Extending reals with logarithm of zero: properties and reference request

Question

If we take logarithmic function, we can see that its real part at zero approaches negative infinity with the same rate and sign from any direction on the complex plane, while the Cauchy main value of the imaginary part averaged over any circle around zero is zero.

This may hint us that in any algebras that include germs or growth rates of functions at a point, logarithm of zero quite consistently represents a reasonable negatively-infinite constant.

This extension of real numbers would be analytic as opposed to algebraic (one can oppose such extension on algebraic grounds, because logarithm of zero necessarily breaks some algebraic properties of logarithm, but they do not hold on the complex plane anyway).

Let us denote it as $\lambda=\ln 0$ and sum up some of its properties.

• The Maclaurin series of the function $\ln (x+1)$ at $x=-1$ is the Harmonic series with negative sign, thus, we can represent $\lambda=-\sum_{k=1}^\infty \frac1k$. Since the Harmonic series has the regularized value of $\gamma$ (Euler-Mascheroni constant), the regularized value (finite part) of $\lambda$ is $-\gamma$.

• If we define $\ln x=\int_1^x \frac1t dt$, which is the generalization of logarithmic function, we can represent $\lambda$ as divergent integral: $\lambda=-\int_0^1 \frac1t dt$.

• Since $\lim_{n\to\infty}\left(\sum_{k=1}^n \frac1k-\int_1^n \frac1tdt\right)=\gamma$, we can also represent $\lambda=-\int_1^\infty \frac1t dt-\gamma$. It also will tell us that $\int_0^\infty \frac1tdt=-2\lambda-\gamma$.

• We can find other integral representations of this constant: $\int_0^\infty \frac{1-e^{-t}}{t} dt$, $\int_0^\infty \frac{dt}{t + t^2}$, and others.

• Since $\int_0^\infty\frac{\ln t+\frac{\gamma }{2}}{t}=-\int_0^\infty\frac{\ln t+\frac{\gamma }{2}}{t}$, it is equal to zero. So, $\int_0^\infty \frac{\ln t}{t} \, dt=-\frac\gamma2\int_0^\infty \frac1tdt=\gamma\lambda+\gamma^2/2$

• If we generalize the notions of periods and $EL$-numbers to our extended set, then $\lambda$ would be both, because it can be represented as $\int_0^1 \frac{-1}t dt$ (integral of an algebraic function over algebraic domain) and $\ln 0$ respectively. On the other hand, $\lambda+\gamma=\int_1^\infty \frac{-1}t dt$ would belong to neither.

• Since we can take logarithm of zero, we also can take logarithms of zero divisors in split-complex numbers: $\ln \left(\frac{a j}{2}+\frac{a}{2}\right)=\frac{j}{2} (\ln a-\lambda)+\frac{1}{2} (\ln a+\lambda)$

• In dual numbers the value of $\varepsilon^\varepsilon$ is usually undefined. But due to the general formula $f(\varepsilon)=f(0)+\varepsilon f'(0)$ and the fact that $(x^x)'=x^x (\ln x+1)$, we can derive $\varepsilon^\varepsilon=1+\varepsilon(1+\lambda)$, which shows a surprising role played by this constant in dual numbers.

That said, I wonder:

• What are some other notable properties of this constant, where can it occur?

• Are there works in which extension of reals with logarithm of zero is seriously considered?

Answer 1

Don't know whether this really matches your query, but casually I've tried to make some sense with this $\log(0)$ thing, via the $\log(1+x)$ series for the argument $x \to -1$. This occured as problem in the inversion of the Carleman-matrix for the function $e^x$ which involves the attempt to do a forbidden matrix-inverse of infinite size.

Let's define the Carleman matrix $B$ such that, with a vector $V(x)=[1,x,x^2,x^3,...]$ we get $$ V(x) \cdot B = V(e^x) \tag 1$$ The infinite-sized matrix $B$ has the top-left edge

  1       1      1        1         1           1        1           1
  0       1      2        3         4           5        6           7
  0     1/2      2      9/2         8        25/2       18        49/2
  0     1/6    4/3      9/2      32/3       125/6       36       343/6
  0    1/24    2/3     27/8      32/3      625/24       54     2401/24
  0   1/120   4/15    81/40    128/15      625/24    324/5   16807/120
  0   1/720   4/45    81/80    256/45    3125/144    324/5  117649/720
  0  1/5040  8/315  243/560  1024/315  15625/1008  1944/35  117649/720

and the definition $$B[r,c]=c^r / r! \tag 2$$.

This matrix is not invertible/reciprocable when infinite size is assumed.
However, a partial procedure for the inversion can be done, and in some situations one can then evaluate a matrix-formula step-by-step with the same result, as if we really had $C=B^{-1}$ - which would be the Carlemanmatrix for $\log(x)$ developed around $0$ (not around $1$ which gives the well known Mercator-powerseries).

The way to go is the separation of $B$ in two triangular matrix-factors to which I assigned named $fS2F$ (indicating factorial scaling of the matrix $S_2$ of Stirlingnumbers $2$'nd kind) and the upper triangular Pascalmatrix $P$ such that $$ B = fS2F \cdot P \tag 3 $$ Here we find, that $fS2F$ is the Carlemanmatrix for the function $x \to e^x-1$ and $P$ is the Carlemanmatrix for the function $x \to x+1$ . This means, we can formulate this in the above style of formal powerseries: $$ \begin{array} {} V(x) \cdot fS2F & = V(e^x-1) \\ V(e^x-1) \cdot P &= V(e^x) \\ &= (V(x) \cdot fS2F) \cdot P \\ & = V(x) \cdot (fS2F \cdot P) \\ & = V(x) \cdot B \end{array} \tag 4$$

This matrix-decomposition and the functional relation between the Stirlingnumbers 2'nd kind and the $e^x-1$ function is in principle all well known, without involving the name of "Carlemanmatrix" I found this for instance in Abramowitz/Stegun as well as in L. Comtet and elsewhere. It is as well easily found by application of the $LR$ (or "$LU$")-matrix-decomposition.

Now, the two matrices $fS2F$ and $P$ are triangular, with $1$ in the diagonals and can thus be inverted even in the case of infinite size which gives the two triangular matrices $ fS1F = fS2F^{-1}$ and $Q=P^{-1}$.

This matrices are consequently the Carlemanmatrices for $x\to\log(1+x)$ and $x\to x-1$ and the inverse of $B$ could be done if the matrixproduct $$ Q \color{red}{ \overset{\text{no!}}{\cdot}} fS1F \tag 5 $$ could be evaluated, but which cannot be done due to singularities in the dotproducts of rows in $Q$ and columns in $fS1F$ .

Interestingly, but no surprise, we can apply the associativity of matrix-multiplication $$ \begin{array}{} \quad V(x) \cdot (Q \color{red}{ \overset{\text{no!}}{\cdot}} fS1F) \\ \quad (V(x) \cdot Q) \color{red}{ \overset{\text{?}}{\cdot}} fS1F \\ =V(x-1) \cdot fS1F \qquad \qquad \text{for } x \ne 0\\ = V(\log(1+(x-1)) \end{array} \tag 6 $$

Formally the dotproduct of the first row in $Q$ and second column in $fS1F$ gives the harmonic series $ 1\cdot 0 - 1/1 -1/2-1/3 - \cdots $ . Similar divergent expressions occur with the dotproducts of the following rows in $Q$ which would give then the derivatives of that $\log(1+(x-1)$-construction. Looking at the next columns in $fS1F$ give similar expressions; if $x \ne 0$ we had simply $\log(x)^c$ where $c$ is the index of the column in $fS1F$.

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I didn't find a smooth solution for insertion of "regularized" values into an attempted Carlemanmatrix $C$ (and do something meaningful with it), and stopped with that exercising. But perhaps this problem is simply one further possible place for application of your shown concept.

Answer 2

I wrote a paper in 2001 that addressed to some extent your second question: https://arxiv.org/abs/math/0112050 Unknown to me at the time is the fact that Albert A. Bennett derived some of the same results in 1915: Bennett, Albert A. "Note on an Operation of the Thrid Grade." The Annals of Mathematics 17.2 (1915): 74-75.