Integration of Real Scalar Fields

Parametric Integrals

Definition: Parametric Integrals

Let $f:D\to \mathbb{R}$ be a real scalar field on a closed rectangle $D=[a;b]\times [c;d]$.

Definition: Parametric Integral (Type I)

Given two real functions $l,u:[a;b]\to [c;d]$, we define a parametric integral $\mathcal{X}:[a;b]\to \mathbb{R}$ as

$$\mathcal{X}(x)\stackrel{\text{def}}{=}{\int }_{l(x)}^{u(x)}f(x,y)\mathrm{dy}$$

Note

For any particular value $x^{\ast }$ of $x$, the functions $l$ and $u$ result in concrete real numbers $l(x^{\ast })$ and $u(x^{\ast })$ and the expression $f(x^{\ast },y)$ is a concrete real function of $y$.

EXAMPLE

For $x=3$, we get

$$\mathcal{X}(3)={\int }_{l(3)}^{u(3)}f(3,y)\mathrm{dy},$$

which is just a regular definite integral.

Definition: Parametric Integral (Type II)

Given two real functions $l,u:[c;d]\to [a;b]$, we define a parametric integral $\mathcal{Y}:[c;d]\to \mathbb{R}$ as

$$\mathcal{Y}(y)\stackrel{\text{def}}{=}{\int }_{l(y)}^{u(y)}f(x,y)\mathrm{dx}$$

Note

For any particular value $y^{\ast }$ of $y$, the functions $l$ and $u$ result in concrete real numbers $l(y^{\ast })$ and $u(y^{\ast })$ and the expression $f(x,y^{\ast })$ is a concrete real function of $x$.

EXAMPLE

For $y=3$, we get

$\mathcal{Y}(3)={\int }_{l(3)}^{u(3)}f(x,3)\mathrm{dx},$

which is just a regular definite integral.

Theorem: Continuity of Parametric Integrals

Let $f:R\subset {\mathbb{R}}^2\to \mathbb{R}$ be a real scalar field on $R=[a;b]\times [c;d]$.

If $f$ is continuous on $R$, then:

• the parametric integral ${\int }_c^{d}f(x,y)\mathrm{dy}$ is continuous on $[a;b]$.

• the parametric integral ${\int }_a^{b}f(x,y)\mathrm{dx}$ is continuous on $[c;d]$.

PROOF

TODO

Theorem: Leibniz Integral Rule

Let $f:R\subset {\mathbb{R}}^2\to \mathbb{R}$ be a real scalar field on $R=[a;b]\times [c;d]$.

If $f$ is continously partially differentiable and $l,u$ are continuously differentiable, then the derivatives of $f$‘s parametric integrals are given by

$$\frac{\mathrm{d}}{\mathrm{d}x}{\int }_{l(x)}^{u(x)}f(x,y)\mathrm{dy}={\int }_{l(x)}^{u(x)}\frac{\partial }{\partial x}f(x,y)\mathrm{dy}+f(x,u(x))\frac{\mathrm{d}}{\mathrm{d}x}u(x)-f(x,l(x))\frac{\mathrm{d}}{\mathrm{d}x}l(x)$$ $$\frac{\mathrm{d}}{\mathrm{d}y}{\int }_{l(y)}^{u(y)}f(x,y)\mathrm{dx}={\int }_{l(y)}^{u(y)}\frac{\partial }{\partial y}f(x,y)\mathrm{dx}+f(u(y),y)\frac{\mathrm{d}}{\mathrm{d}y}u(y)-f(l(y),y)\frac{\mathrm{d}}{\mathrm{d}x}l(y)$$

PROOF

TODO

Lebesgue Integrals

Definition: Integration of Real Scalar Fields

Let $f:\mathcal{D}\subseteq {\mathbb{R}}^n\to \mathbb{R}$ be a measurable real scalar field on a Lebesgue-measurable subset $\mathcal{D}$ and let $S\subseteq {\mathbb{R}}^n$ also be Lebesgue-measurable.

The (Lebesgue) integral of $f$ over $S$ is $f$‘s Lebesgue integral over $S$ with respect to the Lebesgue measure $\mu$ on ${\mathbb{R}}^n$:

$${\int }_{S}f\mathrm{d\mu }$$

We say that $f$ is integrable on $S$ if ${\int }_{S}f\mathrm{d\mu }$ is finite.

NOTATION

$${\int }_{S}f(\mathbf{x}){\mathrm{d}}^n\mathbf{x}{\int }_{S}f\mathrm{dS}$$

When $n=2$ or $n=3$, we use the terms double integral or triple integral, respectively. In such cases, we also use the notations

$${\iint }_{S}f\mathrm{dS}{\iiint }_{S}f\mathrm{dS}$$

Theorem: Linearity of the Integral

Let $f$ and $g$ be real scalar fields on $\mathcal{D}\subseteq {\mathbb{R}}^n$.

If $f$ and $g$ are integrable on $S$, then so is the function $\alpha f+\beta g$ for all $\alpha ,\beta \in \mathbb{R}$. Furthermore,

$${\int }_S\alpha f(\mathbf{x})+\beta g(\mathbf{x}){\mathrm{d}}^{\mathrm{n}}\mathbf{x}=\alpha {\int }_{S}f(\mathbf{x}){\mathrm{d}}^{\mathrm{n}}\mathbf{x}+\beta {\int }_{S}g(\mathbf{x}){\mathrm{d}}^{\mathrm{n}}\mathbf{x}$$

PROOF

TODO

Theorem: Double Integrals via Iterated Integrals

Let $f:\mathcal{D}\subseteq {\mathbb{R}}^2\to \mathbb{R}$ be a real scalar field.

If $f$ is continuous and $\mathcal{D}$ can be expressed as $D=\left\{\left[\begin{array}{c}x\\ y\end{array}\right]\in {\mathbb{R}}^2\mid a\le x\le b,l(x)\le y\le u(x)\right\}$, where $l$ and $u$ are continuous real functions, then the double integral of $f$ over $\mathcal{D}$ can be calculated via iterated parametric integrals:

$$\underset{\mathcal{D}}{\iint }f\mathrm{dD}={\int }_a^b{\int }_{l(x)}^{u(x)}f(x,y)\mathrm{dy}\mathrm{dx}$$

If $f$ is continuous and $\mathcal{D}$ can be expressed as $\mathcal{D}=\left\{\left[\begin{array}{c}x\\ y\end{array}\right]\in {\mathbb{R}}^2\mid l(y)\le x\le u(y),a\le y\le b\right\}$, where $l$ and $u$ are continuous real functions, then the double integral of $f$ over $\mathcal{D}$ can be calculated via iterated parametric integrals:

$$\underset{\mathcal{D}}{\iint }f\mathrm{dD}={\int }_a^b{\int }_{l(y)}^{u(y)}f(x,y)\mathrm{dx}\mathrm{dy}$$

PROOF

TODO

Line Integrals

Definition: Line Integral of a Scalar Field

Let $f:\mathcal{D}\subseteq {\mathbb{R}}^n\to \mathbb{R}$ be a real scalar field and let $\gamma :[a;b]\to {\mathbb{R}}^n$ be a vector-valued function which is continuously differentiable on $(a;b)$ and such that $\gamma ([a;b])\subseteq \mathcal{D}$.

The line integral of $f$ along $\gamma$ is the integral

$${\int }_a^{b}f(\gamma (t))||\dot{\gamma }(t)||\mathrm{dt}$$

NOTATION

$${\int }_{\gamma }f{\int }_{\gamma }f\mathrm{ds}$$

If $\gamma$ is a closed parametric curve, we write

$${\oint }_{\gamma }f{\oint }_{\gamma }f\mathrm{ds}$$

Theorem: Linearity of the Scalar Line Integral

The scalar line integral is linear:

$${\int }_{\gamma }(\lambda f+\mu g)\mathrm{ds}=\lambda {\int }_{\gamma }f\mathrm{ds}+\mu {\int }_{\gamma }g\mathrm{ds}$$

PROOF

$$\begin{array}{rl}{\int }_{\gamma }(\lambda f+\mu g)\mathrm{ds}& ={\int }_a^b[\lambda f(\gamma (t))+\mu g(\gamma (t))]||\dot{\gamma }(t)||\mathrm{dt}\\ & ={\int }_a^b\lambda f(\gamma (t))||\dot{\gamma }(t)||+\mu g(\gamma (t))||\dot{\gamma }(t)||\mathrm{dt}\\ & =\lambda {\int }_a^{b}f(\gamma (t))||\dot{\gamma }(t)||\mathrm{dt}+\mu {\int }_a^{b}g(\gamma (t))||\dot{\gamma }(t)||\mathrm{dt}\\ & =\lambda {\int }_{\gamma }f\mathrm{ds}+\mu {\int }_{\gamma }g\mathrm{ds}\end{array}$$

Mean value theorem for Scalar Line Integrals

Let $f:\mathcal{D}\subseteq {\mathbb{R}}^n\to \mathbb{R}$ be a real scalar field and let $\gamma :[a;b]\to {\mathbb{R}}^n$ be a vector-valued function which is continuously differentiable on $(a;b)$ and such that $\gamma ([a;b])\subseteq \mathcal{D}$.

If $f$ is continuous, then there exists some $\mathbf{p}\in \gamma ([a;b])$ such that the scalar line integral of $f$ along $\gamma$ is

$${\int }_{\gamma }f=f(\mathbf{p})\cdot L,$$

where $L$ is the length of $\gamma$.

PROOF

TODO

Theorem: Line Integrals and Equivalence

Let $\gamma :[a;b]\to {\mathbb{R}}^n$ and $\phi :[c;d]\to {\mathbb{R}}^n$ be vector valued.

If $\gamma$ and $\phi$ are continuously differentiable on $(a;b)$ and $(c;d)$, respectively, and are equivalent up to a continuously differentiable reparametrization, then the scalar line integrals of $f$ along $\gamma$ and $\phi$ are equal.

$${\int }_{\gamma }f={\int }_{\phi }f$$

PROOF

TODO

Surface Integrals

Definition: Scalar Surface Integral

Let $f:{\mathcal{D}}_f\subseteq {\mathbb{R}}^n\to \mathbb{R}$ be a real scalar field and let $\varphi :{\mathcal{D}}_{\varphi }\subseteq {\mathbb{R}}^2\to {\mathbb{R}}^n$ be a differentiable parametric surface such that $S({\mathcal{D}}_{\varphi })\subseteq {\mathcal{D}}_f$.

The (scalar) surface integral of $f$ over $S$ is the double integral

$${\iint }_{{\mathcal{D}}_S}(f\circ S)\sqrt{\det ((D\varphi {)}^{\mathsf{T}}D\varphi )}\mathrm{d}{\mathcal{D}}_S,$$

where $\det ((D\varphi {)}^{\mathsf{T}}D\varphi )$ is the determinant of the product between $\varphi$‘s Jacobian $D\varphi$ and its transpose.

NOTATION

$${\iint }_{S}f\mathrm{dS}$$

If $S$ is a closed surface, then a circle can be put through the two integral signs.

Theorem: Surface Integrals in 3D

Let $f:{\mathcal{D}}_f\subseteq {\mathbb{R}}^n\to \mathbb{R}$ be a real scalar field and let $\varphi :{\mathcal{D}}_{\varphi }\subseteq {\mathbb{R}}^2\to {\mathbb{R}}^n$ be a differentiable parametric surface such that $S({\mathcal{D}}_{\varphi })\subseteq {\mathcal{D}}_f$.

If $n=3$, then the surface integral of $f$ over $\varphi$ is equal to the double integral

$${\iint }_{S}f\mathrm{dS}={\iint }_{{\mathcal{D}}_S}(f\circ S)||\mathbf{N}||\mathrm{d}{\mathcal{D}}_S,$$

where $\mathbf{N}$ is the canonical surface normal of $S$.

PROOF

We need to show that $||\mathbf{N}||$ is equal to $\det ((D\varphi {)}^{\mathsf{T}}D\varphi )$.

By definition,

$$\mathbf{N}=\frac{\partial S}{\partial x}\times \frac{\partial S}{\partial y}=\left[\begin{array}{}\end{array}\right]$$

TODO