# 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?

• Since $n$ does not change as $k$ goes from $0$ to $n,$ one has $\displaystyle I(n) = n!\int_0^n \frac{\mathrm dk}{k!(n-k)!}. \qquad$ Sep 26, 2018 at 22:43
• How did you find it experimentally? Using some tools? Sep 26, 2018 at 22:51
• @Dilworth I myself was able to get a closed form for $n=0$, and was able to manipulate the integrand into a form that Mathematica could integrate using $n=1$. Saw the sine integrals, recognized the coefficients, conjectured the above formula and then verified with numerical integration. Sep 26, 2018 at 22:54
• I see, thanks! - Sep 26, 2018 at 22:56

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.

• Thank you so much for the story to go along with your answer. Rest assured I spend much more time playing around with math on my own than I spend asking about it here! I just haven't actually had a formal calculus class so integrals with special functions always tend to scare me off :) But this was too cool to ignore. I'm currently trying to teach myself contour integration, so I'll search for the proof you mention! Once again, thank you. Sep 26, 2018 at 22:46
• I would be much interested to know your proof of $I(s)=2^s$ Nov 3, 2020 at 11:37
• @GCab: This user hasn't visited the site since this answer. Perhaps, the problem is worth posting as a dedicated question (if it's not done already by someone else). This would give someone (like me ;) an opportunity to provide an answer, although the ideas above (as well as the problem itself) are already a good brain teaser to try it by oneself ;) Jan 3, 2021 at 19:08
• @metamorphy: thanks indeed for your interest and suggestion. Actually I already posted here a bit wider question on what is $$f(z,r) = \int_0^\infty {\binom{ r,t} z^{\,t} dt}$$ and .. you have been the only one to provide a partial answer :). I would be glad if you can add more! Jan 3, 2021 at 19:51
• @GCab: I've appended a proof to that answer (it looks contextually better there). Jan 5, 2021 at 9:33