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"](https://mathoverflow.net/questions/253703/integral-of-a-sin-omial-coefficients-binomial/253965#253965). 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.



Binomial again, and again
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 $n\ge1$  

$$0< R\_n< R\_0=1.$$


