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Chapter 5.5 of Concrete Mathematics discusses generalizing binomial coefficient identities to the Gamma function. It doesn't discuss the two integrals you mention, though.

Doing a bit of thinking on my own, if $n$ is a positive integer then $$\int_{z=0}^n \binom{n}{z} dz = \int_{z=0}^n \frac{n! dz}{\Gamma(1+z) \Gamma(n+1-z)}$$ $$\int_{z=0}^{n} \frac{n! dz}{(n-z)(n-1-z) \cdots (1-z) \Gamma(1-z) \Gamma(1+z)}.$$

We have $\Gamma(1+z) \Gamma(1-z) = \pi z/\sin (\pi z)$, if I haven't made any dumb errors, so this is $$\int_{0}^n \frac{ n! \sin (\pi z) \ dz}{z dz}{\pi z (n-z)(n-1-z) \cdots (1-z)}.$$

I suspect this integrand does not have an elementary anti-derivative, because it reminds me of $\int \sin t \ dt/t$. But there might be some special trick which would let you compute the integral between these specific bounds.

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Chapter 5.5 of Concrete Mathematics discusses generalizing binomial coefficient identities to the Gamma function. It doesn't discuss the two integrals you mention, though.

Doing a bit of thinking on my own, if $n$ is a positive integer then $$\int_{z=0}^n \binom{n}{z} dz = \int_{z=0}^n \frac{n! dz}{\Gamma(1+z) \Gamma(n+1-z)}$$ $$\int_{z=0}^{n} \frac{n! dz}{(n-z)(n-1-z) \cdots (1-z) \Gamma(1-z) \Gamma(1+z)}.$$

We have $\Gamma(1+z) \Gamma(1-z) = \pi z/\sin (\pi z)$, if I haven't made any dumb errors, so this is $$\int_{0}^n \frac{ n! \sin (\pi z) \ dz}{z (n-z)(n-1-z) \cdots (1-z)}.$$

I suspect this integrand does not have an elementary anti-derivative, because it reminds me of $\int \sin t \ dt/t$. But there might be some special trick which would let you compute the integral between these specific bounds.