The Matérn spectral integral, normalization, and radial formula

Since $2s>d$, polar coordinates [F4] and comparison with $r^{d-1-2s}$ for $r\ge 1$ show ${\int }_{{\mathbb{R}}^d}(a^2+4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^{-s}d\xi <\mathrm{\infty }$; the integrand is bounded near $0$. Since $A>0$ by [F1] and the positive parameters, $w>0$ and $w\in L^1$.

For one dimension let $I_t(z)={\int }_{\mathbb{R}}{e}^{-4{\pi }^{2}t{\xi }^2}{e}^{2\pi iz\xi }d\xi$. Differentiation under the integral is justified by [F6], since $2\pi \mathrm{\mid }\xi \mathrm{\mid }{e}^{-4{\pi }^{2}t{\xi }^2}$ is integrable. Apply [F8] with $u(\xi )=e^{-4{\pi }^{2}t{\xi }^2}$ and $v(\xi )=e^{2\pi iz\xi }$: both are $C^1$, $u^{\mathrm{\prime }}v$ and $u{v}^{\mathrm{\prime }}$ are integrable Gaussian multiples, and $uv\to 0$ at both ends. Thus $-8{\pi }^{2}t\int \xi uv=-2\pi iz{I}_t(z)$, which gives $I_t^{\mathrm{\prime }}(z)=-z{I}_t(z)\mathrm{/}(2t)$. The initial value from [F2] and scaling is $I_t(0)=(4\pi t{)}^{-1\mathrm{/}2}$. Set $J(z)=I_t(z)e^{z^2\mathrm{/}(4t)}$; the differential equation gives $J^{\mathrm{\prime }}(z)=0$. The real and imaginary parts of $J$ are constant by the mean value theorem [F5], so $J(z)=J(0)=I_t(0)$ and $I_t(z)=(4\pi t{)}^{-1\mathrm{/}2}{e}^{-z^2\mathrm{/}(4t)}$.

For $B>0$, [F1] and the substitution $u=Bt$ give $B^{-s}=\mathrm{\Gamma }(s{)}^{-1}{\int }_0^{\mathrm{\infty }}{t}^{s-1}{e}^{-Bt}dt$. Apply this with $B=a^2+4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2$ and use Tonelli [F3]. The product Gaussian integral in [F2] is $(4\pi t{)}^{-d\mathrm{/}2}$ by scaling and the Borel product identity in [F3]. A second substitution $u=a^{2}t$ gives $\int w=\frac{A}{\mathrm{\Gamma }(s)}(4\pi {)}^{-d\mathrm{/}2}{\int }_0^{\mathrm{\infty }}{t}^{\nu -1}{e}^{-a^{2}t}dt=\frac{A\mathrm{\Gamma }(\nu )}{\mathrm{\Gamma }(s)(4\pi {)}^{d\mathrm{/}2}{a}^{2\nu }}=1.$

The integrand in $d$ dimensions is a product of the one-dimensional integrands in step 1.2. Absolute integrability and [F3] therefore give the stated $d$-dimensional Gaussian Fourier identity.

Absolute Fubini is valid in the gamma representation of $w$ because its absolute double integral is the finite normalization integral in steps 1.1 and 2.1. Substitute that representation into $k(z)$ and apply step 2.2; the result is $k(z)=\frac{A}{\mathrm{\Gamma }(s)}(4\pi {)}^{-d\mathrm{/}2}{\int }_0^{\mathrm{\infty }}{t}^{s-d\mathrm{/}2-1}{e}^{-a^{2}t-\mathrm{\mid }z{\mathrm{\mid }}^2\mathrm{/}(4t)}dt=\frac{a^{2\nu }}{\mathrm{\Gamma }(\nu )}{\int }_0^{\mathrm{\infty }}{t}^{\nu -1}{e}^{-a^{2}t-\mathrm{\mid }z{\mathrm{\mid }}^2\mathrm{/}(4t)}dt\mathrm{.}$ The same absolute bound proves convergence for every $z$.

From $w\in L^1$, dominated convergence [F6] makes $k$ continuous. The integral formula in step 3.1 is real and depends on $z$ only through $\mathrm{\mid }z\mathrm{\mid }$, so $k$ is real, even, and radial. Step 2.1 gives $k(0)=1$.

In step 3.1 substitute $u=a^{2}t$. For $b=a\mathrm{\mid }z\mathrm{\mid }>0$, then substitute $v=b^2\mathrm{/}(4u)$ in the resulting integral; [F7] gives ${\int }_0^{\mathrm{\infty }}{u}^{\nu -1}{e}^{-u-b^2\mathrm{/}(4u)}du=2^{1-\nu }{b}^{\nu }{K}_{\nu }(b)$. Thus the displayed Bessel expression follows for $z\mathrm{\ne }0$.