Nörlund–Rice integral

Remarks

The proof is the residue theorem. Inside $\gamma$ the integrand has simple poles only at $z = k$ with $\alpha \le k \le n$, and the residue there is $f(k)\, n! / \prod_{j \ne k} (k - j) = (-1)^{n-k} \binom{n}{k} f(k)$. Multiplying by $(-1)^n$ gives the terms $(-1)^k \binom{n}{k} f(k)$. For $\alpha = 0$ the left side is $(-1)^n \Delta^n f(0)$, an $n$-th forward difference, so the formula turns a high-order difference into a contour integral. Example: for $f = 1$ and $\alpha = 0$, the left side is $(1 - 1)^n = 0$. On the right, the integrand is $O(\abs{z}^{-2})$, and the integral over a circle does not depend on its radius once $\gamma$ encloses $0, \dots, n$. So it equals its limit as the radius grows, which is $0$.

Going deeper

For a student

The integrand is $f(z)$ times the kernel $n!/(z(z-1)\cdots(z-n)) = \Gamma(n+1)\Gamma(z-n)/\Gamma(z+1)$, so its size away from the integers is controlled by the gamma function. To get asymptotics one extends $f$ meromorphically, enlarges the contour, and collects the residues at the poles of $f$ that are crossed. This method is a standard tool in the analysis of algorithms, for example for digital trees and related recursive structures. It is closely related to the Mellin transform method for harmonic sums.