Leibniz Integration Formula

Prerequisites: Calculus.

• Intuitive
• Concrete
• Abstract
• Why is it interesting?

Intuitive

Picture the area under a curve between two fence posts. Now let something change: the curve may rise or fall, and the fence posts may move. The Leibniz integration formula tells us how fast the enclosed area changes, and it says there are exactly two sources of change.

The first source is the curve itself. If the curve moves up a little everywhere between the posts, the area grows by the sum of all those small rises.

The second source is the posts. If the right post moves to the right, a thin strip is added at the right edge. Its width is how far the post moved, and its height is the height of the curve at the post. The left post works the other way: moving it to the right cuts a strip off the region. That is why the left boundary term comes with a minus sign.

Adding the interior contribution and the two edge contributions gives the full rate of change. This is the reason the formula has one integral and two boundary terms.

Concrete

This tool gives a nice formula to the derivative of a one dimensional integral with dependencies to the varied quantity every where, with $f, a$ and $b$ having the right conditions and $a = a(0)$, $b = b(0)$,

$$\left.\frac{\partial}{\partial \varepsilon} \int_{a(\varepsilon)}^{b(\varepsilon)}f(\varepsilon, x) d x \right|_{\varepsilon=0} = \int_{a}^{b} \left.\frac{\partial}{\partial \varepsilon} f(\varepsilon, x)\right|_{\varepsilon=0} d x + f(0,b)\left.\frac{\partial}{\partial \varepsilon} b(\varepsilon)\right|_{\varepsilon=0} - f(0,a)\left.\frac{\partial}{\partial \varepsilon} a(\varepsilon)\right|_{\varepsilon=0}$$

Abstract

In one dimension the formula holds, for example, when $f$ and $\partial f/\partial \varepsilon$ are continuous on a rectangle $[\varepsilon_1, \varepsilon_2] \times [x_1, x_2]$ that contains all the points $(\varepsilon, x)$ with $x$ between $a(\varepsilon)$ and $b(\varepsilon)$, and $a$, $b$ are continuously differentiable.

The general form concerns a $k$-form $\omega_t$ that depends on a parameter $t$, integrated over a $k$-dimensional compact oriented submanifold $\Omega_t = \phi_t(\Omega_0)$ (with boundary) of a manifold $M$. Here $\phi_t$ is the flow of a possibly time-dependent vector field $X_t$, so that $\frac{d}{dt}\phi_t(p) = X_t(\phi_t(p))$. Pulling the integral back to the fixed domain $\Omega_0$ and using $\frac{d}{dt}\phi_t^*\omega_t = \phi_t^*\big(\partial_t \omega_t + \mathcal{L}_{X_t}\omega_t\big)$, we get

$$\frac{d}{dt}\int_{\Omega_t} \omega_t = \int_{\Omega_t} \big(\partial_t \omega_t + \mathcal{L}_{X_t}\omega_t\big).$$

With Cartan's formula $\mathcal{L}_X \omega = \iota_X d\omega + d\,\iota_X \omega$ and Stokes' theorem this becomes

$$\frac{d}{dt}\int_{\Omega_t} \omega_t = \int_{\Omega_t} \big(\partial_t \omega_t + \iota_{X_t} d\omega_t\big) + \int_{\partial \Omega_t} \iota_{X_t} \omega_t.$$

The one-dimensional formula is the case $M = \mathbb{R}$, $\omega_t = f(t, x)\,dx$, with $t$ in the role of $\varepsilon$ and $X_t$ any vector field whose flow moves the endpoints along $a(t)$ and $b(t)$. A 1-form on $\mathbb{R}$ is a top form, so $d\omega_t = 0$. The boundary of $[a, b]$ consists of the point $b$ with positive and the point $a$ with negative orientation, and $\iota_X(f\,dx) = fX$ with $X(b) = \dot b$, $X(a) = \dot a$, which reproduces $f(t, b)\,\dot b - f(t, a)\,\dot a$.

For $\omega_t = f\, dV$ an $n$-form on an $n$-dimensional region of $\mathbb{R}^n$ we have $\mathcal{L}_X(f\,dV) = \nabla \cdot (f \vec X)\, dV$, and the general formula becomes the Reynolds transport theorem of continuum mechanics,

$$\frac{d}{dt}\int_{\Omega_t} f\, dV = \int_{\Omega_t} \frac{\partial f}{\partial t}\, dV + \oint_{\partial \Omega_t} f\, \vec X \cdot d\vec A.$$

Why is it interesting?

This formula is really nice in one dimensional Variational Calculus and with $a(0) = a$, $b(0) = b$, and for the particle mechanics Noether's Theorem. For integrals over arbitrary manifolds, it generalizes to the Lie derivative of a volume element.