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Revision as of 16:13, 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
Proof
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].