Let's consider the logarithm of the operator
$xD=\left( \begin{array}{ccccccc} 0 & 0 & 0 & 0 & 0 & 0 & . \\ 0 & 1 & 0 & 0 & 0 & 0 & . \\ 0 & 0 & 2 & 0 & 0 & 0 & . \\ 0 & 0 & 0 & 3 & 0 & 0 & . \\ 0 & 0 & 0 & 0 & 4 & 0 & . \\ 0 & 0 & 0 & 0 & 0 & 5 & . \\ . & . & . & . & . & . & . \\ \end{array} \right)$
My take is straightforward. Formally, the computer algebra system gives (as a matrix logarithm)
$\ln(xD)=\left( \begin{array}{ccccccc} -\infty & 0 & 0 & 0 & 0 & 0 & \cdot \\ 0 & 0 & 0 & 0 & 0 & 0 & . \\ 0 & 0 & \ln 2 & 0 & 0 & 0 & . \\ 0 & 0 & 0 & \ln 3 & 0 & 0 & . \\ 0 & 0 & 0 & 0 & \ln 4 & 0 & \cdot \\ 0 & 0 & 0 & 0 & 0 & \ln 5 & . \\ . & . & . & . & . & . & . \\ \end{array} \right)$
I only replaced $\ln 0$ in the first cell with $-\infty$. This expression can be verified by taking exponential: $\exp(\ln (xD))=xD$ if we assume $e^{-\infty}=0$.
But many sources (https://arxiv.org/abs/1105.5978, https://mathoverflow.net/questions/111165/riemann-zeta-function-at-positive-integers-and-an-appell-sequence-of-polynomials and others) give another expression:
$\ln (xD)=\left( \begin{array}{ccccccc} \psi (1) & 0 & 0 & 0 & 0 & 0 & \cdot \\ 0 & \psi (2) & 0 & 0 & 0 & 0 & \cdot \\ 0 & 0 & \psi (3) & 0 & 0 & 0 & \cdot \\ 0 & 0 & 0 & \psi (4) & 0 & 0 & \cdot \\ 0 & 0 & 0 & 0 & \psi (5) & 0 & \cdot \\ 0 & 0 & 0 & 0 & 0 & \psi (6) & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ \end{array} \right)$
It fails the verification. I wonder, what's the point here, if it does not give an inverse function to exponential? Can we call it "logarithm" then?
How such expression could arise at all? I admit, I myself arrived at the same result with a Mathematica code.
What happens here? Is it that some hidden regularization happens that makes the operation non-invertible? Something else? Why the wrong matrix is so often suggested as the operator?
