Difference between revisions of "Leibniz integral rule"
Borishaase (talk | contribs) (Leibniz integral rule) |
Borishaase (talk | contribs) (Leibniz integral rule) |
||
| (2 intermediate revisions by the same user not shown) | |||
| Line 1: | Line 1: | ||
__NOTOC__ | __NOTOC__ | ||
=== Leibniz integral rule === | === 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}, {} | + | For <math>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>s \in {}^{(\omega)}\mathbb{K} \setminus \{{x}_{1}\}</math>, choosing <math>\overset{\rightharpoonup}{a}(x) = a(\overset{\rightharpoonup}{x})</math> aund <math>\overset{\rightharpoonup}{e}(x) = e(\overset{\rightharpoonup}{x})</math>, it holds that |
| − | <div style="text-align:center;"><math>\tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{ | + | <div style="text-align:center;"><math>\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></div> |
| − | === Proof === | + | === Proof: === |
| − | <div style="text-align:center | + | <div style="text-align:center;"><math>\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></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>{} | + | Complex integration allows a [[w:Path (topology)|<span class="wikipedia">path</span>]] whose start and end points are the limits of integration. If <math>\overset{\rightharpoonup}{a}(x) \ne a(\overset{\rightharpoonup}{x})</math>, then multiply the final summand by <math>(\overset{\rightharpoonup}{a}(x) - a(x))/(a(\overset{\rightharpoonup}{x}) - a(x))</math>, and if <math>\overset{\rightharpoonup}{e}(x) \ne e(\overset{\rightharpoonup}{x})</math>, then the penultimate summand must be multiplied by <math>(\overset{\rightharpoonup}{e}(x) - e(x))/(e(\overset{\rightharpoonup}{x}) - e(x))</math>. |
== Siehe auch == | == Siehe auch == | ||
Latest revision as of 18:19, 1 May 2024
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]
Remark
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].