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Second, new example in the same vein added.

Integral of power of binomials equal to sum of power of binomials?

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.