Discrete differintegral using Fourier transform?

Question

Using Fourier transform we can give a formula for a differintegral:

$$f^{(a)}(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty} e^{- i \omega x}{(-i\omega)}^{a} \int_{-\infty}^{+\infty}f(t)e^{i\omega t}dt \, d\omega$$

I wonder, is there a similar formula for a discrete differintegral (finite difference/indefinite sum)?

Answer 1

Well,

$$(e^D-1)^a f(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty} e^{- i \omega x}(e^{-i\omega}-1)^a \int_{-\infty}^{+\infty}f(t)e^{i\omega t}dt \, d\omega$$