For [math]\displaystyle{ f: {}^{(\omega)}\mathbb{K}^{n+1} \rightarrow {}^{(\omega)}\mathbb{K}, a, b: {}^{(\omega)}\mathbb{K}^{n} \rightarrow {}^{(\omega)}\mathbb{K}, \curvearrowright B x := {(s, {x}_{2}, ..., {x}_{n})}^{T} }[/math] and [math]\displaystyle{ s \in {}^{(\omega)}\mathbb{K} \setminus \{{x}_{1}\} }[/math], choosing [math]\displaystyle{ \curvearrowright D a(x) = a(\curvearrowright B x) }[/math] and [math]\displaystyle{ \curvearrowright D b(x) = b(\curvearrowright B x) }[/math],
[math]\displaystyle{ \frac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}Dt} \right)={\uparrow}_{a(x)}^{b(x)}{\frac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}Dt}+\frac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright Bx,b(x))-\frac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright Bx,a(x)). }[/math]
Beweis
[math]\displaystyle{ \begin{aligned}\frac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}Dt} \right) &={\left( {\uparrow}_{a(\curvearrowright Bx)}^{b(\curvearrowright Bx)}{f(\curvearrowright Bx,t){\downarrow}Dt}-{\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}Dt} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\left( {\uparrow}_{a(x)}^{b(x)}{(f(\curvearrowright Bx,t)-f(x,t)){\downarrow}Dt}+{\uparrow}_{b(x)}^{b(\curvearrowright Bx)}{f(\curvearrowright Bx,t){\downarrow}Dt}-{\uparrow}_{a(x)}^{a(\curvearrowright Bx)}{f(\curvearrowright Bx,t){\downarrow}Dt} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\uparrow}_{a(x)}^{b(x)}{\frac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}dDt}+\frac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright Bx,b(x))-\frac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright Bx,a(x)).\square\end{aligned} }[/math]
Integrating happens in the complex plane over a path whose start and end points are the limits of integration. If [math]\displaystyle{ \curvearrowright D a(x) \ne a(\curvearrowright B x) }[/math], then the final summand must be multiplied by [math]\displaystyle{ (\curvearrowright D a(x) - a(x))/(a(\curvearrowright B x) - a(x)) }[/math], and if [math]\displaystyle{ \curvearrowright D b(x) \ne b(\curvearrowright B x) }[/math], then the penultimate summand must be multiplied by [math]\displaystyle{ (\curvearrowright D b(x) - b(x))/(b(\curvearrowright B x) - b(x)) }[/math].
Siehe auch