Newton series is the following expansion of a function:

$$f(x)=\sum_{k=0}^\infty \binom{x}k \Delta^k [f]\left (0\right)=\sum_{n=0}^{\infty} {x\choose n} \sum_{k=0}^n{n\choose k}(-1)^{k-n}f(k)$$

Now from this paper we have the following identities for Zeta and Digamma functions:

$$\zeta (s)={\frac 1{s-1}}\sum _{{n=0}}^{\infty }{\frac 1{n+1}}\sum _{{k=0}}^{n}{n \choose k}{(-1)^{k} \frac {1}{(k+1)^{{s-1}}}}$$

$$\psi(z)=\sum_{n=0}^\infty \frac1{n+1} \sum_{k=0}^n {n\choose k}(-1)^k\ln(z+k)$$

They seem to be strikingly similar to Newton's series, especially if we re-write them the following way:

$$\zeta (s)={\frac 1{s-1}}\sum _{{n=0}}^{\infty }{-1 \choose n}(-1)^{-n}{\frac 1{n+1}}\sum _{{k=0}}^{n}{n \choose k}{(-1)^{k} \frac {1}{(k+1)^{{s-1}}}}$$

$$\psi(z)=\sum_{n=0}^\infty {-1 \choose n}(-1)^{-n}\frac1{n+1} \sum_{k=0}^n {n\choose k}(-1)^{k}\ln(z+k)$$

Given the noticed similarity between Newton's series and Fourier transform this seems similar to the Fourier differintegral:

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

Thus I wonder whether the given formulas can be interpreted as doing "newtonian transform", then dividing by $(n+1)$ which is analogous (?) with dividing with $(-i \omega)$ of Fourier transform and then making the "inverse Newtonian transform".

Particularly, I wonder, to which operation on straight function corresponds dividing its "Newtonian image" by $(n+1)$. Is there any table of such operations on images, similar to Fourier transform?