Leibniz integral rule
For [math]\displaystyle{ f: {}^{(\omega)}\mathbb{K}^{\grave{n}} \rightarrow {}^{(\omega)}\mathbb{K}, a, b: {}^{(\omega)}\mathbb{K}^{n} \rightarrow {}^{(\omega)}\mathbb{K}, {}^\curvearrowright x := {(s, {x}_{2}, ..., {x}_{n})}^{T} }[/math] and [math]\displaystyle{ s \in {}^{(\omega)}\mathbb{K} \setminus \{{x}_{1}\} }[/math], choosing [math]\displaystyle{ {}^\curvearrowright a(x) = a({}^\curvearrowright x) }[/math] and [math]\displaystyle{ {}^\curvearrowright b(x) = b({}^\curvearrowright x) }[/math], it holds that
[math]\displaystyle{ \tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}t} \right)={\uparrow}_{a(x)}^{b(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}t}+\tfrac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f({}^\curvearrowright x,b(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f({}^\curvearrowright x,a(x)). }[/math]
Proof
[math]\displaystyle{ \begin{aligned}\tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}t} \right) &={\left( {\uparrow}_{a({}^\curvearrowright x)}^{b({}^\curvearrowright x)}{f({}^\curvearrowright x,t){\downarrow}t}-{\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}t} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\left( {\uparrow}_{a(x)}^{b(x)}{(f({}^\curvearrowright x,t)-f(x,t)){\downarrow}t}+{\uparrow}_{b(x)}^{b({}^\curvearrowright x)}{f({}^\curvearrowright x,t){\downarrow}t}-{\uparrow}_{a(x)}^{a({}^\curvearrowright x)}{f({}^\curvearrowright x,t){\downarrow}t} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\uparrow}_{a(x)}^{b(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}t}+\tfrac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f({}^\curvearrowright x,b(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f({}^\curvearrowright x,a(x)).\square\end{aligned} }[/math]
Complex integration allows a path whose start and end points are the limits of integration. If [math]\displaystyle{ {}^\curvearrowright b(x) \ne b({}^\curvearrowright x) }[/math], then multiply the final summand by [math]\displaystyle{ ({}^\curvearrowright a(x) - a(x))/(a({}^\curvearrowright x) - a(x)) }[/math], and if [math]\displaystyle{ {}^\curvearrowright b(x) \ne b({}^\curvearrowright x) }[/math], then the penultimate summand must be multiplied by [math]\displaystyle{ ({}^\curvearrowright b(x) - b(x))/(b({}^\curvearrowright x) - b(x)) }[/math].
Siehe auch