- #1
amcavoy
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Does anyone know how to prove Fubini's Theorem? In multivariable calc., I just accepted it as true but never learned the formal proof. Any ideas?
Thanks for your help.
Thanks for your help.
Hurkyl said:Well, you can always try proving it yourself.
Hurkyl said:It ought to be easier to prove
[tex]
\iint_R f = \int_a^b \int_c^d f(x, y) \, dy \, dx
[/tex]
than to prove
[tex]
\int_a^b \int_c^d f(x, y) \, dy \, dx
= \int_c^d \int_a^b f(x, y) \, dx \, dy
[/tex]
quasar987 said:For your personal amusement apmcavoy, this thm of diff. cal. is called Clairaut's Theorem (also sometimes called Schwartz's Theorem...but Schwartz already has his inequality, so let's be fair and give this one to Clairaut, shall we :P)
Castilla said:Apmcavoy, do you know the proof of the Leibniz rule (differentiation under the integral sign)? From said proof is rather easy to obtain the statement wrote by Quasar.
Put "Leibniz rule" in google and you will see it.
Castilla.
apmcavoy said:Yes I know it (worked through it on PF a few months back!)
quasar987 said:Could you post the link please? I can't find the thread.
As for the statement, Castilla was probably referring to
[tex]\int_a^b \int_c^d f(x, y) \, dy \, dx= \int_c^d \int_a^b f(x, y) \, dx \, dy[/tex]
quasar987 said:I tought Fubini's theorem was just
[tex]\int_a^b \int_c^d f(x, y) \, dy \, dx
= \int_c^d \int_a^b f(x, y) \, dx \, dy[/tex]
Castilla said:This equation. If this is what you want to proof, it can be considered a corolary from Leibniz rule (differentiation under the integral sign).
quasar987 said:I think Castilla is saying that fubini is a corolary, or at least that the proof is similar, to Leibniz rule (the theorem stated at proved by homology in post #15: https://www.physicsforums.com/showpost.php?p=264381&postcount=15)
Castilla said:Fact: the function with domain [tex] [a,b]x [c,d] [/tex] is continuous.
Let be [tex]g(x,z) = \int_c^zf(x,y)dy.[/tex] Then, by the Fundamental Theorem of Calculus, [tex]\frac{\partial g}{\partial z}= f. [/tex]
Let be
[tex] G(z) = \int_a^bg(x,z)dx[/tex], so by differentiation under the integral sign
[tex] G'(z) = \int_a^bf(x,z)dx[/tex].
Assume that G' is continuous. Then
[tex] \int_c^d\int_a^bf(x,z)dxdz = \int_c^dG' = G(d) - G(c) = G(d) =
\int_a^b\int_c^df(x,y)dydx [/tex].
Castilla said:I think that the version developed is known as "Baby Fubini's theorem". There is a second version which includes double integrals (not only iterated, as the "Baby"), and Hurkyl refers to a third more "mature" version.
Fubini's Theorem is a mathematical theorem that states that for a function of two variables defined on a rectangular region, the double integral of the function can be evaluated as an iterated integral, where the inner integral is with respect to one variable and the outer integral is with respect to the other variable.
Proving Fubini's Theorem is important because it allows us to interchange the order of integration when evaluating double integrals. This can greatly simplify calculations and make solving certain problems more efficient.
Some tips for proving Fubini's Theorem include carefully understanding the definition of the theorem, using the properties of integrals and limits, and breaking down the problem into smaller, more manageable parts.
Yes, Fubini's Theorem can be extended to functions of more than two variables. This is known as the Tonelli's Theorem and it states that for a function of n variables, the multiple integral can be evaluated as an iterated integral with n nested integrals.
One common mistake to avoid when proving Fubini's Theorem is assuming that the integrals are interchangeable without properly justifying it. It is important to carefully follow the steps and criteria outlined in the theorem to ensure a correct proof.