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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?

Anixx
  • 9,302
  • I am not sure about all of the details, but I would imagine that there's some way to use ultrafilters to formalize this rigorously. Since every continuous function from $\Bbb R \to \Bbb R$ is determined only by its values on $\Bbb Q$, you could look, for instance, at the set of all functions $\Bbb Q \to \Bbb R$, noting that every continuous function from $\Bbb R \to \Bbb R$ will be in there. Since the integral of any integral function is continuous, then all integrals will also be in there, and in particular all integrals of the form $\int_0^xf(u)du$. (1/2) – Mike Battaglia Oct 18 '22 at 18:00
  • You would like to assign some kind of extended real "value" to $\int_0^\infty f(u) du$. To do this we can just let that "value" be the whole function $\int_0^x f(u) du$, treated as an element in our set, and then look at ways to put a total order on the set. This is not much different than what we do with hyperreal numbers, except we're looking at functions $\Bbb Q \to \Bbb R$ instead of $\Bbb N \to \Bbb R$, but we can still build an ultrafilter and so on. (2/2) – Mike Battaglia Oct 18 '22 at 18:08
  • As a last point, I'm not really sure if it's really necessary to go to $\Bbb Q \to \Bbb R$ at all; the usual hyperreals with $\Bbb N \to \Bbb R$ may be alright. If we just say that $\omega$ is equal to the equivalence class with the sequence $(0,1,2,3,4,...)$ in it, we can certainly look at $\log(\omega)$, or the digamma function $\psi(\omega)$, or the n'th harmonic number evaluated at $H_\omega$, and I think we will get the same basic results as if we do things with a domain of $\Bbb Q$. In particular, $\log(\omega) - \psi(\omega)$ will be some infinitesimal. – Mike Battaglia Oct 18 '22 at 18:14
  • @MikeBattaglia this may be relevant: https://mathoverflow.net/questions/115743/an-algebra-of-integrals/342651#342651 – Anixx Oct 19 '22 at 03:23
  • The logarithm base $b$, $log_b(x)$ gives the number of digits required to represent the number $x$ in base $b$ (negative means counting decimals). Probably infinites in functions represent finite objects with different dimension than the variable $x$ used. (This is purely an speculative idea intended to stimulate creative thinking). – tutizeri Oct 31 '23 at 02:34

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