TITLE: Is every vector field locally integrable up to a rescaling?
QUESTION [7 upvotes]: Let $\Omega$ be a region of the plane (simply connected, say, if it matters), and $v(x,y)$ a smooth vector field on $\Omega$.
Does there always exist a scalar field $s:\Omega\to\mathbb{R}$ so that $sv$ is locally integrable: so that for every point $(x_0,y_0)\in\Omega$, in some neighborhood of that point,
$$s(x,y)v(x,y) = \nabla f(x,y)$$
for some $f:\Omega\to\mathbb{R}$?
Equivalently, does there always exist an $s$ with $\nabla \times (sv) =0$?
EDIT: As pointed out below, we want $s$ not identically zero, as this is clearly a trivial solution. The below answer also gives a counterexample, though it seems to rely fundamentally on $v$ having a singularity of nontrivial index; is the claim true if we require $v$ to be nowhere-vanishing?

REPLY [2 votes]: As s.harp has pointed out this is not always possible, as illustrated by his example where the integral curves of $v$ swirl around some point. However it is always possible to find such a factor provided $v$ is non-zero. As you pointed out since the domain is simple connected it suffices to find $s$ such that $\nabla \times (sv) =0$, which is equivalent to 
$$ \partial_y(sv^1)-\partial_x(sv^2)=v^1\partial_ys-v^2\partial_xs+(\partial_yv^1-\partial_xv^2)s=0$$
which we write as
$$v^\perp\cdot\nabla s + (\partial_yv^1-\partial_xv^2)s=0$$
This is a linear transport equation for $s$, and can be solved by the method of characteristics. The characteristics are the integral curves of $v^\perp=(-v^2,v^1)$, and its easy to see we can solve it provided these curves don't loop around. So long as $v(x_0,y_0)\neq 0$ then certainly you can solve it in a neighbourhood of $(x_0,y_0)$, and I think probably you can even argue somehow that it can be solved globally provided $v$ is nowhere vanishing (or making some other statement ensuring the existence of a Cauchy surface for the integral curves of $v^\perp$) .
Added in note: For global existence it turns out that a stronger condition than nowhere-vanishing $v$ is needed. See  the comments below.