Parametric Integrals

Definition: Parametric Integrals

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

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 some rectangle $R=[a;b]\times [c;d]$.

If $f$ is continuously 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