Notation. Here I will denote Bernoulli umbra (its moments are Bernoulli numbers $B_n$) as $B_-$, $B_-+1$ as $B_+$ (an umbra with moments being Bernoulli numbers except $B_1=1/2$).
I will denote the index-lowering (evaluation) operator as $\operatorname{eval}$.
A general formula holds: $\operatorname{eval}f(B_-+x)=\frac{D}{e^D-1} f(x)$.
I am very much puzzled why umbral calculus is so much similar to the split-complex numbers, and Bernoulli umbra has common properties with a zero divisor (namely, $j/2-1/2$).
Particularly,
The evaluation $\operatorname{eval} B_-=-1/2$, $\operatorname{eval} B_+=1/2$, this is similar to zero divisors $j/2-1/2$ and $j/2+1/2$, the second one being idempotent and the first one being negative of an idempotent.
The $\operatorname{eval}\frac1{B_-+a}=\psi^{(1)}(a)$ ($\psi$ is the polygamma function). This function has a pole at $a=0$, so we cannot divide by Bernoulli umbra.
The evaluation of logarithm of Bernulli umbra is $\operatorname{eval}\ln (B_-+a)=\psi(a)$. This function also has a pole at $a=0$, so we cannot take logarithm of it either.
The regularized value of the logarithm of the zero divisor $j/2+1/2$ is $j\gamma/2-\gamma/2$. The Cauchy main value (e.g. value after removal of the pole) of the evaluation of logarithm of $B_-$ is $-\gamma$, the evaluation of logarithm of $B_+$ is $-\gamma$ as well.
The powers of the umbras $B_-^n$ and $B_+^n$ play the role of diagonal basis is split-complex numbers. At the same time, $e_n=B_+^n-B_-^n$ plays the role of straight basis. It contains the multiplicative unity $B_+-B_-=1$ and $B_+^2-B_-^2=B_-+B_+$, which plays the role, similar to $j$ in split-complex numbers. In split-complex numbers we also have $1=(j/2+1/2)-(j/2-1/2)$ and $j=(j/2+1/2)^2-(j/2-1/2)^2=(j/2+1/2)+(j/2-1/2)$.
If we define umbral-style multiplication on divergent integrals, the set becomes isomorphic to a subset of umbral calculus with regularization playing the role of evaluation. In this setting, $B_-=\int_{-1/2}^\infty dx=\sum_{k=0}^\infty1$ and $B_+=\int_{1/2}^\infty dx=\sum_{k=1}^\infty1$.The role of $j$ is played by $B_-+B_+=\int_{-\infty}^\infty dx$.
In straight basis there are formulas for divergent integrals: $\int_0^\infty x^n dx=\frac{B_+^{n+2}-B_-^{n+2}}{(n+1)(n+2)}=\frac{e_{n+2}}{(n+1)(n+2)}$ and $\int_0^\infty \frac1{x^n} dx=\frac{B _+^{n}-B _-^{n}}{(n-1)n!}=\frac{e_n}{(n-1)n!}$.
It seems, the straight basis is good for working with integrals while diagonal basis is good for working with series.
Thus my question is: what makes Bernoulli umbra so much similar to zero divisors in split-complex numbers, even though it is not a zero divisor? Why we cannot divide by or take logarithm of Bernoulli umbra? Why there is straight and diagonal basis?
Also, trivial multiplication of divergent integrals (in the style of Hardy or Levi-Civita fields) is analogous to the dual numbers (in the sense that regularized part of the product is the product of regularized parts, similar to real part in dual numbers). So, we have one system similar to dual numbers, one system similar to split-complex numbers, is there a definition of multiplication of divergent integrals or umbras similar to complex numbers in this sense?
