Can ln(0) be defined if there are nonzero a,b with ab=0 (zero divisors?)

Question

My "proof" would be as follows:

ε2 = 0
ln 0 = ln ε2
ln x2 = 2 ln x
ln 0 = 2 ln ε

(a=b=ε) isn't really necessary.

 ln 0 = ln ab
ln ab = ln a + ln b

If ln(0) is always undefined, then it implies ln(ε) must be as well, right? (Or the sum ln(a)+ln(b) is undefined, which technically leaves room for defined ln(a) and ln(b) without defined addition, whatever that means.)

I think I know from the definition of exp(xε) that ln(ε) should be undefined. My confusion comes from the fact that there are all different definitions of multiplication that have implications for the addition of exponents, e.g. exp(i+j) ≠ exp(j+i) in quaternions. So... couldn't there be all different definitions of addition and its identities? Is exp necessarily nonzero no matter how you define zero?

I tried to work with c := ln(0); exp(c) = 0 but I couldn't really reason about its properties.

Answer 1

The property $\ln x^a=a\ln x$ is not universal: for instance, if $x=-1, a=2$, we have $\ln (-1)^2=\ln 1=0\ne 2\ln(-1)=2i\pi$

You definitely can extend real numbers with $\ln 0$. Notice though that this will not be a real number.

One way of doing so is via divergent integrals or germs.

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

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

• 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 $-\gamma$.

• 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 $\int_0^\infty \frac1tdt=-2\lambda-\gamma$. Thus, $\lambda$ is the germ at infinity of the function $-\ln x - \gamma$.

• 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 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.

• Since $-\lambda-\gamma=\int_1^\infty \frac1t dt$ corresponds to the germ of the function $\ln x$ at infinity, the germ of the function $f(x)=x$ at infinity, often denoted as $\omega$ is $e^{-\lambda-\gamma}$.

• Via Laplace transforms we can see that $\omega=\int_0^\infty dx=\int_0^\infty\frac1{x^2}dx$, so the germ of function $\frac1x$ at zero (from the positive side) is also $\omega$.

• As such, we can write down the expression $\ln \varepsilon=\lambda-\varepsilon \omega=\lambda-\varepsilon e^{-\lambda-\gamma}=\ln 0-\varepsilon e^{-\ln 0-\gamma}$.