Intuition for formula in proof of Poincaré Lemma

Question

In Poincaré lemma for star shaped domain, the potential $f$ of a vector field ${\bf F}$ is given explicitly for $$f(x) = \int_0^1 {\bf F}(tx)\cdot x\,dt.$$ Can be given some (physical...) intuition for this formula?

Answer 1

Consider $f \colon \mathbb{R} \to \mathbb{R}$. The statement of the Poincare lemma also assumes that $f(0) = 0$. Then, $f(x) = \int_0^x \frac{df}{dx}(x)dx$, which can be interpreted as an infinite sum that, informally, adds all infinitesimal changes $df(z)$ of $f$ over all $z$ in the interval $[0,x]$ and, thus, obtains the total change $f(x) - f(0) = f(x)$. Analogously, when $f$ maps $\mathbb{R}^n$ to $\mathbb{R}$, the integral $\int_0^1 \ldots dt$, informally, sums over all points $z = tx$ on the line segment connecting $0$ and $x$; the infinitesimal change at each point $z$ is given by the inner product of the gradient and the infinitesimal direction: $\nabla f(z) \cdot d(tx)$ (recall that $f(z + \delta) - f(z) \approx \nabla f(z) \cdot \delta$ for any small direction $\delta$). Thus, the integral $\int_0^1 \nabla f(tx)\cdot d(tx)$ sums all these infinitesimal changes and obtains the total change $f(x) - f(0) = f(x)$. It can also be expressed as the line integral: $$ f(x) = \int_{[0,x]} \nabla f(z) \cdot d\ell. $$ It remains to put the vector field $\mathbf{F}$ in place of $\nabla f$.