Integral of Binomial Coefficient

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.

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.