Leibniz integral rule
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],
Proof
Remark
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].