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This 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 i
by Bahamut 6y ago
This 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]...