Hilbert's invariant integral

Remarks

The integrand is a closed differential form on $D$ because the field consists of extremals. Since $D$ is simply connected, the form is exact, so $I^*(\gamma)$ depends only on the end points of $\gamma$.

Along a curve of the field one has $dy = p \, dx$, so $I^*(\gamma) = \int L(x, y, p) \, dx = J$ on that curve. For the graph of any piecewise $C^1$ function $y$ in $D$, subtracting $I^*$ from $J$ gives $$J(y) - I^*(y) = \int_a^b E\big(x, y(x), p(x, y(x)), y'(x)\big) \, dx,$$ with $E$ the Weierstrass excess function. If $y_0$ is a field curve and $y$ has the same end points, then $I^*(y) = J(y_0)$, which gives the Weierstrass formula $J(y) - J(y_0) = \int_a^b E \, dx$.

Example: for $L = p^2$ and the field $y = x + c$ with $p = 1$, the integrand is $(1 - 2) \, dx + 2 \, dy = d(2y - x)$, which is exact.

Non-example: with the slope function $p = 1$ and $L = p^2 + y^2$, the curves $y = x + c$ are not extremals, and the form $(y^2 - 1) \, dx + 2 \, dy$ is not closed, since $\partial_y (y^2 - 1) = 2y$ is not identically equal to $\partial_x 2 = 0$.