Having established existence of the derivative of a SRVF, stochastic integrals can be defined as a mean-square Riemann integration. Let $\widehat{\mathcal{U}}_{i}(x)$ be a RVF on a domain $\mathbb{D}\subset\mathbb{R}^{3}$ and for all $x^{i},y^{j}\in\mathbb{D}$, let $\psi(x,y)$ be a deterministic continuous and bounded function such that $\psi:\mathbb{R}^{3}\times\mathbb{R}^{3}\rightarrow\mathbb{R}$. The function can also be complex such that $\psi:\mathbb{R}^{3}\otimes\mathbb{R}^{3}\rightarrow\mathbb{C}$. The stochastic volume integral is the mean-square Riemann integral

$\widehat{\mathcal{R}}_{i}(y)={\displaystyle\int_{\mathbb{D}}}\psi(y,x)\otimes\widehat{\mathcal{U}}_{i}(x,\omega)d^{3}x$

The integral exists if the limit of the Riemann sum exists
$\widehat{\mathcal{R}}_{i}^{(m)}(x)=\sum_{\zeta=1}^{m}\psi(y,x_{\zeta})\otimes \widehat{\mathcal{U}}_{i_{\zeta}}(x_{\xi},\omega)\Delta_{vol}(x_{\zeta})$

where $\Delta(x_{\zeta})$ is a volume element centered on $x_{\xi}$ and $\sum_{\zeta}\Delta(x_{xi})=vol(\mathbb{D})$. The integral then exists if $\widehat{\mathcal{R}}_{i}(y)=\lim_{m\rightarrow \infty} \mathcal{I}^{(m)}_{i}(x)$
Since $\mathbf{E}(\widehat{\mathcal{U}}(x))=0$ then $\mathbf{E}\langle\widehat{\mathcal{Z}}(y)\rangle=0$. When $\psi(x,y)=1$ then

$\widehat{\mathcal{R}}_{i}(y)={\displaystyle \int_{\mathbb{D}}}\widehat{\mathcal{U}}_{i}(x,\omega)d^{3}x$

This definition leads to the following corollary (See Gilrod ref…).
A SRVF $\widehat{\mathcal{U}}_{i}(x)$ is mean-square Riemann integrable iff

$\mathbf{E}\langle\widehat{\mathcal{R}}_{i}(y)\otimes\widehat{\mathcal{R}}_{j}(y')\rangle$=

${\displaystyle \int_{\mathbb{D}}} {\displaystyle \int_{\mathbb{D}}} \psi(x, y) \psi(x', y')\mathbf{C}ov_{ij}(x,x')d^{3} x d^{3} x'$ < $lim_{m\rightarrow\infty} \sum_{\xi=1}^{m} \sum_{\eta=1}^{m} \psi(y, x_{\xi})\psi(y',x_{\eta}')$  $\mathbf{E} (\widehat{\mathcal{U}}_{i_{\xi}}(x_{\xi})\otimes \widehat{\mathcal{U}}_{j_{\eta}}(x'))\Delta_{vol}(x_{\xi}) \Delta_{vol}(x_{\eta}))$

For a Gaussian RVF this is equivalent to

$\mathbf{E}(\widehat{\mathcal{R}}_{i}(y)\otimes\widehat{\mathcal{R}}_{j}(y'))={\displaystyle \int_{\mathbb{D}}}{\displaystyle \int_{\mathbb{D}}}\psi(x,y)\psi(x',y')\mathbf{E}(\widehat{\mathcal{U}}(x)\otimes\widehat{\mathcal{U}}(x')) d^{3}xd^{3}x'<\infty$

The definition can be extended to include line integrals. Let $\widehat{\mathcal{U}}_{i}(x)$ be a GRVF on a domain $\mathbb{D}\subset\mathbb{R}^{3}$ and for all $x^{i}\in\mathfrak{S}\subset\mathbb{D}$ and $y^{j}\in\bar{\mathfrak{S}}\subset\mathbb{D}$. Let $\psi(x,y)$ be a deterministic continuous and bounded function such that $\psi:\mathbb{R}^{3}\times\mathbb{R}^{3}\rightarrow\mathbb{R}$.

$\widehat{\mathcal{R}}(y)={\displaystyle \int_{\mathfrak{S}}}\psi(y,x)\otimes\widehat{\mathcal{U}}_{i}(x)dx^{i}$

The integral exists if the limit of the Riemann sum exists
$\widehat{\mathcal{I}}^{(m)}(x)=\sum_{\xi=1}^{m}\psi(y,x_{\xi})\otimes\widehat{\mathbb{U}}_{i_{\xi}}(x_{\xi})\Delta x_{\xi}$,

where $\mathcal{R}x_{\xi}$ is a line element centered on $x_{\xi}$ and $\sum_{\xi}\Delta x_{xi}$ is the length of the line. The stochastic line integral then exists if

$\widehat{\mathcal{R}}(y)=\lim_{m\rightarrow \infty}\widehat{\mathcal{R}}^{(m)}(x)$. If $\psi(x,y)=1$ then

$\widehat{\mathcal{L}}(y)={displaystyle\int_{\mathfrak{S}}}\widehat{\mathcal{U}}_{i}(x)dx^{i}$

$\widehat{\mathcal{L}}(y)={\displaystyle \oint_{\mathfrak{S}}}\widehat{\mathcal{U}}_{i}(x)dx^{i}$

For a GRVF, the linking $\gamma\bigcap\gamma'$ of two curves or knots is described by the correlations

$\mathbf{E}\langle\widehat{\mathcal{L}}(x)\otimes\widehat{\mathcal{L}}(y)\rangle={\displaystyle{\int_{\mathfrak{S}}}{\displaystyle\int_{\bar{\mathfrak{S}}}}\mathbf{E}\langle\widehat{\mathcal{U}}_{i}(x) \otimes\widehat{\mathcal{U}}_{i}(y)\rangle dx^{i}dy^{j}$

$\mathbf{E}\langle\widehat{\mathcal{L}}(x)\otimes\widehat{\mathcal{L}}(y)\rangle =\oint_{\mathfrak{C}} \oint_{\bar{\mathfrak{\gamma}}}\mathbf{E}\langle \widehat{\mathcal{U}}_{i}(x)\otimes \widehat{\mathcal{U}}_{i}(y)\rangle dx^{i}dy^{j}$

The n-point correlation for a link comprised of n curves or knots

$\gamma_{1}\bigcap\gamma_{2}\bigcap...\gamma_{n-1}\bigcap\gamma_{n}$

is

$\langle\widehat{\mathcal{L}}(x_{1})\otimes...\otimes\widehat{\mathcal{L}}(x_{n})\rangle$ = $\int_{\mathfrak{S}_{1}}...\int_{\mathfrak{S}_{1}}d^{3} x_{1}...d^{3} x_{n} \mathbf{E} \langle\widehat{\mathcal{U}}_{i_{1}}(x_{1})\otimes...\otimes \widehat{\mathcal{U}}_{i_{1}}(x_{1})\rangle$

which can be expressed in a path integral form as

$\langle\widehat{\mathcal{L}}(x_{1})\otimes...\otimes\widehat{\mathcal{L}}(x_{n})\rangle= \int D_{n}x\mathbf{E}\langle\widehat{\mathcal{U}}_{i_{1}}(x_{1})\otimes...\otimes\widehat{\mathcal{U}}_{i_{1}}(x_{1})\rangle$

We can now examine the spectral properties of Gaussian random vector field.
Given the stochastic integral $\widehat{\mathcal{Z}}(y)=\int_{\mathbb{R}^{3}}d^{3}x f(x,y)\otimes\widehat{\mathcal{U}}_{i}(x)$ of (-), then setting $y=k$ with $k \in\mathbb{R}^{3}$ and $f(x,k)=\frac{1}{2\pi}^{3/2}\exp[-ik_{i}x^{i}]$ gives a stochastic Fourier integral.

$\widehat{\mathcal{U}}_{i}(k)=\frac{1}{2\pi^{3/2}}\int d^{3}k\widehat{\mathcal{U}}_{i}(x)\exp[ik_{i}x^{i}]$

If there exists such a random vector field $\widehat{\mathcal{U}}_{i}(k)$ then the random vector field
$\widehat{\mathcal{U}}_{i}(k)$ is harmonizable. The integral exists in the mss iff

$\mathbf{E}(\widehat{\mathcal{U}}_{i}(x)\otimes\widehat{\mathcal{U}}_{j}(y))$

$=\frac{1}{2\pi^{3}} \int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}d^{3}x d^{3} y$ $exp[-i (x_{i} k^{i})]$ $exp[-ik_{j}y^{j}] \mathcal{C}ov_{ij}(x,y)<\infty$

and $\mathcal{C} ov_{ij}(x, y) = \mathbf{E} (\mathcal{U}_{i}(x) \otimes \mathcal{U}_{j}(y) )$  for a Gaussian RVF. Equation (-) defines the cross spectral density function, provided the integral exists.

Given the transform pair

$\widetilde{\mathcal{U}}_{i}(x)=(2\pi)^{-3/2}\int_{\mathbb{R}^{3}}d^{3}x \widetilde{\mathcal{U}}_{i}({x}) exp[-i k x ]$

$\widetilde{\mathcal{U}}_{i}(k)= (2\pi)^{-3} \int_{\mathbb{R}^{3}} d^{3}k \widetilde{\mathcal{U}}_{i}({k}) exp[+i k x]$

there is a technical difficulty in that the integrals do not actually converge when integrated over all space. The problem is that the random function $\widetilde{\mathcal{B}}({x},t)$ extends over all space and does not decay to zero as $|{x}|\rightarrow \infty$, as is required for the existence of a classical Fourier transform. The resolution of this difficulty is essentially to define the transform of $\widetilde{\mathcal{U}}_{i}({x})$ over a finite volume or domain $\mathbb{D}$ of 3-space so that $\mathbb{D}\subset \mathbb{R}^{3}$ and consider the limit as the volume grows large and approaches infinity.

A finite-range transform $\widetilde{\mathcal{U}}_{i}^{[l]}({k})$ and its
complex conjugate $\widehat{\mathcal{U}}_{i}^{[l]*}({k})$ can be defined as

$\widehat{\mathcal{U}}_{i}^{[l]}({k}) = (2\pi)^{-3}\int _{\mathbb{D}_{l}}d^{3}x~\widetilde{\mathcal{U}}_{i}({x},t) exp[-i k x]$

$\widehat{\mathcal{U}}_{i}^{[l]*}({k})= (2\pi)^{-3}\int _{\mathfrak{D}_{l}} d^{3}y \widehat{\mathcal{U}}_{i}({y})exp[+i{k}.{y}]$

with $(x,y)\in\mathbb{D}$ and where $\mathbb{D}_{l}\subset \mathbb{R}^{3}$ is a cubic volume defined by $x_{1}\in [-l,l]$, $x_{2}\in [-l,l]$, $x_{3}\in [-l,l]$ and $2 l$ defines the sides of the cube. Taking the product of the transform and its conjugate then averaging gives the spectral moment

$\mathbf{E}(\widehat{\mathcal{U}}^{[l]}({k})\circ\widehat{\mathcal{U}}^{l*}({k})) =(2\pi)^{-6}\int_{\mathfrak{D}_{l}}d^{3}x\int_{\mathbb{D}_{l}}d^{3}y G_{ij}({x}-{y},t)\exp[-i{k}.({x}-{y})]$

We then consider the behavior of (36) in the limit as $\mathbb{D}_{l}\rightarrow\infty$ or $l\rightarrow \infty$. The correlation $G_{ij}(x,y)$ is a function of $\arrowvert{x}-{y}\arrowvert$ owing to its homogeneity, and there is a decorrelation for $\arrowvert{x}-{y}\arrowvert\rightarrow \infty$. The length scale over which $G_{ij}(x-y)$ has significant non-zero values is $\mid\!\! {x}-{y}\!\!\mid =O(\ell_{*})$ with $\ell_{*}$ being the correlation length. Let $r=|x-y|$ be a vector separation of two points. Replacing $y$ as an integration variable in (36) gives

$\mathbf{E}(\widetilde{\mathcal{U}}_{i}^{[l]}({k})\otimes\widetilde{\mathcal{U}}_{j}^{[l*]}({k}))=(2\pi)^{-6} \int_{\mathbb{D}_{l}} d^{3}x \int_{\mathbb{D}_{l}'} d^{3}r G_{ij}({r},t) exp[-i{k}.{r}]$

where $\mathbb{D}_{l}'$ depends on on $x$ and is defined by $r_{1}\in [x_{1}-l,x_{2}+l]$, $r_{2}\in [x_{2}-l,x_{2}+l]$, $r_{3}\in[x_{3}-l,x_{3}+l]$. The cube $\mathbb{D}_{l}$ has sides $2L$ and is centred on $r= x$. The integral is only significantly different from zero when $| r |=O(l)$ and this region lies well within $\mathbb{D}_{l}'$, provided that $l$ is many correlation lengths from the boundary of $\mathbb{D}_{l}$.Then as $l\rightarrow \infty$ for most $x\in \mathbb{D}_{l}$, the integration over $r$ can be carried to infinity.

$\mathbf{E}(\widetilde{\mathcal{U}}_{i}({k})\times\widetilde{\mathcal{U}}_{j}({k}))) \sim (2\pi)^{-3} \int_{\mathbb{D}_{l}}d^{3}x \psi_{ij}({k},t)=(l/\pi)^{3}\psi_{ij}(k)$

The approximation of the integral over $r$ by $\psi_{ij}$ applies everywhere except when $x$ is of the order of a correlation length from the boundary of $\mathbb{D}_{l}$.

We are most interested in the case when (38) has different values of $k$. The spectral correlation is where the integration variable has been changed from $y$ to $r=x-y$. Again. provided that $|x|$ is many correlation lengths from the boundary of $\mathbb{D}_{l}$ the integral over $r$ can be
extended to infinity so that

$\mathbf{E}((\widetilde{\mathcal{U}}_{i}^{[l]}({k},t)\circ\widetilde{\mathcal{U}}_{j}^{[l*]}(k)]] )\sim\psi_{ij}(k',t)\Xi_{\ell}(k-k')$

where
$\Xi_{l}(k-k')= \frac{sin[(k_{1}-k_{1}')l]sin[(k_{2}-k_{2}')l]sin[(k_{3}-k_{3}')l]}{\pi^{3} (k_{1}-k_{1}')(k_{2}-k_{2}')(k_{3}-k_{3}')}$

The function $\Xi_{\ell}({k}-{k}')$ has a maximum peak at $k=k'$ and becomes increasingly sharply peaked about $k=k'$ as $l\rightarrow\infty$ with the integral $\int d^{3}k\Xi_{l}(k)=1$ always satisfied. It follows that $\Xi_{l}(k-k')$ has all the properties of a delta function so that $\Xi_{l}(k-k')\rightarrow \delta^{3}(k-k')$ as $l\rightarrow\infty$. This then leads
to the important result

$\mathbf{E}(\widetilde{\mathcal{U}}_{i}({k})\circ \widetilde{\mathcal{U}}_{j}({k}')) =\psi_{ij}({k},t)\delta^{3}(k-k')$

We have replaced $\psi_{ij}(\vec{k}')$ with $\psi_{ij}({k},t)$ since the delta function is effectively zero when $x\ne {k}'$. Since $\widetilde{\mathcal{S}}_{i}(x,t)$ is real the complex conjugate of (-) gives $\widetilde{\mathcal{U}}_{i}^{*}(k')=\widetilde{\mathcal{U}}_{j}(-k')$ so that the spectral correlation (in the infinite $l$ limit) becomes

$\mathbf{E}({\mathcal{U}}_{i}(k)\circ\widetilde{\mathcal{U}}_{j}(k)) =\psi_{ij}(k,t)\delta^{3}(k+k')$

Spectral analysis of turbulence for example, involves Fourier analysis of a stochastic Navier-Stokes flow on $\mathbb{R}^{3}$ so that $\widetilde{\mathcal{U}}_{i}(x)=U_{i}(x)+\widetilde{\mathcal{V}}_{i}(x)$ The essential idea is to decompose the turbulent fluctuations about the mean flow into sinusoidal components and study the distribution of turbulent energy amongst the different wave vectors representing the different scales of turbulence.

The spectral function $\psi_{ij}(k-k',t)$ is given by the Fourier transform
$\mathcal{F}_{T}:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}$ of the stress tensor $S_{ij}(x,y;t)$.

$\psi_{ij}(k,t)=(2\pi)^{-3}\int_{\mathbb{R}^{3}}d^{3}x S_{ij}(x,y,t)\exp[-k.y-x]=(2\pi)^{-3}$ *

$\int d^{3}x\mathbf{E}(\arrowvert\widetilde{\mathcal{U}}_{i}(x,t)\circ\widetilde{\mathcal{U}}_{j}(y,t)\arrowvert exp[-ik.r]$

and depends on the wavenumber vector $k$ and $t$ as the turbulence evolves. The inverse transform is $R_{ij}(\vec{r},t)=\int d^{3}k \psi_{ij}({k},t)exp(i {k}.{r})$, which is an integral over all 3-dimensional wavenumber space. Setting $\vec{r}=0$ and i=j in (-) gives the stress tensor of (-)

$\frac{1}{2}\mathbf{E}(\widetilde{\mathcal{U}}_{i}(x)\circ\widetilde{\mathcal{U}}_{j}({x}))=\frac{1}{2}S_{ij}(0,t)=\frac{1}{2}\int_{\mathbb{R}^{3}} d^{3}k \psi_{ij}(k,t)$

The ‘turbulent kinetic energy’ can therefore be expressed as an integral over $k$. In turbulence theory for example, where $\widehat{\mathcal{U}}_{i}(x)$ is a random Navier-Stokes flow, the function $\psi_{ij}({k},t)/2$ can be interpreted as the distribution of turbulent energy over the various wavenumbers. The stresses and spectral functions have the symmetries $S_{ij}(r)=S_{ij}(-r)$ and $\psi_{ij}(k,t)=\psi_{ji}(-k,t)$. Also $\psi_{ij}^{*}(k,t)=\psi_{ji}(-k,t)$ leading to $\psi_{ij}^{*}(k,t)=\psi_{ji}(k,t)$, which implies that $\psi_{ij}$ is a Hermitian matrix. Each of the diagonal terms is therefore real as should be the case for a quantity interpreted as the distribution of energy with wavenumber. The magnitude of the wavenumber vector $k=\mid {k}\mid$ has dimensions of inverse length. One can interpret the length scale $\ell=(k^{-1})$ as the spatial scale represented by wavenumber $k$. For example if $\ell$ is an integral scale of turbulence the wavevectors $\ell^{-1}$ represent the scales of turbulence that contain most of the kinetic energy. The spectral function then has its maximum values for $\mid {k}\mid= O(l^{-1})$ and decays to zero as $|k| \rightarrow \infty$. The stress tensor can then be represented as the inverse Fourier transform

$S_{ij}(x,y,t)=\mathbf{E}(\widetilde{\mathcal{U}}_{i}(x)\times\widetilde{\mathcal{U}}_{j}(y))$ $=\int_{\mathbb{R}^{3}} d^{3}k\int_{\mathbb{R}^{3}} d^{3}k' \psi_{ij}(k)\delta^{3}(k-\beta')\exp[i k'.(x-y)]$

where $\beta= - k$ and the integration range is taken over all $\mathbb{R}$ space since we take the spectral function to be such that the Fourier integrals now converge. Equation (44) becomes

$S_{ij}(x,y,t)=\mathbf{E}(\widetilde{\mathcal{U}}_{i}(x,t) \times\widetilde{\mathcal{U}}_{j}(y,t)) = (2\pi)^{-3}\int_{\mathbb{R}} d^{3}k \psi_{ij}({k},t)e^{i{\beta}.(x-y)}$

Now $exp[i{\beta}.(x-y)]=exp[-ik.(x-y)]=\exp[ik.(y-x)]$ so that

$S_{ij}(x,y,t)=\mathbf{E})\lbrace\widetilde{\mathcal{U}}_{i}(x,t) \otimes\widetilde{\mathcal{U}}_{j}({y},t) \rbrace=(2\pi)^{-3}\int_{\mathbb{R}^{3}} d^{3}k~\psi_{ij}(k,t)exp[i {k}.(y-x)]$

Finally, for homogeneity and isotropy it is required that

$\frac{d}{d\xi_{j}}\mathbf{E}(\widetilde{\mathcal{U}}_{i}(x,t)\otimes\widetilde{\mathcal{U}}_{j}({x}+{\xi},t)) = (2\pi)^{-3}\int_{\mathbb{R}^{3}} d^{3}k ~ik_{j}\psi_{ij}(k,t)exp[ik_{j}\xi_{j}]=0$

so that the spectral function must satisfy $k_{j}\psi{ij}({k})=0$. The most general form is then

$\psi_{ij}(k)=\left(\delta_{ij}- \frac{k_{i}k_{j}}{k^{2}}\right)f(k)$

The 2-point correlations then depend on the ansatzes for the spectral function.
Let $\psi_{ij}(k)=(\delta_{ij}-\frac{k_{i}k_{j}}{k^{2}}$ be the spectral function corresponding to the 2-point function

$\mathbf{E}(\widetilde{\mathcal{U}}_{i}(x,t)\times\widetilde{\mathcal{U}}_{j}(y,t))= \frac{1}{(2\pi)^{3/2}}\int_{\mathbb{R}^{3}}d^{3}k \psi_{ij}(k)\exp[ik^{i}(x-y)_{i}]$

The ansatz for the spectral function and relationship between the wave vectors $k_{i}$ then determines the form of the correlation for the GRVFS $\widehat{\mathcal{U}}_{i}(x)$ such that

[1] If $k_{i}k_{j}=0$ then $\psi_{ij}(k)=\delta_{ij}$ and

$\mathbf{E}(\widetilde{\mathcal{U}}_{i}(x)\times\widetilde{\mathcal{U}}_{j}(y))=\frac{1}{(2\pi)^{3/2}}\int_{\mathbb{R}^{3}}d^{3}k \delta_{ij}(k)\exp[ik^{i}(x-y)_{i}]=\delta_{ij}\delta^{(3)}(x-y)$

This correlation then describes a white-in-space noise.

[2]  If $k_{i}k_{j}=\epsilon_{ijk}k^{k}$ and $\psi_{ij}(k)=\frac{\epsilon_{ijk}k^{k}}{k^{2}}$ then

$\mathbf{E}(\widetilde{\mathcal{U}}_{i}({x})\circ\widetilde{\mathcal{U}}_{j}(y))= \frac{1}{(2\pi)^{3/2}}\int_{\mathbb{R}^{3}}d^{3}k \epsilon_{ijk}\frac{k^{k}}{k^{2}} exp[ik^{i}(x-y)_{i}]=\frac{1}{4\pi}\epsilon_{ijk}|x-y|^{k}\|x-y\|^{-3}$

which is of the same mathematical form as the propagator in a Chern-Simons theory(ref). The following lemma on the possible 2-point correlations is also stated.

Let $\widehat{\mathcal{F}}_{i}(x)$ be a Gaussian random field defined for all $x\in\mathbb{D}$ with $\mathbf{E}[\widehat{\mathcal{F}}_{i}(x)]=0$. Let $x^{i}\in\mathfrak{S}\subset\mathbb{D}$ and $y^{j}\in\bar{\mathfrak{S}}\subset\mathbb{D}$. If the 2-point function is a ‘white-in-space noise’ of the form

$\mathbf{E}[\widehat{\mathcal{F}}_{i}(x)\times\widehat{\mathcal{F}}_{i}(x)]= \delta_{ij}\delta^{3}(x-y)$

then
[1] There exists a random Gaussian field $\widehat{\mathcal{B}}_{i}(x)$ for all $x\in\mathbb{D}$ with the 2-point correlation
$\mathbf{E}[\widehat{\mathcal{B}}_{i}(x)\times\widehat{\mathcal{B}}_{i}(y)]= \delta_{ij}|x-y|^{-1}$
[2] There exists a random Gaussian field $\widehat{\mathcal{G}}_{i}(x)$ for all $x\in\mathbb{D}$ with the 2-point correlation

$\mathbf{E}[\widehat{\mathcal{G}}_{i}(x)\times\widehat{\mathcal{G}}_{i}(y)]= \frac{1}{4\pi}\epsilon_{ijk}(x-y)^{k}|x-y|^{-3}$

Let $\mathbf{E}[\widehat{\mathcal{R}}_{i}(x)\otimes\widehat{\mathcal{R}}_{j}(y)]=K_{ij}(x-y)$ then

$K_{ij}(x-y)=-\frac{\delta_{ij}}{4\pi}\nabla_{(x)}^{2}(|x-y|^{-1})= \delta_{ij}\delta^{(3)}(x-y)$

$K_{ij}(x-y) = -\frac{\delta_{ij}}{4\pi}\nabla_{(x)}^{2} G(x-y)=\delta_{ij}\delta^{(3)}(x-y)$

which follows from the Poisson equation

$-\frac{1}{4\pi} \nabla_{(x)}^{2}|x-y|^{-1}=\delta^{(3)}(x-y)$

Hence

$\mathbf{E}[\widehat{\mathcal{B}}_{i}(x)\otimes\widehat{\mathcal{B}}_{j}(x)]\equiv G_{ij}(x-y)=\delta_{ij}|x-y|^{-1}$

or $G(x-y)=|x-y|^{-1}$. Next, given $G(x-y)$ there is a Greens function $C_{ij}(x-y)$ such that

$C_{ij}(x-y)=\frac{1}{4\pi}\epsilon_{ijk}\nabla^{k}G(x-y)= \frac{1}{4\pi}\epsilon_{ijk}\nabla^{k}|x-y|^{-1}$

so that

$C(x-y)=\mathbf{E}[\widehat{\mathcal{G}}_{i}(x)\otimes\widehat{\mathcal{G}}_{j}(x)] =\frac{1}{4\pi}\epsilon_{ijk}(x-y)^{k}|x-y|^{-3}$

Let $(x,y)\in\mathbb{D}\subset\mathbb{R}^{3}$ and $\widehat{\mathcal{U}}_{i}(x)$ and $\widehat{\mathcal{U}}_{i}(x)$ are GRVFs on $\mathbb{D}$. As in (-), the spectral function choice $\psi_{ij}(k)=\delta_{ij}$ gives a 2-point correlation for white noise.

$\mathbf{E}(\widetilde{\mathcal{U}}_{i}({x})\times\widetilde{\mathcal{U}}_{j}({y})) =\frac{1}{(2\pi)^{3/2}}\int_{\mathbb{R}^{3}}d^{3}k \delta_{ij}(k)\exp[ik^{i}(x-y)_{i}]=\delta_{ij}\delta^{(3)}(x-y)$

Let $\mathbb{B}\subset\mathbb{D}$ be a ball of radius $\xi$ and $vol(\mathbb{B})=\tfrac{4}{3}\pi\xi^{3}$. If $(x,y)\in\mathbb{B}$ the delta-function is ‘smeared-out’ over $\mathbb{B}$ into a very narrow and highly peaked rectangular or top-hat function of width $\xi$. Then

$\delta_{ij}\delta(x-y)\rightarrow \xi^{-2}\Pi[\xi^{-1}(|x-y|-\tfrac{1}{2}\xi)]$

where

$\xi^{-2}\Pi(\xi^{-1}|x-y|-\tfrac{1}{2}\xi)=1$~for $|x-y|\le \xi$

$\xi^{-2}\Pi(\xi^{-1}|x-y|-\tfrac{1}{2}\xi)0$ for $x>\xi$

so that the correlation vanishes outside $\mathbb{B}$. This is equivalent to choosing

$\psi_{ij}(k)=\tfrac{1}{2\pi}^{1/2}\delta_{ij}sinc[k/2] = \tfrac{1}{2\pi}^{1/2}k^{-1}\sin[k/2]$

for the spectral function so that the Fourier transform is

$\mathscr{F}[sinc[k/2]]=\mathbf{E}(\widetilde{\mathcal{U}}_{i}({x})\times\widetilde{\mathcal{U}}_{j}({y}))$

$=\frac{1}{(2\pi)^{3/2}}\int_{\mathbb{R}^{3}}d^{3}k\delta_{ij}~sinc[k/2] \exp[ik^{i}(x-y)_{i}]=\delta_{ij}\Pi[x-y]$

and then making the rescaling $|x-y|\rightarrow [|x-y|-\tfrac{1}{2}\xi]\xi^{-1}$.

$\mathbf{E}(\widetilde{\mathcal{U}}_{i}({x})\times\widetilde{\mathcal{U}}_{j}({y}))= \frac{\alpha}{\xi^{2}}\delta_{ij}\Pi[\xi^{-1}(|x-y|-\tfrac{1}{2}\xi)]$