Euclidean Gaussian transform with the 2π normalization

For $c>0$ and integer $m\ge 0$, F3 bounds $\mathrm{\mid }x{\mathrm{\mid }}^m{e}^{-c{x}^2\mathrm{/}2}$ on the tails, and it is bounded on a compact middle interval by continuity. Thus $\mathrm{\mid }x{\mathrm{\mid }}^m{e}^{-c{x}^2}\le C{e}^{-c{x}^2\mathrm{/}2}$. F1, F2 on both half-lines (reflect the negative half), and F7 give integrability of these majorants and $\int e^{-\pi x^2}dx=1$. In several dimensions bound a polynomial by a finite sum of monomials and factor the Gaussian; successive nonnegative integration gives the product of the finite one-dimensional bounds.

Put $G(\xi )=\int e^{-\pi x^2}{e}^{-2\pi ix\xi }dx$ in dimension one. F4 applies on every frequency interval with derivative majorant $2\pi \mathrm{\mid }x\mathrm{\mid }{e}^{-\pi x^2}$ from step 1.1. Hence $G^{\mathrm{\prime }}(\xi )=-2\pi i\int x{e}^{-\pi x^2}{e}^{-2\pi ix\xi }dx$. Apply F5 to $u=e^{-\pi x^2}$ and $v=e^{-2\pi ix\xi }$: both derivative products are integrable by step 1.1 and $uv\to 0$ at both ends. Since $u^{\mathrm{\prime }}=-2\pi xu$ and $v^{\mathrm{\prime }}=-2\pi i\xi v$, it follows that $\int xuv=-i\xi G(\xi )$, and therefore $G^{\mathrm{\prime }}=-2\pi \xi G$.

The product rule gives $(e^{\pi {\xi }^2}G(\xi ){)}^{\mathrm{\prime }}=0$. Applying the finite-interval complex FTC to its real and imaginary parts shows this product is constant, equal to $G(0)=1$. Thus $G(\xi )=e^{-\pi {\xi }^2}$. F6 tensors this formula in $n$ coordinates; absolute integrability is supplied by step 1.1. Finally substitute $y=\sqrt{t}x$ using F7; the Jacobian is $t^{n\mathrm{/}2}$ and the frequency becomes $\xi \mathrm{/}\sqrt{t}$. This gives exactly the claimed formula. Countable choice is inherited from F2 and F7, and the argument uses neither later Schwartz theory nor a Fourier inversion theorem.