| definiendum | $\mathrm{ln}:\mathbb R_+^*\to \mathbb R$ |
| postulate | $\mathrm{ln}=\mathrm{exp}^{-1}$ |
$\int_1^y \frac {1 } {x} {\mathrm d}x = \ln(y) $
$\int_0^{y} \frac {1 } {1+x } {\mathrm d}x = \ln(1+y) $
Log[a] == Log[b] + Integrate[1/(t+b)-1/(t+a),{t,0,Infinity}]
The function $x\mapsto\frac{x}{x-1}\log(x)$ is one without bad behaviours (singularities) on $[0,\infty)$.