Binomial again, and again Let $\lceil a\rceil=$ the smallest integer $\geq a$, otherwise known as the ceiling function. When the arguments are real, interpret $\binom{a}b$ using the Euler's gamma function, $\Gamma$.
Recently, I posted a problem on MO about "sin-omials". Here is another one in the same spirit. 

QUESTION. Numerical evidence suggest that 
  $$\left\lceil \int_0^n\binom{n}x dx\right\rceil=\sum_{k=0}^n\binom{n}k.$$
  If true, how do we prove this?

REMARK. The RHS evaluates to $2^n$; I only want to stress parallelism between the integral and sum.
 A: The $2^n$ result in Noam Elkies answer generalizes to $n=s$ for $real(s)>-1$, as he notes in MSE. In this range, the generalized binomial coefficient $ \binom{s}{\alpha}$ is a Fourier band-limited fct. in $\alpha$, or, equivalently a sinc fct. interpolation of the sequence $\binom{s}{n}$. See my notes "Fractional calculus and interpolation of generalized binomial coefficients" for simple derivations and my contribution to the MO-Q Generalizing a problem to make it easier for the integral of the sinc fct. over infinte limits, also simply derived.
The Wikipedia article on the classic Gibbs phenomena explores the extrema of $Si(x)$, easily visualized as a convolution of the sinc with the rectangle function, and, therefore, bounds on the residual of the ceiling.
The sinc function interpolation is an important illustration of the Nyquist-Shannon sampling theorem and a special case of a generalized Chu-Vandermonde identity using Euler's reflection formula for the gamma/factorial function. (I remember long ago reading somewhere that Ramanujan noted this. Anyone know a ref? This would be simple check for his Master Theorem/Formula.)
A: Following the hint by Noam D. Elkies, we just need to show that the remainder $$R_n:=\int_{-\infty}^0{n!\over\Gamma(n+1-x)\Gamma(1+x)}dx+ \int_n^{+\infty} {n!\over\Gamma(n+1-x)\Gamma(1+x)}dx $$ satisfies $$0\le R_n<1.$$ The integrand writes
$${n!\over\Gamma(n+1-x)\Gamma(1+x)}={n! \over  (n-x)(n-1-x)\dots(1-x)\Gamma(1-x)x\Gamma(x)}={n!\over \pi}{ \sin \pi x \over (n-x)(n-1-x)\dots(1-x)x}.$$
With a linear change of variables it is easy to see that the two integrals above coincide,  so that the remainder takes the  form
$$R_n= {2n!\over\pi}\int_{0}^{+\infty} {{ \sin \pi x \over x(x+1)\dots(x+n)}}\, dx  $$
so in particular $R_0=1$. If we further write the integral as a Leibnitz alternating sum of the integrals over unit intervals, we find
$$ R_n= {2n!\over\pi}\sum_{k=0}^{\infty}\; (-1)^k\int_{0}^{1} {{ \sin \pi x \over (x+k)\dots(x+k+n)}}\, dx$$
$$ ={2n!\over\pi}\int_{0}^{1}\sin \pi x\bigg[ \sum_{k=0}^{\infty}\;  {{ 1 \over (x+2k)\dots(x+2k+n)}}-{{ 1 \over (x+2k+1)\dots(x+2k+1+n)}}\bigg]\, dx $$
$$ ={2\over\pi}\int_{0}^{1}\sin \pi x \bigg[\sum_{k=0}^{\infty}\;  {{ (n+1)! \over (x+2k)\dots(x+2k+1+n)}} \bigg]\, dx. $$
This way it is apparent that $R_n$ is positive and strictly decreasing w.r.to $n$, and we conclude that for $n\ge1$
$$0< R_n< R_0=1.$$
A: Here is a proof which doesn't use the identity $\int_{-\infty}^\infty {n \choose x}\,dx= 2^n$:
Using the representation ${ n \choose x}=\frac{1}{2\pi}\int_{-\pi}^\pi e^{-ixt}\left(1+e^{it}\right)^n\,dt$ (valid for $n,x\in\mathbb{R}, n>-1$) we find
\begin{align*}
\int_0^n {n \choose x}\,dx&=\frac{1}{2\pi} \int_0^n\int_{-\pi}^\pi e^{-ixt}\left(1+e^{it}\right)^n\,dt\,dx \\
&=\frac{1}{2\pi} \int_{-\pi}^\pi \frac{1-e^{-int}}{it}\left(1+e^{it}\right)^n\,dt\\
&=\frac{1}{2\pi} \int_{-\pi}^\pi \sum_{k=0}^n {n \choose k} \frac{e^{ikt}-e^{-i(n-k)t}}{it}\,dt\\
&=\frac{1}{2\pi} \int_{-\pi}^\pi \sum_{k=0}^n {n \choose k} \frac{e^{ikt}-e^{-ikt}}{it}\,dt\\
&=\frac{1}{\pi} \int_{-\pi}^\pi \sum_{k=0}^n {n \choose k} \frac{\sin(kt)}{t}\,dt\\
&=\frac{2}{\pi} \sum_{k=0}^n {n \choose k} \int_{0}^\pi\frac{\sin(kt)}{t}\,dt\\
&=\frac{2}{\pi} \sum_{k=0}^n {n \choose k} \mathrm{Si}(k\pi) \;\;\;\;\;\;\;\; (*)
\end{align*}
where $\mathrm{Si}(x)=\int_0^x \frac{\sin(t)}{t}\,dt$ denotes the sine integral. For $a\geq 0$ the sine integral has the representation
$$\mathrm{Si}(a)=\frac{\pi}{2} - \cos(a)\,\int_{0}^\infty \frac{e^{-at}}{1+t^2}\,dt - \sin(a)\,\int_{0}^\infty \frac{t\,e^{-at}}{1+t^2}\,dt$$
Plugging this into $(*)$ (and using $\sin(k\pi)=0,\, \cos(k\pi)=(-1)^k$) we arrive at
$$\int_0^n {n \choose x}\,dx=2^n - \frac{2}{\pi} \int_0^\infty \frac{(1-e^{-\pi t})^n}{1+t^2} \,dt=:2^n - R_n$$
Clearly the sequence $(R_n)$ is nonnegative, strictly decreasing and $R_0=1$.
A: 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.
