Positive geometry and log singularities In order to define a positive geometry it is a requirement that has to be a logarithmic singularities on the boundaries, for example for an interval (endpoints $a$ and $b$) the canonical form is
$$\frac{dx(b-a)}{(b-x)(x-a)}$$ My question is: Why do we need that logarithm singularities requirement in order to get that differential form?
 A: I'm not sure I understood exactly what you want to know.
Anyhow, a positive geometry is an oriented geometry that has a canonical form. A canonical form is a differential form with dlog divergences, which is like d$x/x$ for a boundary $x=0$, on all the boundaries of the geometry. Moreover, the residue of the canonical form on a boundary must be equal to the canonical form of the boundary. Finally, the canonical form of a point is equal to 1 for a positively oriented point and to -1 for a negatively oriented point.
Your example is the canonical form of a segment $a<x<b$ with an orientation form equal to d$x$. Following the definition, to verify that the expression is indeed correct we need to verify that the residue in $x=a$ is equal to 1, since $x=a$ is a positively oriented point, and the residue in $x=b$ is equal to -1, since $x=b$ is a negatively oriented point.
Your expression for the canonical form can also be rewritten as
$$\Omega(x)= \text{dlog}(x-a)-\text{dlog}(x-b)=\frac{\text{d} x}{x-a}-\frac{\text{d} x}{x-b}$$
The residue $Res_{x=a}(\Omega(x))$ for example can be computed as
$$Res_{x=a}(\Omega(x))=\lim_{x\to a}(x-a)(\frac{1}{x-a}-\frac{1}{x-b})=1$$
The idea of the canonical form can be generalized to non-logarithmic singularities for polytopes, as proposed in https://arxiv.org/abs/2005.03612, but the topic remains still in development for more general geometries.
I hope this answer was helpful!
