# The Unapologetic Mathematician

## 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 $f:R\rightarrow\mathbb{R}$ over some $n$-dimensional interval $R=[a^1,b^1]\times\dots\times[a^n,b^n]$. 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 $n-1$-dimensional integral inside.

Specifically, we can project the interval $R$ onto the coordinate hyperplane defined by $x^k=0$ just by leaving the coordinates $x^i$ of each point $x\in R$ the same if $i\neq k$ and setting $x^k=0$. We’ll call the resulting interval

$\displaystyle R_k=[a^1,b^1]\times\dots\times\widehat{[a^k,b^k]}\times\dots\times[a^n,b^n]$

where the wide hat means that we just leave out that one factor in the product. We’ll also write $\hat{x}$ to mean the remaining coordinates on $R_k$.

Essentially, we want to integrate first over $R_k$, and then let $x^k$ run from $a^k$ to $b^k$. We have a collection of assertions that parallel those from the two-dimensional case

• $\displaystyle{\int\limits_-}_Rf(x)\,dx\leq{\int\limits_-}_{a^k}^{b^k}{\int\limits^-}_{R_k}f(\hat{x},x^k)\,d\hat{x}\,dx^k\leq{\int\limits^-}_{a^k}^{b^k}{\int\limits^-}_{R_k}f(\hat{x},x^k)\,d\hat{x}\,dx^k\leq{\int\limits^-}_Rf(x)\,dx$
• $\displaystyle{\int\limits_-}_Rf(x)\,dx\leq{\int\limits_-}_{a^k}^{b^k}{\int\limits_-}_{R_k}f(\hat{x},x^k)\,d\hat{x}\,dx^k\leq{\int\limits^-}_{a^k}^{b^k}{\int\limits_-}_{R_k}f(\hat{x},x^k)\,d\hat{x}\,dx^k\leq{\int\limits^-}_Rf(x)\,dx$
• If $\int_Rf(x)\,dx$ exists, then we have
$\displaystyle\int\limits_Rf(x)\,dx=\int\limits_{a^k}^{b^k}{\int\limits_-}_{R_k}f(\hat{x},x^k)\,d\hat{x}\,dx^k=\int\limits_{a^k}^{b^k}{\int\limits^-}_{R_k}f(\hat{x},x^k)\,d\hat{x}\,dx^k$

with a copy of these three for each index $k$ between ${1}$ and $n$. 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 $n$ variables and split it off as the outermost integral, we’re left with an $n-1$-dimensional integral on the inside. We can pick any one of these variables and split it off, leaving an $n-2$-dimensional integral on the inside, and so on. For each of the $n!$ orderings of the original $n$ variables, we get a way of writing the $n$-dimensional integral over $R$ as a sequence of $n$ 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 $m$-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 $f$ defined on a three-dimensional interval $R=[a^1,b^1]\times[a^2,b^2]\times[a^3,b^3]$, we can write (up to) six different iterated integrals, assuming that all the integrals in sight exist.

\displaystyle\begin{aligned}\iiint\limits_R&=\int\limits_{a^3}^{b^3}\int\limits_{a^2}^{b^2}\int\limits_{a^1}^{b^1}f(x,y,z)\,dx\,dy\,dz=\int\limits_{a^2}^{b^2}\int\limits_{a^3}^{b^3}\int\limits_{a^1}^{b^1}f(x,y,z)\,dx\,dz\,dy\\&=\int\limits_{a^3}^{b^3}\int\limits_{a^1}^{b^1}\int\limits_{a^2}^{b^2}f(x,y,z)\,dy\,dx\,dz=\int\limits_{a^1}^{b^1}\int\limits_{a^3}^{b^3}\int\limits_{a^2}^{b^2}f(x,y,z)\,dy\,dz\,dx\\&=\int\limits_{a^2}^{b^2}\int\limits_{a^1}^{b^1}\int\limits_{a^3}^{b^3}f(x,y,z)\,dz\,dx\,dy=\int\limits_{a^1}^{b^1}\int\limits_{a^2}^{b^2}\int\limits_{a^3}^{b^3}f(x,y,z)\,dz\,dy\,dx\end{aligned}

December 21, 2009 Posted by | Analysis, Calculus | 4 Comments