The Unapologetic Mathematician

Mathematics for the interested outsider

Iterated Integrals V

Our iterated integrals worked out nicely over n-dimensional intervals because these regions are simple products of one-dimensional intervals. When working over more general sets, though, it’s not so nice. Still, we can often do almost as well.

Unlike our earlier approach, though, we won’t peel off an integral from the outside first, but from an inside. That is, instead of writing


we’ll write


Underlying all of these equations is the assumption that each of the integrals we write down exists, or at least the inner ones don’t fail to exist often enough to cause the outer ones to also fail to exist.

Okay, so let’s say that we’ve got some bounded region S contained in an n-dimensional interval R. Just like before, we project S into the plane with x^k=0 to get the n-1-dimensional region S_k. This will be contained in the projection R_k of R. Again, we write \hat{x} for the remaining n-1 variables.

Here’s where our restriction comes in: assume that S is contained between the graphs of two integrable functions on S_k. That is, we want

\displaystyle \mathrm{Cl}(S)=\left\{\left(x^1,\dots,x^n\right)\big\vert\hat{x}\in\mathrm{Cl}(S_k),g_1(\hat{x})\leq x^k\leq g_2(\hat{x})\right\}

where g_1:S_k\rightarrow\mathbb{R} and g_2:S_k\rightarrow\mathbb{R} are two integrable functions defined on S_k.

Now we can try to evaluate the integral


We first replace R_k by S_k because \chi_S(x) is identically zero wherever \hat{x} is outside of S_k, and thus so will the whole inner integral. Then we replace the limits of integration for the inner integral by g_1(\hat{x}) and g_2(\hat{x}) because for any fixed \hat{x} the function \chi_S(x) is zero for x^k below the former and above the latter.

If we’re lucky, S_k itself can be written in a similar form, as the region between the graphs of two integrable functions in another coordinate direction, and so on. Each inner integral’s integrand depends on the remaining variables of later integrals, as before. But now the bounds of integration can also depend on later variables, which adds new complication. However, once a given variable has been integrated away, nothing further out ever depends on it again, and so this whole procedure is still well-defined.

December 31, 2009 Posted by | Analysis, Calculus | 9 Comments