Inspired by this MO question about integrating binomial coefficients and the answers, I was wondering whether integrating powers of binomial coefficients also relates to the respective sums. And indeed I have strong numerical evidence that $$\int_{-\infty}^{\infty} \binom{n}{x}^2 dx =\sum_{k=0}^n\binom{n}{k}^2$$. (The latter expression is of course $\binom{2n}{n}$.)

So I'm wondering for which $l$ the following identity holds: $$\int_{-\infty}^{\infty} \binom{n}{x}^l dx =\sum_{k=0}^n\binom{n}{k}^l$$.

(And furthermore one could conjecture that there are similar examples, where sum over binomials is identical to integral of the same expression over real numbers.)

EDIT: Regarding the last sentence here an example: I conjecture (and have numerical evidence) that $$\int_{-\infty}^{\infty} x\binom{n}{x} dx =\sum_{k=0}^n k\binom{n}{k},$$ the latter expression being of course $2^{n-1}n$.

EDIT2: And for Vandermonde's identity it seems also (by numerical evidence) to work analogously: $$ \int_{-\infty}^{\infty} \binom{m}{x} \binom{n}{r-x} dx = \sum_{k=0}^r \binom{m}{k} \binom{n}{r-k},$$ the latter expression being of course $\binom{m+n}{r}$.

I dare to conjecture that one can still find more examples.

  • $\begingroup$ Did you check any other values of $\ell$? $\endgroup$
    – Lucia
    Nov 27, 2016 at 21:02
  • 1
    $\begingroup$ Your last statement is just an easy corollary: $k\binom{n}{k}$ is just $n \binom{n-1}{k-1}$, and similarly for $x\binom{n}{x}$. $\endgroup$
    – Lucia
    Nov 27, 2016 at 21:20
  • $\begingroup$ the $l$-dependent identity only seems to hold for $l=1,2$ $\endgroup$ Nov 27, 2016 at 21:23
  • $\begingroup$ BTW, the RHS sum fails to have a closed form beginning with $\ell=3$. Also, I don't expect the generalization to hold true. $\endgroup$ Nov 27, 2016 at 21:58
  • $\begingroup$ Your second identity follows also from Parseval/Plancherel as in my answer. $\endgroup$
    – Lucia
    Nov 27, 2016 at 22:25

2 Answers 2


The generalization looks like this $$ \int_{-\infty}^{\infty} \binom{n}{\alpha x}^l dx =\sum_{k=-\infty}^\infty\binom{n}{\alpha k}^l,\quad 0<\alpha\le 2/l,~l\in\mathbb{N}\tag{1} $$ where $n$ need not be an integer. The general theorem is given for example in the paper Surprising sinc sums and integrals. Below I give the general outline of the proof which is based on the well known fact that the following function is band limited (its Fourier transform has limited spectrum) $$g(x)=\binom{n}{x}=\frac{1}{2\pi}\int_{-\pi}^{\pi} (1+e^{i \omega})^n e^{- ix\omega} d\omega$$ One can see that Fourier transform is limited to frequencies $|\omega|<\pi$. Whenever spectrum of a function $f(x)$ is limited to frequencies $|\omega|<2\pi$ one expects that $$ \int_{-\infty}^{\infty} f(x) dx =\sum_{k=-\infty}^\infty f(k). $$

Now the Fourier transform of $g(\alpha x)^l$ has a spectrum limited in the band $|\omega|<\pi\alpha l$. This is easy to see calculating Fourier transform $$ \int_{-\infty}^{\infty}g(\alpha x)^le^{-ikx}dx $$ with the help of $\int_{-\infty}^{\infty}e^{-ikx}dx=2\pi\delta(k)$, where $\delta$ is delta function. The general theorem from the paper cited above now states that $$ \int_{-\infty}^{\infty} g(\alpha x)^l dx =\sum_{k=-\infty}^\infty g(\alpha k)^l,\quad 0<\pi\alpha l\le 2\pi,~l\in\mathbb{N}, $$ which is equivalent to (1).

All this analysis also explains that when $\alpha=1$ the proposed identity holds for $l=1,2$ but not for larger $l\in\mathbb{N}$.


I don't know if there is a relation for values of $\ell$ apart from $1$ and $2$ (that would be very interesting, and surprising to me), but here is a unified way to look at what's going on for exponents $1$ and $2$.

Consider the function on ${\Bbb R}$ defined by $$ f(x) = (1+e^{2\pi i x})^n $$ for $-1/2 \le x \le 1/2$ and $f(x) = 0$ if $|x| >1/2$. We can compute its Fourier transform: $$ {\widehat f}(\xi) = \int_{-\infty}^{\infty} f(x) e^{-2\pi i x\xi} dx, $$ and this turns out to be $$ \binom{n}{\xi} = \frac{n!}{\Gamma(\xi+1) \Gamma(n-\xi + 1)}. $$ (Check for $n=1$, and then use $\binom{n}{\xi} = \binom{n-1}{\xi} + \binom{n-1}{\xi -1}$). So now Fourier inversion gives $$ f(0) = 2^n = \int_{-\infty}^{\infty} \binom{n}{\xi} d\xi, $$ which is the case $\ell =1$.

Next Plancherel gives $$ \int_{-1/2}^{1/2} |f(x)|^2 dx = \int_{-\infty}^{\infty} \binom{n}{\xi}^2 d\xi. $$ Now instead of $f$ consider the $1$-periodic function $F$ defined by $F(x) = f(x)$ on $(-1/2,1/2)$ and extended periodically. The Fourier coefficients of $F$ are $$ {\widehat F}(k) = \int_{-1/2}^{1/2} f(x) e^{-2\pi ik} dx = \binom{n}{k}, $$ and so Parseval gives $$ \sum_{k} {\widehat F}(k)^2 =\sum_{k=0}^{n} \binom{n}{k}^2 = \int_{-1/2}^{1/2} f(x)^2 dx. $$ This gives the $\ell=2$ identity.

  • $\begingroup$ Many thanks. That's very nice. (I will wait for other answers, but will probably accept your answer.) $\endgroup$ Nov 27, 2016 at 21:19
  • $\begingroup$ There is a relation for values of $ℓ$ apart from 1 and 2 $\endgroup$
    – Nemo
    Dec 24, 2016 at 7:18

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