38. Leibniz Integral Rule
38. Leibniz 积分法则
此法则关于对积分求导，在连续时间模型中频繁使用。
定理 38.1（Leibniz 积分法则） 在正则性条件下，下式成立： $$ > \frac{d}{dx}\left(\int_{a(x)}^{b(x)}f(x,t)\,dt\right)=f(x,b(x))\frac{db(x)}{dx}-f(x,a(x))\frac{da(x)}{dx}+\int_{a(x)}^{b(x)}\frac{\partial f(x,t)}{\partial x}\,dt > $$
证明 令 \(\psi(x)=\int_{a(x)}^{b(x)}f(x,t)\,dt\)。则 $$ > \begin{aligned} > d\psi(x)&=\psi(x+dx)-\psi(x)=\int_{a(x+dx)}^{b(x+dx)}f(x+dx,t)\,dt-\int_{a(x)}^{b(x)}f(x,t)\,dt\\ > &=-\int_{a(x)}^{a(x+dx)}f(x+dx,t)\,dt+\int_{a(x)}^{b(x)}f(x+dx,t)\,dt+\int_{b(x)}^{b(x+dx)}f(x+dx,t)\,dt-\int_{a(x)}^{b(x)}f(x,t)\,dt > \end{aligned} > $$ 由中值定理，第一项 \(\exists\xi_1\in(a(x),a(x+dx))\) s.t. \(-\int_{a(x)}^{a(x+dx)}f(x+dx,t)\,dt=-[a(x+dx)-a(x)]f(x+dx,\xi_1)\)，第三项 \(\exists\xi_2\in(b(x),b(x+dx))\) s.t. \(\int_{b(x)}^{b(x+dx)}f(x+dx,t)\,dt=[b(x+dx)-b(x)]f(x+dx,\xi_2)\)。于是 $$ > \begin{aligned} > \lim_{dx\to0}\frac{d\psi(x)}{dx}=&\lim_{dx\to0}\underbrace{\frac{-[a(x+dx)-a(x)]}{dx}}_{\to-a'(x)}\underbrace{f(x+dx,\xi_1)}_{\to f(x,a(x))}+\int_{a(x)}^{b(x)}\lim_{dx\to0}\underbrace{\frac{f(x+dx,t)-f(x,t)}{dx}}_{\to\frac{\partial f(x,t)}{\partial x}}dt\\ > &+\lim_{dx\to0}\underbrace{\frac{[b(x+dx)-b(x)]}{dx}}_{\to b'(x)}\underbrace{f(x+dx,\xi_2)}_{\to f(x,b(x))}\\ > =&f(x,b(x))\frac{db(x)}{dx}-f(x,a(x))\frac{da(x)}{dx}+\int_{a(x)}^{b(x)}\frac{\partial f(x,t)}{\partial x}\,dt > \end{aligned} > $$ 证毕。\(\blacksquare\)
例 38.1 考虑积分方程 \(P(t)=\int_{s=t}^{+\infty}D(s)\cdot e^{-r(s-t)}\,ds\)。证明 \(\dot P\equiv\dfrac{dP(t)}{dt}=rP-D\)。
证明 $$ > \begin{aligned} > \dot P=\frac{dP(t)}{dt}&=\frac{d}{dt}\int_{s=t}^{+\infty}D(s)\cdot e^{-r(s-t)}\,ds\\ > \overset{\text{Leibniz}}{=}&\int_{s=t}^{+\infty}\frac{d}{dt}\big(D(s)e^{-r(s-t)}\big)\,ds-D(t)\cdot e^{-r(t-t)}\\ > =&\int_{s=t}^{+\infty}rD(s)e^{-r(s-t)}\,ds-D(t)=r\int_{s=t}^{+\infty}D(s)e^{-r(s-t)}\,ds-D(t)=rP-D > \end{aligned} > $$ （注意 \(a(t)=t\) 给出下限项 \(-f(t,a(t))a'(t)=-D(t)e^{0}\cdot1=-D(t)\)，上限 \(+\infty\) 项为零。）\(\blacksquare\)
38. Leibniz Integral Rule
This rule is about taking derivative to an integral, and it is frequently used in continuous time models.
Theorem 38.1 (Leibniz integral rule) Under regularity conditions, the following equality holds: $$ > \frac{d}{dx}\left(\int_{a(x)}^{b(x)}f(x,t)\,dt\right)=f(x,b(x))\frac{db(x)}{dx}-f(x,a(x))\frac{da(x)}{dx}+\int_{a(x)}^{b(x)}\frac{\partial f(x,t)}{\partial x}\,dt > $$
Proof Let \(\psi(x)=\int_{a(x)}^{b(x)}f(x,t)\,dt\). Then, $$ > \begin{aligned} > d\psi(x)&=\psi(x+dx)-\psi(x)=\int_{a(x+dx)}^{b(x+dx)}f(x+dx,t)\,dt-\int_{a(x)}^{b(x)}f(x,t)\,dt\\ > &=-\int_{a(x)}^{a(x+dx)}f(x+dx,t)\,dt+\int_{a(x)}^{b(x)}f(x+dx,t)\,dt+\int_{b(x)}^{b(x+dx)}f(x+dx,t)\,dt-\int_{a(x)}^{b(x)}f(x,t)\,dt > \end{aligned} > $$ Consider the first term (i.e. \(-\int_{a(x)}^{a(x+dx)}f(x+dx,t)\,dt\)) and the third term (i.e. \(\int_{b(x)}^{b(x+dx)}f(x+dx,t)\,dt\)). By mean value theorem, \(\exists\xi_1\in(a(x),a(x+dx))\) s.t. \(-\int_{a(x)}^{a(x+dx)}f(x+dx,t)\,dt=-[a(x+dx)-a(x)]f(x+dx,\xi_1)\), and \(\exists\xi_2\in(b(x),b(x+dx))\) s.t. \(\int_{b(x)}^{b(x+dx)}f(x+dx,t)\,dt=[b(x+dx)-b(x)]f(x+dx,\xi_2)\). So, $$ > \begin{aligned} > \lim_{dx\to0}\frac{d\psi(x)}{dx}=&\lim_{dx\to0}\underbrace{\frac{-[a(x+dx)-a(x)]}{dx}}_{\to-a'(x)}\underbrace{f(x+dx,\xi_1)}_{\to f(x,a(x))}+\int_{a(x)}^{b(x)}\lim_{dx\to0}\underbrace{\frac{f(x+dx,t)-f(x,t)}{dx}}_{\to\frac{\partial f(x,t)}{\partial x}}dt\\ > &+\lim_{dx\to0}\underbrace{\frac{[b(x+dx)-b(x)]}{dx}}_{\to b'(x)}\underbrace{f(x+dx,\xi_2)}_{\to f(x,b(x))}\\ > =&f(x,b(x))\frac{db(x)}{dx}-f(x,a(x))\frac{da(x)}{dx}+\int_{a(x)}^{b(x)}\frac{\partial f(x,t)}{\partial x}\,dt > \end{aligned} > $$ which completes the proof. \(\blacksquare\)
Example 38.1 Consider the following integral equation, \(P(t)=\int_{s=t}^{+\infty}D(s)\cdot e^{-r(s-t)}\,ds\). Show that \(\dot P\equiv\dfrac{dP(t)}{dt}=rP-D\).
Proof $$ > \begin{aligned} > \dot P=\frac{dP(t)}{dt}&=\frac{d}{dt}\int_{s=t}^{+\infty}D(s)\cdot e^{-r(s-t)}\,ds\\ > \overset{\text{Leibniz}}{=}&\int_{s=t}^{+\infty}\frac{d}{dt}\big(D(s)e^{-r(s-t)}\big)\,ds-D(t)\cdot e^{-r(t-t)}\\ > =&\int_{s=t}^{+\infty}rD(s)e^{-r(s-t)}\,ds-D(t)=r\int_{s=t}^{+\infty}D(s)e^{-r(s-t)}\,ds-D(t)=rP-D > \end{aligned} > $$ (Here the lower limit \(a(t)=t\) gives the boundary term \(-f(t,a(t))a'(t)=-D(t)e^{0}\cdot1=-D(t)\), and the upper limit \(+\infty\) term is zero.) \(\blacksquare\)