ is an integer, as long as $n>-1$.
(When $n$ is an integer, the sum $\sum_{k=0}^n {n \choose k}$ is also
$\sum_{k=-\infty}^\infty {n \choose k}$, which is a Riemann sum 
with mesh 1 for the integral.)
I refer to formula 6.414#2 in Gradshteyn and Ryzhik (which in turn cites "ET II 297(5)",
where ET II = Erdélyi et al., Tables of Integral Transforms II
[New York: McGraw Hill, 1954]):
$$
\int_{-\infty}^\infty \frac{dx}{\Gamma(\alpha+x)\,\Gamma(\beta-x)}
= \frac{2^{\alpha+\beta-2}}{\Gamma(\alpha+\beta-1)},
$$
which holds for all complex $\alpha,\beta$ with ${\rm Re}(\alpha+\beta) > 1$.
Your integrand is obtained by taking $(\alpha,\beta) = (1,n+1)$
and multiplying through by $\Gamma(n+1)$.
Added later: In fact $2^n = \sum_{k=-\infty}^\infty {n \choose w+k}$
holds for all real $-$ or even complex $-$ $w$
as long as $n > -1$ (more generally ${\rm Re}(n) > -1$ for $n \in \bf C$),
where the terms $n \choose w+k$ are defined
using the Gamma function as suggested.  Integration with respect to $w$
from $w=0$ to $w=1$ then gives the integral
$\int_{x=-\infty}^\infty {n \choose x} \, dx = 2^n$.
We outline a complex-analytic proof of the formula for
$F_n(w) := \sum_{k=-\infty}^\infty {n \choose w+k}$
for ${\rm Re}(n) > 0$, for which the sum converges absolutely;
to extend to ${\rm Re}(n) > -1$ we could group the terms in pairs
to get absolute convergence and then argue similarly.
Using the Stirling's approximation to the complex Gamma function
we find that $F_n(w)$ converges to an analytic function of period $1$
in $w$.  [When $-1 < {\rm Re}(n) \leq 0$ an extra step is needed
to prove $F_n(w+1) = F_n(w)$.]  Moreover, on a period strip such as
$0 \leq {\rm Re}(w) \leq 1$ the Stirling estimate shows that
$|F_n(w)|$ grows no faster than some power of $w$ times
$\exp \pi\left|{\rm Im}(w)\right|$.
This implies that $F_n$ is constant by a standard argument(*).
But we already saw that $F_n(0) = 2^n$.  Hence $F_n(w) = 2^n$
for all $w$, QED.
(*) An analytic function of $w$ has period $1$ iff it is
an analytic function of $q := e^{2\pi i w}$; now use Laurent
expansions about $q=0$ and $q=\infty$, where $F_n$ is
$O(|q|^{\pm(1/2+\epsilon)})$, to show that this function
is entire and bounded, and thus constant by Liouville.
P.S. The case $n=0$ of the formula
$$
\int_{-\infty}^\infty \frac{\Gamma(n+1)}{\Gamma(n-x+1)\,\Gamma(x+1)} dx = 2^n
$$
is equivalent (via the identity 
$\Gamma(z)\,\Gamma(1-z) = \frac\pi{\sin \pi z}$)
with the famous definite integral $\int_{-\infty}^\infty \sin t \, dt/t = \pi$.
