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>one needs the notions of Lebesgue integral and measure. The Riemann integral from calculus courses is just not robust enough. definitely not the case. leibniz
by anon_tor_12345 6y ago
>one needs the notions of Lebesgue integral and measure. The Riemann integral from calculus courses is just not robust enough.
definitely not the case. leibniz's rule
https://en.wikipedia.org/wiki/Leibniz_integral_rule#Proof_of_basic_form https://en.wikipedia.org/wiki/Leibniz_integral_rule#Proof_of...
only requires fubini's theorem for exchanging order of integration
https://en.wikipedia.org/wiki/Fubini%27s_theorem#Riemann_integrals https://en.wikipedia.org/wiki/Fubini%27s_theorem#Riemann_int...
which i'm pretty sure everyone learns in multivariable calc.
i personally learned it from apostol's calc (not analysis) books.
- Bahamut 6y agoThis is not true - it's not necessarily obvious that things should hold when you're dealing with infinity as one of the bounds. In fact, there are cases where interchanging the order is not allowed when you're in that situation depending on the integrand.
- anon_tor_12345 6y agoi'm at a loss. >In fact, there are cases where interchanging the order is not allowed when you're in that situation depending on the integrand. yup exactly in the cases where fubini's theorem doesn't hold and therefore those cases for which the integral theorem doesn't hold. Leibniz Theorem: Let f(x, t) be a function such that both f(x, t) and its partial derivative fx(x, t) are continuous in t and x in some region of the (x, t)-plane, including a(x) ≤ t ≤ b(x), x0 ≤ x ≤ x1. Also suppose that the functions a(x) and b(x) are both continuous and both have continuous derivatives for x0 ≤ x ≤ x1... Fubini's Theorem: If f(x,y) is a *continuous function* on a rectangle R=[a,b]×[c,d]...