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}, \overset{\rightharpoonup}{x} := {(s, {x}_{2}, ..., {x}_{n})}^{T} }[/math] and [math]\displaystyle{ s \in {}^{(\omega)}\mathbb{K} \setminus \{{x}_{1}\} }[/math], choosing [math]\displaystyle{ \overset{\rightharpoonup}{a}(x) = a(\overset{\rightharpoonup}{x}) }[/math] aund [math]\displaystyle{ \overset{\rightharpoonup}{e}(x) = e(\overset{\rightharpoonup}{x}) }[/math], it holds that
[math]\displaystyle{ \tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{e(x)}{f(x,t){\downarrow}t} \right)={\uparrow}_{a(x)}^{e(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}t}+\tfrac{{\downarrow} e(x)}{{\downarrow} {{x}_{1}}}f(\overset{\rightharpoonup}{x},e(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\overset{\rightharpoonup}{x},a(x)). }[/math]
Proof:
[math]\displaystyle{ \begin{aligned}\tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{e(x)}{f(x,t){\downarrow}t} \right) & ={\left( {\uparrow}_{a(\overset{\rightharpoonup}{x})}^{e(\overset{\rightharpoonup}{x})}{f(\overset{\rightharpoonup}{x},t){\downarrow}t}-{\uparrow}_{a(x)}^{e(x)}{f(x,t){\downarrow}t} \right)}/{{\downarrow} {{x}_{1}}}\; \\ & ={\left( {\uparrow}_{a(x)}^{e(x)}{(f(\overset{\rightharpoonup}{x},t)-f(x,t)){\downarrow}t}+{\uparrow}_{e(x)}^{e(\overset{\rightharpoonup}{x})}{f(\overset{\rightharpoonup}{x},t){\downarrow}t}-{\uparrow}_{a(x)}^{a(\overset{\rightharpoonup}{x})}{f(\overset{\rightharpoonup}{x},t){\downarrow}t} \right)}/{{\downarrow} {{x}_{1}}}\; \\ & ={\uparrow}_{a(x)}^{e(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}t}+\tfrac{{\downarrow} e(x)}{{\downarrow} {{x}_{1}}}f(\overset{\rightharpoonup}{x},e(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\overset{\rightharpoonup}{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{ \overset{\rightharpoonup}{a}(x) \ne a(\overset{\rightharpoonup}{x}) }[/math], then multiply the final summand by [math]\displaystyle{ (\overset{\rightharpoonup}{a}(x) - a(x))/(a(\overset{\rightharpoonup}{x}) - a(x)) }[/math], and if [math]\displaystyle{ \overset{\rightharpoonup}{e}(x) \ne e(\overset{\rightharpoonup}{x}) }[/math], then the penultimate summand must be multiplied by [math]\displaystyle{ (\overset{\rightharpoonup}{e}(x) - e(x))/(e(\overset{\rightharpoonup}{x}) - e(x)) }[/math].
Siehe auch