Integral of Binomial Coefficient 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? 
 A: Properties of the $\Gamma$-function, together with partial fraction expansion, do the trick. We have
$$\Gamma(x+1)\Gamma(n-x+1)=x\Gamma(x)\Gamma(1-x)\prod_{k=1}^{n}(k-x)=\frac{(-1)^n\pi}{\sin\pi x}\prod_{k=0}^{n}(x-k),$$
so our integral is $I(n)=\displaystyle\frac{(-1)^n n!}{\pi}\int_{0}^{n}\frac{\sin\pi x\,\mathrm{d}x}{\prod_{k=0}^{n}(x-k)}$. Doing partial fractions, we have
$$\prod_{k=0}^{n}(x-k)^{-1}=\sum_{k=0}^{n}\frac{a_k}{x-k},\quad a_m=\prod_{\substack{0\leq k\leq n\\k\neq m}}(m-k)^{-1}=\frac{(-1)^{n-m}}{m!(n-m)!}$$
(say, multiplying by $x-m$ and letting $x\to m$). Thus we get
$$I(n)=\frac{1}{\pi}\int_{0}^{n}\sum_{k=0}^{n}(-1)^k\binom{n}{k}\frac{\sin\pi x}{x-k}\,\mathrm{d}x=\frac{1}{\pi}\sum_{k=0}^{n}\binom{n}{k}\int_{-k\pi}^{(n-k)\pi}\frac{\sin t}{t}\,\mathrm{d}t.$$
Simplification of this, using the sine integral function, gives the expected result.
A: 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. 
