TITLE: Integral of Binomial Coefficient
QUESTION [25 upvotes]: We all know the famous theorem that:
$$\sum_{i=1}^n\binom{n}{i}=2^n$$
This theorem got me wondering about a similar formula - the properties of the following function:
$$I(n)=\int_{0}^{n} \binom{n}{k}\,\,\mathrm{d}k$$
where $n$ is any positive integer and the definition of binomial coefficient is "extended" by way of gamma functions (i.e the integrand is really $\frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(n-k+1)}$). 
What I found, experimentally, is pretty cool. It seems that the following is true:
$$I(n)=\frac{2}{\pi} \sum_{i=1}^n \binom{n}{i}\operatorname{SinInt}(\pi i)$$
Where $\operatorname{SinInt}(x)$ is the Sine Integral, or $\int_0^x \frac{\sin t}{t}dt$. 
To me, this is quite interesting as the Sine Integral tends to $\pi/2$ so the above formula will tend to $2^n$, so the integral is just the sum with some error term. But how would I go about proving it?

REPLY [14 votes]: When I was your age (15 or so), I also played around with a very similar integral. My take was slightly different.
First of all, for suitable real numbers $s$ (say with positive real part) one actually has the sum
$$\sum_{-\infty}^\infty \binom{s}{k} = 2^s.$$
So a more natural integral to consider might be
$$I(s):=\int_{-\infty}^\infty \frac{\Gamma(s+1)}{\Gamma(x+1) \Gamma(s+1-x)} \, dx.$$
You can prove $I(n) = 2^n$ for non-negative integers $n$. The point here is that one can use the reflection formula for the $\Gamma$ function. For example, in the easiest case when $s = 0$, the integral becomes
\begin{align}
& \int_{-\infty}^\infty \frac{1}{\Gamma(x+1) \Gamma(1-x)} \, dx \\[10pt]
= {} & \int_{-\infty}^\infty \frac{1}{x \Gamma(x) \Gamma(1-x)} \, dx \\[10pt]
= {} & \int_{-\infty}^\infty \frac{\sin(\pi x)}{x \pi} \, dx = 1.
\end{align}
For larger integers $n$, you can do pretty much the same thing and then use partial fractions. The same argument will work with your integral, except now the sin integral will go from $0$ to $n$ rather than $\infty$, and hence you pick up the corresponding functions.
(and looks like someone has done that).
This doesn't work for general $s$, however. It took me quite a while to work out, but eventually I found a few different arguments to prove that $I(s) = 2^s$, including a fairly clean proof by contour integration.
Of course, all this was in the days before the internet, so I had to figure it out on my own by thinking about it (even though it was surely known before and also not that hard in the end --- there are very few truly deep definite integrals). I feel a little sad that you can just ask the question and someone will come and answer it.