natural logarithm

The natural logarithm of a number is the logarithm in base of Euler’s number $e$. It can be defined as the map $\ln :{ℝ}_{+}\to ℝ$ satisfying

$$\ln (x):={\int }_1^x\frac{1}{t}𝑑t.$$(1)

Figure 1 shows the graph of $\ln$.

Figure 1: The graph of $\ln (x)$.

Instead of $\ln$ many mathematicians write $\log$, physicists (and calculators) however consider $\log$ as the symbol for the logarithm in base 10.One can show that the function defined in this way is the inverse of the exponential function. Indeed with equation (1) we have

$$\frac{d}{dx}\ln (e^x)=e^x\cdot \frac{1}{e^x}=1,$$

so there exists $C\in ℝ$ such that

$$\ln (e^x)=x+C.$$

Since $\ln (e^0)=0$ we have $C=0$.One can also prove that the above integral has the defining properties of a logarithm. For example if $x,y\in {ℝ}_{+}$ we have

$\ln (xy)$$=$${\displaystyle {\int }_1^x}{\displaystyle \frac{1}{t}}\mathrm{𝑑𝑡}+{\displaystyle {\int }_x^{xy}}{\displaystyle \frac{1}{t}}\mathrm{𝑑𝑡}$

$=$$\ln (x)+{\displaystyle {\int }_x^{xy}}{\displaystyle \frac{1}{t}}𝑑t.$

Now applying the substitution law with $u:=\frac{t}{x}$ we have

$${\int }_x^{xy}\frac{1}{t}\mathrm{𝑑𝑡}={\int }_1^y\frac{1}{u}\mathrm{𝑑𝑢}=\ln (y),$$

so we have

$$\ln (xy)=\ln (x)+\ln (y).$$

The natural logarithm can also be represented as a power series around $1$. For we have

$$\ln (1+x)=\sum _{k=1}^{\mathrm{\infty }}\frac{{(-1)}^{k+1}}{k}{x}^k.$$(2)

This series is divergent at $x=-1$ but for $x=1$ we have convergence due to Leibniz’s theorem and obtain

$$\ln (2)=\sum _{k=1}^{\mathrm{\infty }}\frac{{(-1)}^{k+1}}{k}.$$

In real analysis there is no reasonable way to extend the logarithm to negative numbers. In complex analysis the situation is a bit more complicated. Basically one can use the Euler relation to write a non-zero complex number $z$ in the form $z=R{e}^{i\phi }$ with $R,\phi \in ℝ$.We could try to define the complex logarithm $\ln (z)$ to be $\ln (R)+i\phi$. However $\phi$ is unique only up to addition of a multiple of $2\pi$.While at first glance this does not appear to be very problematic, it actually prevents one from setting up a continuous logarithm on the complex plane (without 0, where the logarithm should be infinite). Say for example that we let the imaginary part of our logarithm take values from $-\pi$ to $\pi$. Then

$$\underset{\phi \searrow -\pi }{lim}\ln e^{i\phi }=-i\pi$$

and

$$\underset{\phi \nearrow \pi }{lim}\ln e^{i\phi }=i\pi$$

while $e^{i\phi }\to -1$ for both limits. Therefore the logarithm we defined is not continuous at -1. The same argument allows one to show that it is not continuous on the negative real numbers. In fact you can only define a continuous complex logarithm on a sliced plane, i.e. the complex plane with a half-line starting at 0 removed.