The Gaussian Fourier transform from improper integrals

For $a>0$ and integer $m\ge 0$, choose a positive integer $M\ge m$. When $\mathrm{\mid }x\mathrm{\mid }\ge 1$, $\mathrm{\mid }x{\mathrm{\mid }}^m\le \mathrm{\mid }x{\mathrm{\mid }}^{2M}$; applying [F2] to $u=x^2$ with exponent $M$ and parameter $a\mathrm{/}2$ gives $\mathrm{\mid }x{\mathrm{\mid }}^{2M}{e}^{-a{x}^2\mathrm{/}2}\to 0$ as $\mathrm{\mid }x\mathrm{\mid }\to \mathrm{\infty }$. Hence $\mathrm{\mid }x{\mathrm{\mid }}^m{e}^{-a{x}^2\mathrm{/}2}\to 0$ and, by continuity on bounded intervals, it is bounded by some $C$. Thus $\mathrm{\mid }x{\mathrm{\mid }}^m{e}^{-a{x}^2}\le C{e}^{-a{x}^2\mathrm{/}2}$. For $\mathrm{\mid }x\mathrm{\mid }\ge 1$, $e^{-a{x}^2\mathrm{/}2}\le e^{-a\mathrm{\mid }x\mathrm{\mid }\mathrm{/}2}$, and [F11] together with [F4] gives ${\int }_R^S{e}^{-ax\mathrm{/}2}dx=(2\mathrm{/}a)(e^{-aR\mathrm{/}2}-e^{-aS\mathrm{/}2})$ for $1\le R<S$, with a tail tending to zero as $R\to \mathrm{\infty }$; the negative tail is identical by reflection. Thus each $\mathrm{\mid }x{\mathrm{\mid }}^m{e}^{-a{x}^2}$ is absolutely improperly integrable and its tails tend to zero.

For fixed $R>0$ in dimension one, put $G_R(\xi ):={\int }_{-R}^R{e}^{-\pi x^2}{e}^{-2\pi ix\xi }dx$ and $H_R(\xi ):={\int }_{-R}^R(-2\pi ix)e^{-\pi x^2}{e}^{-2\pi ix\xi }dx$. On each compact frequency interval, continuity of the integrands in $(x,\xi )$ and uniform continuity on the compact product, followed by [F10] on their real and imaginary parts, show $G_R$ and $H_R$ are continuous. For $g(x,\xi ):=e^{-\pi x^2}{e}^{-2\pi ix\xi }$, [F6] and the complex chain rule give ${\mathrm{\partial }}_{\xi }g=-2\pi ixg$; this derivative is continuous. On any compact frequency interval containing $\xi$ in its interior, [F7] makes ${\mathrm{\partial }}_{\xi }g$ uniformly continuous on the compact product with $[-R,R]$. For sufficiently small real $h$, the finite-interval fundamental theorem [F4], applied to real and imaginary parts as functions of frequency, gives $[g(x,\xi +h)-g(x,\xi )]\mathrm{/}h={\int }_0^1{\mathrm{\partial }}_{\xi }g(x,\xi +sh)ds$. Uniform continuity makes these difference quotients converge uniformly in $x$ to ${\mathrm{\partial }}_{\xi }g(x,\xi )$. Their real and imaginary integrals therefore converge by [F10], so $G_R^{\mathrm{\prime }}(\xi )={\int }_{-R}^R{\mathrm{\partial }}_{\xi }g(x,\xi )dx=H_R(\xi )$.

In dimension $n$, the Gaussian and each coordinate-weighted absolute Gaussian factor into one-dimensional factors on every box $[-R,R{]}^n$; product rectangular Riemann sums give the corresponding product of one-dimensional integrals. Step 1.1 makes each factor finite and its tail tend to zero. For $S>R$, the part of $[-S,S{]}^n$ outside $[-R,R{]}^n$ lies in the union of the $n$ coordinate-tail slabs; the integral on slab $j$ factors as one one-dimensional tail integral times $n-1$ bounded one-dimensional integrals. Therefore the integral on the complement tends to zero as $R\to \mathrm{\infty }$, uniformly in $S>R$. Each integrand is continuous, so [F19] gives local Riemann integrability on every compact Jordan set. For any such set $K$, choose $S$ with $K\subset [-S,S{]}^n$; extend the nonnegative integrand by zero from $K$ to a containing rectangle as in [F12], and use monotonicity [F15] to bound its integral by that over $[-S,S{]}^n$. The supremum in [F13] is therefore finite, and [F14] identifies the improper integral with the cube exhaustion limit. This proves absolute improper integrability for $e^{-\pi t\mathrm{\mid }x{\mathrm{\mid }}^2}$ and $\mathrm{\mid }{x}_j\mathrm{\mid }{e}^{-\pi \mathrm{\mid }x{\mathrm{\mid }}^2}$.

Step 1.1 bounds the tails of $G_R$ and $H_R$ uniformly in $\xi$, because $\mathrm{\mid }{e}^{-2\pi ix\xi }\mathrm{\mid }=1$ by [F5]; hence $G_R\to G$ and $H_R\to H$ uniformly, where $G(\xi ):={\int }_{\mathbb{R}}{e}^{-\pi x^2}{e}^{-2\pi ix\xi }dx$ and $H(\xi ):={\int }_{\mathbb{R}}(-2\pi ix)e^{-\pi x^2}{e}^{-2\pi ix\xi }dx$. Since each $G_R,H_R$ is continuous by step 1.2, [F16] makes both limits continuous.

By step 1.2 and [F4], $G_R(\xi )-G_R(0)={\int }_0^{\xi }{H}_R(s)ds$ for every finite $R$, applying the real theorem to each component. The uniform convergence from step 3.1 and [F10] on the finite interval from $0$ to $\xi$ let $R\to \mathrm{\infty }$ in this identity, giving $G(\xi )-G(0)={\int }_0^{\xi }H(s)ds$. On any compact interval containing $0$ and the evaluation frequency in its interior, both real components of $H$ are continuous by step 3.1 and hence Riemann integrable by [F8]. For each component, its integral from $0$ differs from the integral function based at the interval's left endpoint by a constant. Thus [F17] applies at every evaluation frequency and yields $G^{\mathrm{\prime }}=H$.

On $[-R,R]$, integration by parts [F3] applied to the real and imaginary parts gives ${\int }_{-R}^R(-2\pi x)e^{-\pi x^2}{e}^{-2\pi ix\xi }dx=[e^{-\pi x^2}{e}^{-2\pi ix\xi }{]}_{-R}^R+2\pi i\xi G_R(\xi )\mathrm{.}$ The boundary term tends to zero by step 1.1 and the phase modulus in [F5]; the finite-interval integrands are continuous and hence integrable by [F8], and the improper integrals converge absolutely by steps 1.1-2.1. Since $G_R\to G$ and the polynomial-Gaussian tails converge by step 1.1, taking $R\to \mathrm{\infty }$ gives $-2\pi {\int }_{\mathbb{R}}x{e}^{-\pi x^2}{e}^{-2\pi ix\xi }dx=2\pi i\xi G(\xi )$. By step 4.1, $G^{\mathrm{\prime }}=H$, and the definition of $H$ therefore gives $G^{\mathrm{\prime }}=-2\pi i\int x{e}^{-\pi x^2}{e}^{-2\pi ix\xi }dx=-2\pi \xi G$.

The complex exponential chain rule [F6] gives $(e^{\pi {\xi }^2}{)}^{\mathrm{\prime }}=2\pi \xi e^{\pi {\xi }^2}$. Apply the real product rule [F18] separately to the real and imaginary parts of $e^{\pi {\xi }^2}G(\xi )$; step 5.1 gives derivative zero, so both parts are constant by [F4]. By [F1] and the finite-interval substitutions $u=\sqrt{\pi }x$ followed by the improper limit, $G(0)={\int }_{\mathbb{R}}{e}^{-\pi x^2}dx=1$ using [F9] and the exhaustion convention [F14]. Thus $G(\xi )=e^{-\pi {\xi }^2}$.

For $t>0$, the one-dimensional substitution $u=\sqrt{t}x$ using [F9] gives ${\int }_{\mathbb{R}}{e}^{-\pi t{x}^2}{e}^{-2\pi ix\xi }dx=t^{-1\mathrm{/}2}G(\xi \mathrm{/}\sqrt{t})=t^{-1\mathrm{/}2}{e}^{-\pi {\xi }^2\mathrm{/}t}$. In $n$ dimensions the integrand is the product of these one-dimensional factors. On each box $[-R,R{]}^n$ its integral factors by finite rectangular Riemann sums, whose continuous summands are integrable by [F8]. Step 2.1 bounds the absolute tail outside the box by a sum of products of one-dimensional Gaussian tails; [F14]-[F15] therefore identify the improper integral with the product of the one-dimensional limits. Multiplying the factors gives $t^{-n\mathrm{/}2}{e}^{-\pi \mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{/}t}$.