📂Vector Analysis

Integration of Multivariable Functions

Definition¹

Let $I^{k}$ be a k-cell, and assume $\mathbf{x} \in I^{k}$.

$$\mathbf{x} = (x_{1},\dots,x_{k}),\quad a_{i} \le x_{i} \le b_{i} (i=1,\dots,k)$$

Suppose $f: I^{k} \to \mathbb{R}$ is continuous. Then, since it is integrable, let us set it as $f=f_{k}$, and define $f_{k-1} : I^{k-1} \to \mathbb{R}$ as follows:

$$f_{k-1} (x_{1}, \dots, x_{k-1}) = \int_{a_{k}}^{b_{k}} f_{k}(x_{1}, \dots, x_{k}) dx_{k}$$

Then, by the Leibniz Rule, $f_{k-1}$ is continuous in $I^{k-1}$. Repeating this process $k$ times yields the constant $f_{0}\in \mathbb{R}$. This is called the integral of $f$ over $I^{k}$, and is denoted as follows:

$$\begin{equation} \int_{I^{k}} f(\mathbf{x}) d\mathbf{x} \quad \text{or} \quad \int_{I^{k}}f \label{eq1} \end{equation}$$

Explanation

Such a definition of integration might seem to depend on the order of integration. However, in reality, the integral value of $f$ does not depend on the order of integration.

Theorem

Let $\eqref{eq1}$ be $L(f)$. Let another integral of some $f$ with a different order of integration be $L^{\prime}(f)$. Then, for all $f$ that are continuous in $I^{k}$, the following holds:

$$L(f) = L^{\prime}(f)$$

Proof

Let us set the function $h$ as follows:

$$h(\mathbf{x}) = h_{1}(x_{1}) h_{2}(x_{2}) \cdots h_{k}(x_{k}),\quad h_{j}\in \href{https://freshrimpsushi.github.io/posts/support-and-classes-of-continuous-functions/}{C([a_{j},b_{j}])}$$

Then, the following holds:

$$L(h) = \prod \limits_{i=1}^{k} \int_{a_{i}}^{b_{i}}h_{i}(x_{i})dx_{i} = L^{\prime}(h)$$

Now, let $\mathscr{A}$ be the set of all finite sums of such $h$s.

$$\mathscr{A} = \left\{ \text{all of finite sums of such functions h} \right\}$$

Since integration is linear, the following holds:

$$L(g) = L^{\prime}(g),\quad \forall g\in \mathscr{A}$$

Furthermore, $\mathscr{A}$ becomes an algebra over $I^{k}$.

Algebra

The set of functions $\mathscr{A}$ is called an algebra if it satisfies the following for all $f,g \in \mathscr{A}$ and the constant $c$:

• $f + g \in \mathscr{A}$
• $fg \in \mathscr{A}$
• $cf \in \mathscr{A}$

Now, by applying the Stone-Weierstrass Theorem, we obtain the following result:

• Let’s say $V= \prod \limits_{i=1}^{k}(b_{i} - a_{i})$. If $f \in C(I^{k})$ and $\varepsilon >0$ are given, then there exists $g\in \mathscr{A}$ that satisfies the following: $$\| f - g \| < \dfrac{\varepsilon}{V}$$ In this case, the norm is given as $\| f \| = \max | f |$.

Then, the following inequality holds:

$$\begin{align*} |L(f-g)| &\le L(\max|f-g|) = \| f-g \| V < \varepsilon \\ |L^{\prime}(f-g)| &\le L^{\prime}(\max|f-g|) = \| f-g \| V < \varepsilon \end{align*}$$

Also, the following equation holds:

$$L(f) - L^{\prime}(f) = L(f) - L(g) + L^{\prime}(g) - L(f) = L(f-g) + L^{\prime}(g-f)$$

Therefore, we obtain the following equation:

$$| L(f) - L^{\prime}(f) | = | L(f-g) + L^{\prime}(g-f) | < 2\varepsilon$$

Since this is true for all $\varepsilon$, we obtain the following:

$$L(f) = L^{\prime}(f)$$

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