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?

  • $\begingroup$ Since $n$ does not change as $k$ goes from $0$ to $n,$ one has $\displaystyle I(n) = n!\int_0^n \frac{\mathrm dk}{k!(n-k)!}. \qquad$ $\endgroup$ Sep 26, 2018 at 22:43
  • $\begingroup$ How did you find it experimentally? Using some tools? $\endgroup$
    – Dilworth
    Sep 26, 2018 at 22:51
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    $\begingroup$ @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. $\endgroup$
    – Nico A
    Sep 26, 2018 at 22:54
  • $\begingroup$ I see, thanks! - $\endgroup$
    – Dilworth
    Sep 26, 2018 at 22:56

2 Answers 2


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.

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    $\begingroup$ 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. $\endgroup$
    – Nico A
    Sep 26, 2018 at 22:46
  • $\begingroup$ I would be much interested to know your proof of $I(s)=2^s$ $\endgroup$
    – G Cab
    Nov 3, 2020 at 11:37
  • $\begingroup$ @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 ;) $\endgroup$
    – metamorphy
    Jan 3, 2021 at 19:08
  • $\begingroup$ @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! $\endgroup$
    – G Cab
    Jan 3, 2021 at 19:51
  • $\begingroup$ @GCab: I've appended a proof to that answer (it looks contextually better there). $\endgroup$
    – metamorphy
    Jan 5, 2021 at 9:33

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