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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.

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.

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 $2^{n-1}n$.

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.