Being a novice with differential equations, I recently learned that if $\mu$ is an integrating factor for $\frac{dy}{dx}f(x,y)+ g(x,y)=0$, then the corresponding 1-form, $\mu fdy+\mu g dx$, is exact.


My question is, **is there some constructive proof that given a 1-form,
$fdy+gdx$, we can find a scalar function $\mu$ so that $\mu fdy+\mu gdx$ is exact, and under what conditions can we do so?** The only proofs for the existence of integrating factors I've found so far use existence theorems for solutions to ODEs and then prove the there is an integrating factor based on the existence of a solution.


To be clear, I would be very satisfied with an answer that looked like: subdivide a region into squares; linearly interpolate a scalar function to ensure the integral around each square is zero; cut each square into smaller squares; repeat. I would also be satisfied (though less so) with a less-constructive fixed-point theorem method. But, I'd like to stay in the context of 1-forms if possible.



Existance of Integrating Factors, a Constructive Proof
You are basically asking for an «algorithm» deciding whether a first-order ODE (with, say, analytic coefficients $f, g$) has a (meromorphic,say) first-integral. This question (which Poincaré asked when $f, g$ are polynomials) is difficult and still open. I'll try to give here examples of undecidability from the sole knowledge of Taylor series representation of $f$ and $g$.


First, I don't know what you exactly mean by «constructive», as a lot depends on your coefficients field. For instance the $1$-form $$\omega\_\lambda:=\lambda x\mathrm{d}y-y\mathrm{d}x$$ has an integrating factor if, and only if, $\lambda\in \mathbb Q$. But in most reasonable computation models (ex: BSS) you can't decide if a given (complex/real) number is rational.


That being said, even in a «perfect» computational model you still can't determine this fact algorithmically in all generality. Take the example of the focus/center problem: being given the $1$-form $$\omega:=x\mathrm{d}x+y(1+h(x,y))\mathrm{d}y,$$ is it possible to read in a finite jet of $h$ at $(0,0)$ if there exists a *local* integrating factor? The answer is «no» in general. If $h=0$ the foliation integrating the distribution $\ker\omega$ has a center singularity at $(0,0)$ (all orbits are closed loops) with first-integral $x^2+y^2$, for $\mathrm{d}(x^2+y^2)=2\omega$. In general though the singularity is a focus (no closed orbits). To see this, fix the Poincaré section $\Sigma:=\{x<0, y=0\}$) and take the partial orbit $\gamma\_z$ starting from $(z,0)\in\Sigma$, making one turn around the singularity and coming back at $\Sigma$ at a point $(P(z),0)$. Then $\gamma\_z$ is closed if, and only if, $P(z)=z$, that is $$\int\_{\gamma\_z} \frac{yh(x,y)}{x^2+y^2}\mathrm{d}y = 0.~~~~~(\*)$$ Gory computations shows you all the Taylor coefficients of $h$ mixed up with (in general) non-zero coefficients in a big infinite sum. You can always arrange this sum to be $0$ or not, by perturbing a sufficiently far away Taylor term of $h$ (meaning that you can't decide if it sums to $0$ or not without encompassing the whole infinite collection of terms).


If $h$ is polynomial of given degree $d$ (or more generally if $h$ belongs to a finitely-determined functional space) then condition $(\*)$ determines an analytic hypersurface. I'm almost sure that for degrees $d$ as low as $4$ the equation of this hypersurface is not known. I don't know if it is computable (in a reasonable model) or not, but I'm sure brute-forcing the equation is not an option here.


