Leibniz integral rule
Leibniz integral rule
How to take the derivative of the integral?
General form: differentiation under the integral sign
\[\int_{a(x)}^{b(x)} f\left ( x,t \right ) dt\]
Theorem — Let f(x,t) be a function such that both \(f\left ( x,t \right )\) are continuous it t and x in some
region of the xt-plane, including \(a(x)\le t\le b(x),x_{0} \le x\le x_{1}\). Also suppose that the functions \(a(x)\) and
functions \(b(x)\) are both continuous and both have continuous derivatives for \(x_{0} \le x\le x_{1}\).Then for \(x_{0} \le x\le x_{1}\),
\[{\displaystyle {\frac {d}{dx}}\left(\int _{a(x)}^{b(x)}f(x,t)\,dt\right)=f{\big (}x,b(x){\big )}\cdot {\frac {d}{dx}}b(x)-f{\big (}x,a(x){\big )}\cdot {\frac {d}{dx}}a(x)+\int _{a(x)}^{b(x)}{\frac {\partial }{\partial x}}f(x,t)\,dt.}\]
The right hand side may also be written using Lagrange’s notation as:
\[{\textstyle f(x,b(x))\,b^{\prime }(x)-f(x,a(x))\,a^{\prime }(x)+\displaystyle \int _{a(x)}^{b(x)}f_{x}(x,t)\,dt.}\]
another form
\[\Phi(x)=\int_{\alpha(x) }^{\beta(x)} f(x, y) dy\]
\[\Phi(t)=\int_{\alpha(t) }^{\beta(t)} f(t, x) dx\]
\[\Phi(t)=\int_{\alpha(t) }^{\beta(t)} F(x, t) dx\]
\[\begin{split}\begin{align}
\frac{\mathrm{d} \Phi(t)}{\mathrm{d} t}&=\frac{\Phi(t+\Delta t)-\Phi(t)}{\Delta t}\\
&=\int_{\alpha(t) }^{\beta(t)}\frac{ F(x, t+\Delta t)-F(x, t)}{\Delta t} dx\\
&+\frac{1}{\Delta t}\int_{\beta(t) }^{\beta(t+\Delta t)}{ F(x, t+\Delta t)} dx \\
&-\frac{1}{\Delta t}\int_{\alpha(t) }^{\alpha(t+\Delta t)}{ F(x, t+\Delta t)} dx
\end{align}\end{split}\]
\[\begin{split}\begin{align}
\frac{\mathrm{d} \Phi(t)}{\mathrm{d} t}&=\frac{\mathrm{d} }{\mathrm{d} t}\int_{\alpha(t) }^{\beta(t)}F(x, t) dx\\\\
&=\int_{\alpha(t) }^{\beta(t)}\frac{\partial F(x, t)}{\partial t} dx
+[ F(\beta(t), t)\dot{\beta}(t) - F(\alpha(t), t)\dot{\alpha}(t) ]\\
\end{align}\end{split}\]
This can also be written as
\[\cfrac{\mathrm{d}}{\mathrm{d}t}{\int_{g(t)}^{h(t)}F(x,t)\mathrm{d}x }={\int_{g(t)}^{h(t)}\frac{\partial F(x,t)}{\partial t}\mathrm{d}x }+\left \{ F[h(t),t]\dot{h}(t)-F[g(t),t]\dot{g}(t) \right \}\]
One proof runs as follows, modulo precisely stated hypotheses and some analytic
details. Set
\[\Phi(u, v, t)=\int_{u}^{v} F(x, t) d x\]
\(u = g(t)\), and \(v = h(t)\). By the chain rule
\[\frac{d}{d t} \Phi[g(t), h(t), t]=\left(\frac{\partial \Phi}{\partial u} \dot{g}+\frac{\partial \Phi}{\partial v} h\right)+\frac{\partial \Phi}{\partial t}\]
The first two terms are bracketed because they measure all changes due to variation
of the interval of integration [g(t), h(t)], and they are evaluated by applying the
Fundamental Theorem to \(\Phi(u, v, t)=\int_{u}^{v} F(x, t) d x\). The third term measures change due to variation of
the integrand. If enough smoothness is assumed to justify interchange of the integration and differentiation operators, then
\[\frac{\partial \Phi}{\partial t}=\frac{\partial}{\partial t} \int_{u}^{v} F(x, t) d x=\int_{u}^{v} \frac{\partial F(x, t)}{\partial t} d x .\]
Another proof
\[\begin{split}\begin{aligned}
\phi_{t}(u) & =x(u, t), \\
\phi_{t}:[a, b] & \rightarrow\left[\phi_{t}(a), \phi_{t}(b)\right]=[g(t), h(t)]=C_{t} .
\end{aligned}\end{split}\]
By the formula for change of variable in a simple integral
\[\int_{g(t)}^{h(t)} F(x) d x=\int_{\phi_{t}(a)}^{\phi_{t}(b)} F(x) d x=\int_{a}^{b} F[x(u, t)] \frac{\partial x}{\partial u} d u .\]
This transition is excellent, because it has changed the integral over a moving domain
to one over a fixed domain. We pay for this fixed domain with a time-varying integrand. No matter, we like it; we thrive on differentiation under the integral sign:
\[\begin{split}\begin{aligned}
\frac{\mathrm{d} }{\mathrm{d} t} \int_{g(t)}^{h(t)} F(x) d x
&=\frac{\mathrm{d} }{\mathrm{d} t} \int_{a}^{b} F[x(u, t)] \frac{\partial x(u, t)}{\partial u} d u \\
&=\int_{a}^{b}\frac{\partial}{\partial t}\left \{ F[x(u, t)] \frac{\partial x(u, t)}{\partial u}\right \}d u \\
&=\int_{a}^{b}\left \{\frac{\partial F[x(u, t)]}{\partial x}\frac{\partial x(u, t)}{\partial t} \frac{\partial x(u, t)}{\partial u}+F[x(u, t)]\frac{\partial x^{2} (u, t)}{\partial u\partial t}\right \}d u \\
\end{aligned}\end{split}\]
The fixed domain has done its job, and we return to the moving domain. The instantaneous velocity is \(v = v(u, t) = \partial x(u, t)/ {\partial t}\), which we also consider as a function of \(x\)
and \(t\) via the transformation \((u, t) \leftrightarrow (x, t)\). When \(t\) is fixed,
\[\begin{split}\begin{aligned}
\frac{\partial x^{2} (u, t)}{\partial u\partial t}&=\frac{\partial v(u, t)}{\partial u}\\
&=\cfrac{\cfrac{\partial v(u, t)}{\partial u}}{\cfrac{\partial x(u, t)}{\partial u}} {\frac{\partial x(u, t)}{\partial u}} \\
&={\cfrac{\partial v(u, t)}{\partial x(u, t)}}{\cfrac{\partial x(u, t)}{\partial u}}
\end{aligned}\end{split}\]
hence
\[\begin{split}\begin{aligned}
&\int_{a}^{b}\left \{\frac{\partial F[x(u, t)]}{\partial x}\frac{\partial x(u, t)}{\partial t} \frac{\partial x(u, t)}{\partial u}+F[x(u, t)]\frac{\partial x^{2} (u, t)}{\partial u\partial t}\right \}d u \\
=&\int_{a}^{b}\left \{\frac{\partial F[x(u, t)]}{\partial x}\frac{\partial x(u, t)}{\partial t} \frac{\partial x(u, t)}{\partial u}+F[x(u, t)]{\cfrac{\partial v(u, t)}{\partial x(u, t)}}{\cfrac{\partial x(u, t)}{\partial u}}\right \}d u \\
=&\int_{a}^{b}\left \{\frac{\partial F[x(u, t)]}{\partial x}\frac{\partial x(u, t)}{\partial t} +F[x(u, t)]{\cfrac{\partial v(u, t)}{\partial x(u, t)}}\right \}{\cfrac{\partial x(u, t)}{\partial u}}d u \\
=&\int_{\phi_{t}(a)}^{\phi_{t}(b)}\left \{\frac{\partial F[x(u, t)]}{\partial x}\frac{\partial x(u, t)}{\partial t} +F[x(u, t)]{\cfrac{\partial v(u, t)}{\partial x(u, t)}}\right \}d x \\
\end{aligned}\end{split}\]
that is
\[\begin{split}\begin{aligned}
&\int_{\phi_{t}(a)}^{\phi_{t}(b)}\left \{\frac{\partial F[x(u, t)]}{\partial x}\frac{\partial x(u, t)}{\partial t} +F[x(u, t)]{\cfrac{\partial v(u, t)}{\partial x(u, t)}}\right \}d x \\
=&\int_{\phi_{t}(a)}^{\phi_{t}(b)}\frac{\partial }{\partial x} [F[x(u, t)]v(u, t)]d x \\
=&\int_{g(t)}^{h(t)}\frac{\partial }{\partial x} [F[x(u, t)]v(u, t)]d x
\end{aligned}\end{split}\]
Two-dimensional, time-dependent case
We are also given a function \(F(x, y, t)\). The problem is to find
\[\frac{\mathrm{d} }{\mathrm{d} t}\iint_{D(t)}^{} F(x,y,t)dxdy\]
Certainly our first move should be separation of boundary variation from integrand variation. This is easy enough by the chain rule device in the first section
and results in
\[\begin{split}\begin{aligned}
&\frac{\mathrm{d} }{\mathrm{d} t}\iint_{D(t)}^{} F(x,y,t)dxdy{\Bigg|}_{t=t_{0} }\\
=&\frac{\mathrm{d} }{\mathrm{d} t}\iint_{D(t)}^{} F(x,y,t_{0})dxdy{\Bigg|}_{t=t_{0} }+
\iint_{D(t_{0})}^{} \frac{\partial F(x,y,t)}{\partial t}{\Bigg|}_{t=t_{0} }dxdy
\end{aligned}\end{split}\]
This is routine. The essence of the problem is to find
\[\frac{\mathrm{d} }{\mathrm{d} t}\iint_{D(t)}^{} F(x,y)dxdy\]
Let \(\mathbf{v}=\mathbf{v}(x, y, t)\) denote the velocity vector at a boundary point \((x, y)\) of \(Dt\) and let \(n\)
denote the outward unit normal. In the difference
\[\iint_{D(t+\Delta t )}^{} F(x,y)dxdy-\iint_{D(t)}^{} F(x,y)dxdy\]
everything in the overlap of \(D(t)\) and \(D(t+\Delta t)\) cancels; only the thin boundary strip makes
a contribution. From the detail, this contribution is
\[F(x,y)(\mathbf{v}\Delta t)\cdot(\mathbf{n}ds)\]
up to higher order differentials, where \(ds\) is the element of arc length.
Hence
\[\begin{split}\begin{aligned}
&\lim_{\Delta t \to 0} \cfrac{1}{\Delta t}\left \{ \iint_{D(t+\Delta t )}^{} F(x,y)dxdy-\iint_{D(t)}^{} F(x,y)dxdy\right \}\\
=&\lim_{\Delta t \to 0}\cfrac{1}{\Delta t}\int_{\partial D(t)}^{} F(x,y)(\mathbf{v}\Delta t)\cdot(\mathbf{n}ds)
\end{aligned}\end{split}\]
\[\cfrac{\mathrm{d} }{\mathrm{d} t}\iint_{D(t)}^{} F(x,y)dxdy =
\int_{\partial D(t)}^{} F(x,y){\mathbf{v}}\cdot{\mathbf{n}}ds\]
where \(\partial\) a denotes boundary. Before taking limits, we compute \({\mathbf{v}}\cdot{\mathbf{n}}ds\). We rotate the
unit tangent \((dy/ds,-dx/ds)\) backwards through a right angle to obtain \(\mathbf{n}=(dy/ds,-dx/ds)\), hence
\[\mathbf{n}=(dy/ds,-dx/ds)=(u,v)\cdot(dy,-dx)=udy-vdx\]
Therefore
\[\begin{split}\begin{aligned}
\cfrac{\mathrm{d} }{\mathrm{d} t}\iint_{D(t)}^{} F(x,y)dxdy =&
\int_{\partial D(t)}^{} F(x,y){\mathbf{v}}\cdot{\mathbf{n}}ds \\
=&\int_{\partial D(t)}^{} F(x,y)(u,v)\cdot(dy,-dx) \\
=&\int_{\partial D(t)}^{} F(x,y)(udy-vdx)\\
=&\int_{\partial D(t)}^{} (-F(x,y)vdx+F(x,y)udy)
\end{aligned}\end{split}\]
We can transform the boundary integral into an integral over D, by Green’s Theorem.
\[\begin{split}\begin{aligned}
P(x,y)&\equiv -F(x,y)v(x,y)\\
Q(x,y)&\equiv F(x,y)u(x,y)
\end{aligned}\end{split}\]
\[\begin{split}\begin{aligned}
(\cfrac{\partial Q(x,y)}{\partial x}-\cfrac{\partial P(x,y)}{\partial y})=
&(\cfrac{\partial [F(x,y)u(x,y)]}{\partial x}-\cfrac{\partial [-F(x,y)v(x,y)]}{\partial y})\\
=&(\cfrac{\partial [F(x,y)u(x,y)]}{\partial x}+\cfrac{\partial [F(x,y)v(x,y)]}{\partial y})\\
=&(\cfrac{\partial [F(x,y)u(x,y)]}{\partial x}+\cfrac{\partial [F(x,y)v(x,y)]}{\partial y})\\
\end{aligned}\end{split}\]
\[\begin{split}\begin{aligned}
\cfrac{\partial [F(x,y)u(x,y)]}{\partial x}-\cfrac{\partial [-F(x,y)v(x,y)]}{\partial y}=&\\
\cfrac{\partial [F(x,y)u(x,y)]}{\partial x}+\cfrac{\partial [F(x,y)v(x,y)]}{\partial y}=\mathrm{div} \left \{ F(x,y)\mathbf{v(x,y)} \right \} \\
\cfrac{\partial (F u)}{\partial x}+\cfrac{\partial (F v)}{\partial y}=\mathrm{div} ( F\mathbf{v} )
\end{aligned}\end{split}\]
\[\cfrac{\partial (F u)}{\partial x}+\cfrac{\partial (F v)}{\partial y}=\mathrm{div} ( F\mathbf{v} )= \nabla \cdot( F\mathbf{v} )\]
Here
\[\begin{split}\begin{aligned}
\mathrm{div} ( F\mathbf{v} )\equiv\nabla \cdot( F\mathbf{v} )=&\cfrac{\partial (F u)}{\partial x}+\cfrac{\partial (F v)}{\partial y}\\
=&F \nabla \cdot \mathbf{v}+\mathbf{v} \cdot \nabla F \\
=&\nabla F \cdot \mathbf{v} + F \nabla \cdot \mathbf{v}
\end{aligned}\end{split}\]
\[\nabla \cdot(\phi \mathbf{v})=\phi \nabla \cdot \mathbf{v}+\mathbf{v} \cdot \nabla \phi\]
\[\nabla \cdot(F \mathbf{v})=F \nabla \cdot \mathbf{v}+\mathbf{v} \cdot \nabla F\]
\[\begin{split}\begin{aligned}
&\frac{\mathrm{d} }{\mathrm{d} t}\iint_{D(t)}^{} F(x,y,t)dxdy\\
\end{aligned}\end{split}\]
\[\begin{aligned}
\frac{\mathrm{d} }{\mathrm{d} t}\iint_{D(t)}^{} F(x,y,t)dxdy
=\int_{\partial D(t)}^{} F(x,y){\mathbf{v}}\cdot{\mathbf{n}}ds+
\iint_{D(t)}^{} \frac{\partial F(x,y,t)}{\partial t}dxdy
\end{aligned}\]
\[\begin{split}\begin{aligned}
\frac{\mathrm{d} }{\mathrm{d} t}\iint_{D(t)}^{} F(x,y,t)dxdy
=&\iint_{D(t)}^{} \nabla \cdot(F \mathbf{v})dxdy+
\iint_{D(t)}^{} \frac{\partial F(x,y,t)}{\partial t}dxdy\\
=&\iint_{D(t)}^{} \left \{\nabla \cdot(F \mathbf{v})+\cfrac{\partial F(x,y,t)}{\partial t}\right \}dxdy
\end{aligned}\end{split}\]
A space formula.
\[\begin{split}\begin{aligned}
\frac{d}{d t} \iiint_{D_{t}} F(\mathbf{x}, t) d x d y d z & =\iint_{\partial D_{t}} F \mathbf{v} \cdot \mathbf{n} dS+\iiint_{D_{t}} \frac{\partial F}{\partial t} d x d y d z \\
\end{aligned}\end{split}\]
\[\begin{split}\begin{aligned}
\frac{d}{d t} \iiint_{D_{t}} F(\mathbf{x}, t) d x d y d z & =\iint_{\partial D_{t}} F \mathbf{v} \cdot d\mathbf{S} +\iiint_{D_{t}} \frac{\partial F}{\partial t} d x d y d z \\
\end{aligned}\end{split}\]
\[\begin{split}\begin{aligned}
\frac{d}{d t} \iiint_{D_{t}} F(\mathbf{x}, t) d x d y d z & =
\iiint_{D_{t}}\left[\operatorname{div}(F \mathbf{v})+\frac{\partial F}{\partial t}\right] d x d y d z\\
&=\iiint_{D_{t}}\left[\nabla \cdot(F \mathbf{v})+\frac{\partial F}{\partial t}\right] d x d y d z\\
\end{aligned}\end{split}\]