Integral of power of binomials equal to sum of power of binomials? Inspired by this MO question about integrating binomial coefficients and the answers, I was wondering whether integrating powers of binomial coefficients also relates to the respective sums. And indeed I have strong numerical evidence that 
$$\int_{-\infty}^{\infty} \binom{n}{x}^2 dx =\sum_{k=0}^n\binom{n}{k}^2$$. (The latter expression is of course $\binom{2n}{n}$.) 
So I'm wondering for which $l$ the following identity holds: 
$$\int_{-\infty}^{\infty} \binom{n}{x}^l dx =\sum_{k=0}^n\binom{n}{k}^l$$. 
(And furthermore one could conjecture that there are similar examples, where sum over binomials is identical to integral of the same expression over real numbers.) 
EDIT: 
Regarding the last sentence here an example: I conjecture (and have numerical evidence) that 
$$\int_{-\infty}^{\infty} x\binom{n}{x} dx =\sum_{k=0}^n k\binom{n}{k},$$ the latter expression being of course $2^{n-1}n$.
EDIT2:
And for Vandermonde's identity it seems also (by numerical evidence) to work analogously:
$$ \int_{-\infty}^{\infty} \binom{m}{x} \binom{n}{r-x} dx = \sum_{k=0}^r \binom{m}{k} \binom{n}{r-k},$$ the latter expression being of course $\binom{m+n}{r}$. 
I dare to conjecture that one can still find more examples.  
 A: The generalization looks like this
$$
\int_{-\infty}^{\infty} \binom{n}{\alpha x}^l dx =\sum_{k=-\infty}^\infty\binom{n}{\alpha k}^l,\quad 0<\alpha\le 2/l,~l\in\mathbb{N}\tag{1}
$$
where $n$ need not be an integer. The general theorem is given for example in the paper Surprising sinc sums and integrals.
Below I give the general outline of the proof which is based on the well known fact that the following function is band limited (its Fourier transform has limited spectrum)
$$g(x)=\binom{n}{x}=\frac{1}{2\pi}\int_{-\pi}^{\pi} (1+e^{i \omega})^n e^{- ix\omega} d\omega$$
One can see that Fourier transform is limited to frequencies $|\omega|<\pi$. Whenever spectrum of a function $f(x)$ is limited to frequencies $|\omega|<2\pi$ one expects that 
$$
\int_{-\infty}^{\infty} f(x) dx =\sum_{k=-\infty}^\infty f(k).
$$
Now the Fourier transform of $g(\alpha x)^l$ has a spectrum limited in the band $|\omega|<\pi\alpha l$. This is easy to see calculating Fourier transform
$$
\int_{-\infty}^{\infty}g(\alpha x)^le^{-ikx}dx
$$
with the help of $\int_{-\infty}^{\infty}e^{-ikx}dx=2\pi\delta(k)$, where $\delta$ is delta function. The general theorem from the paper cited above now states that
$$
\int_{-\infty}^{\infty} g(\alpha x)^l dx =\sum_{k=-\infty}^\infty g(\alpha k)^l,\quad 0<\pi\alpha l\le 2\pi,~l\in\mathbb{N},
$$
which is equivalent to (1).
All this analysis also explains that when $\alpha=1$ the proposed identity holds for $l=1,2$ but not for larger $l\in\mathbb{N}$.
A: I don't know if there is a relation for values of $\ell$ apart from $1$ and $2$ (that would be very interesting, and surprising to me), but here is a unified way to look at what's going on for exponents $1$ and $2$.  
Consider the function on ${\Bbb R}$ defined by 
$$ 
f(x) = (1+e^{2\pi i x})^n 
$$ 
for $-1/2 \le x \le 1/2$ and $f(x) = 0$ if $|x| >1/2$.  We can compute its Fourier transform: 
$$ 
{\widehat f}(\xi) = \int_{-\infty}^{\infty} f(x) e^{-2\pi i x\xi} dx, 
$$ 
and this turns out to be 
$$ 
\binom{n}{\xi} = \frac{n!}{\Gamma(\xi+1) \Gamma(n-\xi + 1)}. 
$$ 
(Check for $n=1$, and then use $\binom{n}{\xi} = \binom{n-1}{\xi} + \binom{n-1}{\xi -1}$).  So now Fourier inversion gives 
$$ 
f(0) = 2^n = \int_{-\infty}^{\infty} \binom{n}{\xi} d\xi, 
$$ 
which is the case $\ell =1$.  
Next Plancherel gives 
$$ 
\int_{-1/2}^{1/2} |f(x)|^2 dx = \int_{-\infty}^{\infty} \binom{n}{\xi}^2 d\xi. 
$$ 
Now instead of $f$ consider the $1$-periodic function $F$ defined by $F(x) = f(x)$ on $(-1/2,1/2)$ and extended periodically.  The Fourier coefficients of $F$ are 
$$ 
{\widehat F}(k) = \int_{-1/2}^{1/2} f(x) e^{-2\pi ik} dx = \binom{n}{k}, 
$$ 
and so Parseval gives 
$$ 
\sum_{k} {\widehat F}(k)^2 =\sum_{k=0}^{n} \binom{n}{k}^2 = \int_{-1/2}^{1/2} f(x)^2 dx. 
$$ 
This gives the $\ell=2$ identity.  
