| definiendum | $\mathrm{ln}:\mathbb C\setminus {]-\infty,0]} \to \mathbb R$ |
| postulate | $\mathrm{ln}(z):=\mathrm{ln}|z|+i\,\arg(z)$ |
todo: Complex argument
$\lim_{x\to 0}x\ln(x)=0$
$\int \ln(x^n)\,{\mathrm d}x^n=\int \left(x^n\right)'\ln(x^n)\,{\mathrm d}x=x^n\left(\ln(x^n)-1\right)$
At least around $z=0$ (I think for $|z|<1$)
$\ln{\left(\frac{1}{1-z}\right)} = \sum_{n=1}^\infty \frac{z^n}{n}$
or
$\ln{\left(1+z\right)} = -\sum_{n=1}^\infty \frac{(-z)^n}{n}$
From this series in $(-z)^n$, if we always constitutive positive and negative germs, we get
$\ln{\left(1+z\right)} = z - \sum_{n=1}^\infty \left(\dfrac{1}{2n}-\dfrac{z}{2n+1}\right) z^{2n}$
Note that this implies that $\exp$ maps the series on the right to $\frac{1}{1-z}=1+\sum_{n=1}^\infty z^n$.
The following is related to Minus twelve . Note:
$\left(\dfrac{z}{\ln(1+z)}\right)^n=1+\dfrac{n}{2}z+\dfrac{n}{2}\dfrac{3n-5}{12}z^2+\dfrac{n}{2}\dfrac{(n-2)(n-3)}{24}z^3+\dots$.
In particular, this tells us that
$\sum_{k=0}^\infty k\,z^k-\dfrac{1}{\ln(z)^2}=-\dfrac{1}{12}+{\mathcal O}\left((z-1)^1\right)$