Physlib.SpaceAndTime.Space.Integrals.RadialAngularMeasure

On the $d$-dimensional Euclidean space $\text{Space } d$, the radial angular measure is the measure defined by the density function $f(x) = \frac{1}{\|x\|^{d-1}}$ with respect to the standard Lebesgue volume measure, where $\|x\|$ denotes the Euclidean norm of $x$. In terms of the standard volume element $dV$, this measure $\mu$ is given by $d\mu = \frac{1}{\|x\|^{d-1}} dV$.

For any dimension $d \in \mathbb{N}$, the radial angular measure on the $d$-dimensional Euclidean space $\text{Space } d$ is the measure defined by the density function $f(x) = \frac{1}{\|x\|^{d-1}}$ with respect to the standard Lebesgue volume measure, where $\|x\|$ denotes the Euclidean norm of $x$. That is, for a measurable set $A \subseteq \text{Space } d$, the measure $\mu(A)$ is given by $$\mu(A) = \int_A \frac{1}{\|x\|^{d-1}} dV$$ where $dV$ is the standard volume element.

On the 0-dimensional space $\text{Space}(0)$, the radial angular measure is equal to the standard Lebesgue volume measure.

For any dimension $d \in \mathbb{N}$, the radial angular measure on the $d$-dimensional Euclidean space $\text{Space } d$ is $S$-finite. The radial angular measure is defined as the measure with density $f(x) = \frac{1}{\|x\|^{d-1}}$ with respect to the standard Lebesgue volume measure. A measure is $S$-finite if it can be represented as a countable sum of finite measures.

Let $\text{Space } d$ be the $d$-dimensional real inner product space equipped with the radial angular measure $\mu$. For any function $f: \text{Space } d \to F$, the integral of $f$ with respect to $\mu$ is equal to the integral with respect to the standard Lebesgue volume measure of $f$ weighted by the density $\|x\|^{-(d-1)}$: $$\int f(x) \, d\mu(x) = \int \frac{1}{\|x\|^{d-1}} f(x) \, dx,$$ where $\|x\|$ denotes the Euclidean norm on $\text{Space } d$.

Let $d$ be a natural number and let $\text{Space}(d)$ be a $d$-dimensional Euclidean space equipped with its standard Euclidean norm $\|x\|$. For any measurable function $f: \text{Space}(d) \to [0, \infty]$, its integral with respect to the radial angular measure is equal to the integral of $f$ with respect to the standard Lebesgue volume measure weighted by $\|x\|^{-(d-1)}$: $$\int f(x) \, d(\text{radialAngularMeasure}) = \int \frac{1}{\|x\|^{d-1}} f(x) \, dx$$ where the integral on the right is taken with respect to the standard volume measure.

Let $d$ be a natural number and let $f: \text{Space}(d+1) \to [0, \infty]$ be a measurable function. The integral of $f$ with respect to the radial angular measure on $\text{Space}(d+1)$ is equal to the integral of $f(r\omega)$ with respect to the product measure of the standard surface measure $\sigma$ on the unit sphere $S^d$ and the Lebesgue measure on the interval $(0, \infty)$: $$\int_{\text{Space}(d+1)} f(x) \, d(\text{radialAngularMeasure}) = \int_0^\infty \int_{S^d} f(r\omega) \, d\sigma(\omega) \, dr$$ where $r \in (0, \infty)$ is the radial distance and $\omega \in S^d$ is the angular component such that $x = r\omega$.

For any real number $r$, the radial angular measure of the closed ball in $\text{Space } 3$ centered at the origin with radius $r$ is equal to $4\pi r$: $$\mu(\overline{B}_r(0)) = 4\pi r$$ where $\mu$ denotes the radial angular measure and $\overline{B}_r(0)$ denotes the closed ball $\{x \in \text{Space } 3 \mid \|x\| \leq r\}$.

For any real number $r > 0$, the real-valued radial angular measure of the closed ball in $\text{Space } 3$ centered at the origin with radius $r$ is equal to $4\pi r$: $$\mu_{\text{real}}(\overline{B}_r(0)) = 4\pi r$$ where $\mu_{\text{real}}$ denotes the real-valued measure associated with the radial angular measure and $\overline{B}_r(0) = \{x \in \text{Space } 3 \mid \|x\| \leq r\}$ denotes the closed ball.

For any natural number $d$ and any function $f: \text{Space } d \to F$, $f$ is integrable with respect to the radial angular measure if and only if the function $x \mapsto \frac{1}{\|x\|^{d-1}} f(x)$ is integrable with respect to the standard Lebesgue volume measure, where $\|x\|$ denotes the Euclidean norm on $\text{Space } d$.

Let $d$ be a natural number and let $f: \text{Space}(d+1) \to F$ be a strongly measurable function. If the mapping $(r, \omega) \mapsto f(r\omega)$ is integrable with respect to the product measure of the standard surface measure $\sigma$ on the unit sphere $S^d$ and the Lebesgue measure on the interval $(0, \infty)$, then $f$ is integrable with respect to the radial angular measure on $\text{Space}(d+1)$.

For any natural number $d$, the function $x \mapsto (1 + \|x\|)^{-(d+1)}$ is integrable with respect to the radial angular measure on the $d$-dimensional Euclidean space $\text{Space } d$, where $\|x\|$ denotes the Euclidean norm.

For any natural number $d$, the radial angular measure on the $d$-dimensional Euclidean space $\text{Space } d$ has temperate growth. This means that there exists some power $N$ such that the function $x \mapsto (1 + \|x\|)^{-N}$ is integrable with respect to the radial angular measure, where $\|x\|$ denotes the Euclidean norm.