If we take the MacLaurin series for $\ln(x+1)$ and evaluate it at $x=-1$, we will get the Harmonic series with the opposite sign: $-\sum_{k=1}^\infty \frac1x$. Since the regularized sum of the Harmonic series is $\gamma$, it follows that the regularized value of logarithm at zero is $-\gamma$ (Euler-Mascheroni constant). One can see this via other ways as well.
Now, in umbral calculus we can consider index-lowering (evaluation) operator as some kind of "regularization" (particularly because it is exacly regularization if we represent the umbra as formal power series).
Then (because $\operatorname{eval}\ln (B+x)=D\Delta^{-1}f(x))$), $\operatorname{eval}\ln (B+x)=\psi(x)$ (where $B$ is Bernoulli umbra and $\psi(x)$ is digamma function).
This directly gives us that the regularized value of $\ln(B+1)$ is $-\gamma$ (the result is obtained here as well, but they use symbol $B$ for Bernoulli umbra plus $1$).
At $x=0$ the function $\psi(x)$ has a pole, so we need another regularization step: taking Cauchy principal value: $\lim_{h\to0}\frac{\psi(x+h)+\psi(x-h)}2=-\gamma$.
So in cases of $0$, $B$ and $B+1$ we have the regularized values of logarithm equal to $-\gamma$, and in the case of umbral calculus, values of the evaluation of logarithm at other points in which the constant $-\gamma$ appears are also common (for instance, $\operatorname{eval}\ln (B+1/2)=-\gamma-\ln 4$).
On the other hand, the regularized values of logarithm at zero divisors also involve $-\gamma$, for instance, in $\mathbb{R}^2$ and split-complex numbers, where this directly follows from the regularized value of logarithm of zero. For instance, $\operatorname{reg}\ln(j/2+1/2)=j\gamma/2-\gamma/2$ and $\operatorname{reg}\ln(j+1)=j\gamma/2-\gamma/2+(j+1)\ln \sqrt{2}$.
Thus, my question is, what is the intuition, why the regularized values of logarithm at $0$, $B$, $B+1$ is $-\gamma$? What's the analogy between Bernoulli umbra and zero divisors?
