When is the gradient of a line integral equal to the vector field being integrated

Question

Let $\newcommand{\R}{\mathbb{R}}$ $g:\R^n \to \R^n$ be a $C^\infty$ vector field. We know that if $g$ is conservative (or exact as a 1-form) then the gradient theorem (stokes theorem) tells us $$ g = \nabla G \implies \int_0^1 g(tx)\cdot x dt = G(x) - G(0) $$ and so $$ \nabla \int_0^1 g(tx) \cdot x dt = \nabla G(x) = g(x) $$ is there some criteria such that this relationship holds, even in the case where $g$ is not conservative (resp. exact) ? I.e, what conditions do we need to assume on $g$ such that $$ \nabla \int_0^1 g(tx) \cdot x dt = g(x) $$ Is exactness necessary?

Answer 1

Just clarifying my comment: yes we have an equivalence.

Let $g:\Bbb{R}^n\to\Bbb{R}^n$ be a smooth map, and define the smooth $1$-form $\omega=\sum_{i=1}^ng_i\,dx^i$, and the smooth function $G:\Bbb{R}^n\to\Bbb{R}$ as \begin{align} G(x):=\int_0^1\sum_{i=1}^ng_i(tx)x^i\,dt=\sum_{i=1}^n\left(\int_0^1g_i(tx)\,dt\right)x^i \end{align} Then, we have the following results:

1. If $\omega$ is a closed $1$-form (i.e $d\omega=0$, which is equivalent to saying that for all $i,j\in \{1,\dots, n\}$, $\frac{\partial g_j}{\partial x^i}-\frac{\partial g_i}{\partial x^j}=0$), then $\omega = dG$. This is a special case of (the proof of) Poincare's lemma.

2. If $dG=\omega$, then $d\omega=d(dG)=0$. This is a trivial consequence of the fact that mixed partial derivatives of smooth functions are equal (explicitly, our hypothesis is $g_i=\frac{\partial G}{\partial x^i}$, from which it follows that $\frac{\partial g_j}{\partial x^i}-\frac{\partial g_i}{\partial x^j}=\frac{\partial^2 G}{\partial x^i\partial x^j}-\frac{\partial^2 G}{\partial x^j\partial x^i}=0$).

So, indeed, we have an equivalence: $dG=\omega$ if and only if $d\omega=0$. If words rather than formulae: $G$ is a primitive for $\omega$ if and only if $\omega$ is closed.

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More generally, let $\omega$ be a smooth $k$-form on an open set $A\subset\Bbb{R}^n$ which is star-shaped with respect to the origin (meaning for any point $\xi \in A$, the line segment from $0$ to $\xi$ lies in $A$), and define the $(k-1)$-form $P\omega$ on $A$ by setting for each point $\xi\in A$, \begin{align} (P\omega)(\xi)&:=\sum_{i_1<\cdots <i_k}\sum_{\alpha=1}^k(-1)^{\alpha-1}\left(\int_0^1t^{k-1}\omega_{i_1\dots i_k}(t\xi)\,dt\right)\xi^{i_{\alpha}}\\&\qquad (dx^{i_1}\wedge \cdots \wedge \widehat{dx^{i_{\alpha}}}\wedge \cdots \wedge dx^{i_k})(\xi) \end{align} Then, $d(P\omega)=\omega$ if and only if $d\omega=0$. The implication $\implies$ is trivial because of the identity $d^2=0$ (which boils down to equality of mixed partials), while the implication $\impliedby$ is a direct but tedious calculation (performing this calculation is exactly the proof of Poincare's lemma given in Spivak's Calculus on Manifolds, so if you get stuck, you can refer to that). Once again, in words, $P\omega$ is a primitive for $\omega$ if and only if $\omega$ is closed.