Iterated Integrals IV
So we’ve established that as long as a double integral exists, we can use an iterated integral to evaluate it. What happens when the dimension of our space is even bigger?
In this case, we’re considering integrating an integrable function 
over some 
-dimensional interval ![R=[a^1,b^1]\times\dots\times[a^n,b^n]](http://s0.wp.com/latex.php?latex=R%3D%5Ba%5E1%2Cb%5E1%5D%5Ctimes%5Cdots%5Ctimes%5Ba%5En%2Cb%5En%5D&bg=e6e6e6&fg=333333&s=0 "R=[a^1,b^1]\times\dots\times[a^n,b^n]")
. We want something like iterated integrals to allow us to evaluate this multiple integral. We’ll do this by peeling off a single integral from the outside and leaving an integral over an 
-dimensional integral inside.
Specifically, we can project the interval 
onto the coordinate hyperplane defined by 
just by leaving the coordinates 
of each point 
the same if 
and setting . We’ll call the resulting interval
![\displaystyle R_k=[a^1,b^1]\times\dots\times\widehat{[a^k,b^k]}\times\dots\times[a^n,b^n]](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+R_k%3D%5Ba%5E1%2Cb%5E1%5D%5Ctimes%5Cdots%5Ctimes%5Cwidehat%7B%5Ba%5Ek%2Cb%5Ek%5D%7D%5Ctimes%5Cdots%5Ctimes%5Ba%5En%2Cb%5En%5D&bg=e6e6e6&fg=333333&s=0 "\displaystyle R_k=[a^1,b^1]\times\dots\times\widehat{[a^k,b^k]}\times\dots\times[a^n,b^n]")
where the wide hat means that we just leave out that one factor in the product. We’ll also write 
to mean the remaining coordinates on 
.
Essentially, we want to integrate first over , and then let 
run from 
to 
. We have a collection of assertions that parallel those from the two-dimensional case
with a copy of these three for each index 
between 
and . The proofs of these are pretty much identical to the proofs in the two-dimensional case, and so I’ll just skip them.
Anyhow, once we’ve picked one of the variables and split it off as the outermost integral, we’re left with an -dimensional integral on the inside. We can pick any one of these variables and split it off, leaving an 
-dimensional integral on the inside, and so on. For each of the 
orderings of the original variables, we get a way of writing the -dimensional integral over as a sequence of integrals, each over a one-dimensional interval. Now, we may find some of these iterated integrals easier to evaluate than others, but in principle, if each of the 
-dimensional integrals in the sequence exists it doesn’t matter which of the orderings we use.
So, for example, if we’re considering a bounded function 
defined on a three-dimensional interval ![R=[a^1,b^1]\times[a^2,b^2]\times[a^3,b^3]](http://s0.wp.com/latex.php?latex=R%3D%5Ba%5E1%2Cb%5E1%5D%5Ctimes%5Ba%5E2%2Cb%5E2%5D%5Ctimes%5Ba%5E3%2Cb%5E3%5D&bg=e6e6e6&fg=333333&s=0 "R=[a^1,b^1]\times[a^2,b^2]\times[a^3,b^3]")
, we can write (up to) six different iterated integrals, assuming that all the integrals in sight exist.
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December 21, 2009 -
Posted by John Armstrong |
Analysis, Calculus
exists, then we have
I don’t think you should use metaphors like “bigger” when writing about mathematics. It’s shaky enough in science as it is. See e.g. Thomas Kuhng’s “Metaphors in Science.” I think it’s a lecture, but it’s found in several collections as well.
NS
NS, you’re welcome to go read Principia Mathematica and do everything in formal systems. Most humans need some amount of metaphor to get their hands on mathematical concepts, especially when speaking conversation. Forswearing “bigger” when referring to natural numbers would be incredibly stilted and awkward.
But then I’ve read your prose, so..
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