The Unapologetic Mathematician

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

$\displaystyle\int\limits_Rf(x)\,dx=\int\limits_{a^k}^{b^k}\int\limits_{R_k}f(\hat{x},x^k)\,d\hat{x}\,dx^k$

we’ll write

$\displaystyle\int\limits_Rf(x)\,dx=\int\limits_{R_k}\int\limits_{a^k}^{b^k}f(\hat{x},x^k)\,dx^k\,d\hat{x}$

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

\displaystyle\begin{aligned}\int\limits_Sf\,dx&=\int\limits_Rf(x)\chi_S(x)\,dx\\&=\int\limits_{R_k}\int\limits_{a^k}^{b^k}f(x)\chi_S(x)\,dx^k\,d\hat{x}\\&=\int\limits_{S_k}\int\limits_{a^k}^{b^k}f(x)\chi_S(x)\,dx^k\,d\hat{x}\\&=\int\limits_{S_k}\int\limits_{g_1(\hat{x})}^{g_2(\hat{x})}f(x)\chi_S(x)\,dx^k\,d\hat{x}\\&=\int\limits_{S_k}\int\limits_{g_1(\hat{x})}^{g_2(\hat{x})}f(x)\,dx^k\,d\hat{x}\end{aligned}

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

1. I wish you a Happy New Year!

Comment by Américo Tavares | December 31, 2009 | Reply

2. Thanks, but it doesn’t look promising.

Comment by John Armstrong | December 31, 2009 | Reply

3. I admire the verbal precision of “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.”

And it’s too early to prove that 2010 will NOT be a good year for both of us.

Comment by Jonathan Vos Post | January 1, 2010 | Reply

4. Yea, the race goes not always to the swift, nor the battle to the strong, but that’s the way to bet.

Comment by John Armstrong | January 1, 2010 | Reply

5. John,

Américo

Comment by Américo Tavares | January 1, 2010 | Reply

6. And I’m sure it’ll never be written unless I get a stable job.

Comment by John Armstrong | January 1, 2010 | Reply

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