I'd like to characterize rational $k$-forms on a singular scheme $X$ which can be integrated on any (real) bounded submanifold that is locally LCI in $X$ in a real sense (i.e., which is the real preimage of a flat map $X\to \mathbb{A}^{n-k}$).
Specifically, suppose $X$ is a sufficiently nice but possibly singular real affine scheme (more generally, real analytic spaces should work), and let $Z\subset X$ be a subscheme that includes the singular locus. Let $h$ be some polynomial function valued in $\mathbb{R}^n$ which vanishes on $Z$ (not necessarily flat as a family over $\mathbb{R}^n$) which we will use to produce small tubular neighborhoods of $Z$ in $X$, and let $U: = X\setminus Z$ be the smooth complement.
Now suppose $C$ is a real "$k$-cycle", which we define to be an open subset of a $k$-dimensional real subvariety $\tilde{C}\subset X,$ which is bounded (i.e., has compact closure) and such that the boundary of $C$ in $\tilde{C}$ is in $U$ (and in particular in the smooth locus). For example, if $X = \mathbb{R}$ is the affine line and $Z = \{0\}$ is the origin, then a one-cycle is an open bounded subset of $\mathbb{R}$ whose boundary doesn't contain the origin. Then for a rational $k$-form $\omega\in \Omega^k U$ (with singularities on $Z$), we say $\omega$ is integrable on $C$ if the limit $$\lim_{\epsilon\to 0} \int_{x\in C; |h(x)|<\epsilon} \omega$$ of integrals on the complement of a small tubular neighborhood exists and is independent of choice of regular function $h$ defining the tubular neighborhood. Note that in the example above with $X = \mathbb{R}, Z = \{0\},$ a regular one-form on $\mathbb{R}^*$ is integrable on any one-cycle if and only if it has at worst first-order singularity at $0$.
Finally, we say that a cycle $C\subset X$ is an LCI cycle if it is locally cut out by flat equations, i.e., locally open in the fiber of a flat map $X\to \mathbb{R}^{n-k}$.
If $X$ is sufficiently singular, there can be forms on $U$ with arbitrarily large order of singularity which are integrable on any LCI.
Here is an example. Fix $n\ge 2$ and let $X$ be the degree-$n$ "osculating cross", i.e., the scheme obtained by gluing $\mathbb{A}^1$ to itself along the subscheme $\mathrm{Spec}\mathbb{R}[x]/x^n$ (so functions on $X$ are pairs of polynomials $f(x)=\sum f_k x^k, g(y) = \sum g_k y^k$ satisfying $f_0=g_0, \ldots, f_{n-1} = g_{n-1}.$) Then the open locus is $$U = \mathbb{R}^*\sqcup \mathbb{R}^*$$ and a one-form on the open locus is a pair of Laurent forms $$\omega = (f(x,x^{-1}) dx, g(y, y^{-1}) dy)$$ with $f(x,x^{-1}), g(y,y{-1})$ Laurent polynomials in one variable.
In this example a one-dimensional LCI cycle is simply a bounded open subset of the cross with no boundary at the origin, and it's relatively easy to show that a form $\omega = (f(x)dx, g(y) dy)$ as above is integrable on any LCI cycle if and only if the functions $f(x), g(x)$ have degree of singularity $\le n$ and their sum has at worst a first-order singularity. In other words, the pair $(f(x) dx, g(y)dy)$ is integrable on any LCI if and only if we can express $f(x) = f_{-n}x^{-n}+f_{1-n} x^{1-n} + \ldots,$ and $g(x) = g_{-n} x^{-n} + g_{1-n}x^{1-n} + \ldots,$ and these satisfy $$f_{-2} = -g_{-2}, f_{-3} = -g_{-3}, \ldots, f_{-n} = -g_{-n}.$$
My question: does there exist an algebro-geometric characterization of the space of rational forms which are integrable on any (bounded) LCI, in the sense described above, or in a similar sense? I'm hoping for an algebro-geometric answer that gives a notion of sheaf of differential forms for a singular algebraic variety and roughly agrees with the answer given above for the osculating cross $\mathbb{A}^1\cup_{\mathrm{Spec} \mathbb{R}[x]/x^n} \mathbb{A}^1$.
