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?
Best Answer
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.