Let $M$ be a smooth manifold and let $\mathcal D$ be an integrable distribution of codimension one. Then by Frobenius theorem there exists a foliation $\mathcal F^{\mathcal D}$ such that the leaves are integral manifolds to $\mathcal D$.
In terms of differential forms there exists on every trivializing patch $U$ of $\mathcal D$ a one form $\sigma\in \Omega^1(M)$ that annihilates vectors in $\mathcal D$ and satisfies $\sigma\wedge d\sigma=0$. This is another version of Frobenius' integrability condition for co-dimension one foliations.
In general the leaf space $\mathcal L=M/ \mathcal F^{\mathcal D}$ will look horrible. So I wonder under which conditions the quotient topology is Hausdorff. More importantly, are there sufficient conditions to impose on the spanning one-forms $\sigma$ in order to ensure that the quotient is Hausdorff.
My first guess was that if $\sigma$ is closed, then maybe $span(\sigma)$ defines a nice quotient. But this is not true. For the irrational slopes on the torus the defining one-forms may be chosen closed yet the quotient is not Hausdorff. So maybe one should demand it to be exact. But I think according to Frobenius theorem there always exist on every neighborhood exact one-forms that define $\mathcal D$, right? So maybe the condition should be that there exists globally an exact form that defines $\mathcal D$. If there are no local conditions, then maybe one can at least find global ones? Any ideas?