积分符号内取微分

${\displaystyle F(x,a(x),b(x))=\int _{a(x)}^{b(x)}f(x,t)\,dt}$,

${\displaystyle f(x,t)\,}$${\displaystyle {\frac {\partial }{\partial x}}\,f(x,t)\,}$${\displaystyle t\,}$${\displaystyle x\,}$${\displaystyle (t,x)\,}$ 平面连续, ${\displaystyle a(x)\leq t\leq b(x)\,}$, ${\displaystyle x_{0}\leq x\leq x_{1}\,}$, 且若对于${\displaystyle x_{0}\leq x\leq x_{1}\,}$, ${\displaystyle a(x)\,}$${\displaystyle b(x)\,}$ 及其导数连续，

{\displaystyle {\begin{aligned}{\frac {d}{dx}}\,F(x,a(x),b(x))&=\left({\frac {\partial F}{\partial b}}\right){\frac {db}{dx}}+\left({\frac {\partial F}{\partial a}}\right){\frac {da}{dx}}+{\frac {\partial F}{\partial x}}\\&=f(x,b(x))\,b'(x)-f(x,a(x))\,a'(x)+\int _{a(x)}^{b(x)}{\frac {\partial }{\partial x}}\,f(x,t)\;dt\,\end{aligned}}}

${\displaystyle {\frac {d}{dx}}\left(\int _{a}^{b}f(x,t)\,dt\right)=\int _{a}^{b}{\frac {\partial }{\partial x}}f(x,t)\,dt.}$

高维情况

${\displaystyle {\frac {d}{dt}}\int _{D(t)}F({\vec {\textbf {x}}},t)\,dV=\int _{D(t)}{\frac {\partial }{\partial t}}\,F({\vec {\textbf {x}}},t)\,dV+\int _{\partial D(t)}\,F({\vec {\textbf {x}}},t)\,{\vec {\textbf {v}}}\cdot {\vec {\textbf {n}}}\,dA,\,}$

定理的证明

${\displaystyle {\frac {\partial }{\partial b}}\left(\int _{a}^{b}f(x)\;\mathrm {d} x\right)=f(b),\qquad {\frac {\partial }{\partial a}}\left(\int _{a}^{b}f(x)\;\mathrm {d} x\right)=-f(a).}$

{\displaystyle {\begin{aligned}{\frac {\partial }{\partial b}}\left(\int _{a}^{b}f(x)\;\mathrm {d} x\right)&=\lim _{\Delta b\to 0}{\frac {1}{\Delta b}}\left[\int _{a}^{b+\Delta b}f(x)\,\mathrm {d} x-\int _{a}^{b}f(x)\,\mathrm {d} x\right]\\&=\lim _{\Delta b\to 0}{\frac {1}{\Delta b}}\int _{b}^{b+\Delta b}f(x)\,\mathrm {d} x\\&=\lim _{\Delta b\to 0}{\frac {1}{\Delta b}}\left[f(b)\Delta b+{\mathcal {O}}\left(\Delta b^{2}\right)\right]\\&=f(b)\\{\frac {\partial }{\partial a}}\left(\int _{a}^{b}f(x)\;\mathrm {d} x\right)&=\lim _{\Delta a\to 0}{\frac {1}{\Delta a}}\left[\int _{a+\Delta a}^{b}f(x)\,\mathrm {d} x-\int _{a}^{b}f(x)\,\mathrm {d} x\right]\\&=\lim _{\Delta a\to 0}{\frac {1}{\Delta a}}\int _{a+\Delta a}^{a}f(x)\,\mathrm {d} x\\&=\lim _{\Delta a\to 0}{\frac {1}{\Delta a}}\left[-f(a)\,\Delta a+{\mathcal {O}}\left(\Delta a^{2}\right)\right]\\&=-f(a).\end{aligned}}}

${\displaystyle \psi (\alpha )=\int _{a}^{b}f(x,\alpha )\;\mathrm {d} x.}$
${\displaystyle \psi }$ 可以對 ${\displaystyle \alpha }$ 在积分符号内取微分,即
${\displaystyle {\frac {\mathrm {d} \psi }{\mathrm {d} \alpha }}=\int _{a}^{b}{\frac {\partial }{\partial \alpha }}\,f(x,\alpha )\,\mathrm {d} x.\,}$

${\displaystyle |f(x,\alpha +\Delta \alpha )-f(x,\alpha )|<\varepsilon .}$

{\displaystyle {\begin{aligned}\Delta \psi &=\psi (\alpha +\Delta \alpha )-\psi (\alpha )\\&=\int _{a}^{b}f(x,\alpha +\Delta \alpha )\;\mathrm {d} x-\int _{a}^{b}f(x,\alpha )\;\mathrm {d} x\\&=\int _{a}^{b}\left(f(x,\alpha +\Delta \alpha )-f(x,\alpha )\right)\;\mathrm {d} x\\&\leq \varepsilon (b-a)\end{aligned}}}

${\displaystyle \forall x\in [a,b]\quad \left|{\frac {f(x,\alpha +\Delta \alpha )-f(x,\alpha )}{\Delta \alpha }}-{\frac {\partial f}{\partial \alpha }}\right|<\varepsilon .}$

${\displaystyle {\frac {\Delta \psi }{\Delta \alpha }}=\int _{a}^{b}{\frac {f(x,\alpha +\Delta \alpha )-f(x,\alpha )}{\Delta \alpha }}\;\mathrm {d} x=\int _{a}^{b}{\frac {\partial \,f(x,\alpha )}{\partial \alpha }}\,\mathrm {d} x+R}$

${\displaystyle |R|<\int _{a}^{b}\varepsilon \;\mathrm {d} x=\varepsilon (b-a).}$

${\displaystyle \lim _{{\Delta \alpha }\rightarrow 0}{\frac {\Delta \psi }{\Delta \alpha }}={\frac {\mathrm {d} \psi }{\mathrm {d} \alpha }}=\int _{a}^{b}{\frac {\partial }{\partial \alpha }}\,f(x,\alpha )\,\mathrm {d} x.\,}$

${\displaystyle \int _{a}^{b}f(x,\alpha )\;\mathrm {d} x=\varphi (\alpha ),}$

{\displaystyle {\begin{aligned}\Delta \varphi &=\varphi (\alpha +\Delta \alpha )-\varphi (\alpha )\\&=\int _{a+\Delta a}^{b+\Delta b}f(x,\alpha +\Delta \alpha )\;\mathrm {d} x\,-\int _{a}^{b}f(x,\alpha )\;\mathrm {d} x\,\\&=\int _{a+\Delta a}^{a}f(x,\alpha +\Delta \alpha )\;\mathrm {d} x+\int _{a}^{b}f(x,\alpha +\Delta \alpha )\;\mathrm {d} x+\int _{b}^{b+\Delta b}f(x,\alpha +\Delta \alpha )\;\mathrm {d} x-\int _{a}^{b}f(x,\alpha )\;\mathrm {d} x\\&=-\int _{a}^{a+\Delta a}\,f(x,\alpha +\Delta \alpha )\;\mathrm {d} x+\int _{a}^{b}[f(x,\alpha +\Delta \alpha )-f(x,\alpha )]\;\mathrm {d} x+\int _{b}^{b+\Delta b}\,f(x,\alpha +\Delta \alpha )\;\mathrm {d} x.\end{aligned}}}

${\displaystyle \Delta \varphi =-\Delta a\,f(\xi _{1},\alpha +\Delta \alpha )+\int _{a}^{b}[f(x,\alpha +\Delta \alpha )-f(x,\alpha )]\;\mathrm {d} x+\Delta b\,f(\xi _{2},\alpha +\Delta \alpha )}$
${\displaystyle =-\Delta a\,f(\xi _{1},\alpha +\Delta \alpha )+\psi (\alpha +\Delta \alpha )-\psi (\alpha )+\Delta b\,f(\xi _{2},\alpha +\Delta \alpha )}$.

${\displaystyle {\frac {\mathrm {d} \psi }{\mathrm {d} \alpha }}=\int _{a}^{b}{\frac {\partial }{\partial \alpha }}\,f(x,\alpha )\,\mathrm {d} x}$

${\displaystyle {\frac {\mathrm {d} \varphi }{\mathrm {d} \alpha }}=\int _{a}^{b}{\frac {\partial }{\partial \alpha }}\,f(x,\alpha )\,\mathrm {d} x+f(b,\alpha ){\frac {\partial b}{\partial \alpha }}-f(a,\alpha ){\frac {\partial a}{\partial \alpha }}.}$