How Does Fubini's Theorem Relate to the Product of Two Integrals?

[O.J.]

Product of two integrals...

In proving a theorem, my DE textbook uses an unfamiliar approach by stating that
the product of two integrals = double integral sign - the product of two functions - dx dy

i hope my statement is descriptive enough. My question is, what's the proof to this?

 

[CompuChip]

You mean, why is it true that
\left( \int f(x) dx \right) \left( \int g(x) dx \right) = \iint f(x) g(y) dx dy?
Because that's simply by the property that we can put any "constant" (that is, anything which does not depend on the integration variable) outside or inside an integral as we like. We can obviously rewrite
\left( \int f(x) dx \right) \left( \int g(x) dx \right) = \left( \int f(x) dx \right) \left( \int g(y) dy \right)
by renaming a dummy variable in the second integral. Now you see that the first part does not depend on y at all (just on x, but that's a different variable!), so we can take the entire first integral inside the second integration,
\left( \int f(x) dx \right) \left( \int g(x) dx \right) = \int \left( \int f(x) dx \right) g(y) dy
and since g(y) doesn't depend on x we can take that inside the x-integration (and then of course drop the brackets, as they don't clarify anything anymore).

Hope that answers the question.

 

[Gib Z]

Rewriting the dummy variable only yields the same result in Definite integration. I'm sure CompuChip meant this, writing bounds on all those integral signs wouldn't have been fun.

 

[arildno]

It is a property of Fubini's theorem that assuming the double integral over the absolute value of f converges, then it will yield the same value as the "nested" integral computations, irrespective of nesting order.