Difference between revisions of "Leibniz integral rule"
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| − | For <math>f: {}^{(\omega)}\mathbb{K}^{n | + | === Leibniz integral rule === |
| + | For <math>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>s \in {}^{(\omega)}\mathbb{K} \setminus \{{x}_{1}\}</math>, choosing <math>{}^\curvearrowright a(x) = a({}^\curvearrowright x)</math> and <math>{}^\curvearrowright b(x) = b({}^\curvearrowright x)</math>, it holds that | ||
| − | <div style="text-align:center;"><math>\ | + | <div style="text-align:center;"><math>\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></div> |
| − | === | + | === Proof === |
| − | <div style="text-align:center; font-size: 84%;"><math>\begin{aligned}\ | + | <div style="text-align:center; font-size: 84%;"><math>\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></div> |
=== Remark === | === Remark === | ||
| − | + | Complex integration allows a [[w:Path (topology)|<span class="wikipedia">path</span>]] whose start and end points are the limits of integration. If <math>{}^\curvearrowright b(x) \ne b({}^\curvearrowright x)</math>, then multiply the final summand by <math>({}^\curvearrowright a(x) - a(x))/(a({}^\curvearrowright x) - a(x))</math>, and if <math>{}^\curvearrowright b(x) \ne b({}^\curvearrowright x)</math>, then the penultimate summand must be multiplied by <math>({}^\curvearrowright b(x) - b(x))/(b({}^\curvearrowright x) - b(x))</math>. | |
== Siehe auch == | == Siehe auch == | ||
Revision as of 16:00, 29 September 2023
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]
Remark
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].