= Solution
First $\int_{\mathbb R}e^{-x^2/2}\,dx=\sqrt{2\pi}$: square the positive <integral> and use polar coordinates to obtain $2\pi\int_0^\infty re^{-r^2/2}\,dr=2\pi$. Scaling therefore normalizes the one-dimensional <Gaussian function> $g_t$.
Let $I(\xi)=\int_{\mathbb R}(2\pi t)^{-1/2}e^{-x^2/(2t)}e^{-i\xi x}\,dx$. Differentiation under the <integral> is justified by the integrable majorant $|x|g_t(x)$. Since $g_t'(x)=-xg_t(x)/t$, <integration by parts>, with zero endpoint terms, gives $I'(\xi)=-t\xi I(\xi)$. With $I(0)=1$, this yields $I(\xi)=e^{-t\xi^2/2}$. Multiplication of the one-dimensional <integrals> gives
$$
\boxed{\widehat g_t(\xi)=e^{-t|\xi|^2/2}.}
$$
Applying the same computation with reciprocal scale $1/t$ evaluates its inverse <integral>:
$$
(2\pi)^{-d}\int e^{i\xi\cdot x}e^{-t|\xi|^2/2}\,d\xi
=(2\pi t)^{-d/2}e^{-|x|^2/(2t)}=g_t(x).
$$
For $f\in L^1$, <Fubini's theorem> gives $\widehat{f*g_t}=\widehat f\,e^{-t|\xi|^2/2}$, which is integrable because $\widehat f$ is bounded. Its inverse <integral> is
$$
\int_{\mathbb R^d} f(y)\left[(2\pi)^{-d}\int_{\mathbb R^d}e^{i\xi\cdot(x-y)}e^{-t|\xi|^2/2}\,d\xi\right]dy
=\int f(y)g_t(x-y)\,dy.
$$
Absolute integrability of $|f(y)|e^{-t|\xi|^2/2}$ justifies this second application of <Fubini's theorem>. Thus \b[<Fourier inversion> holds for every Gaussian <convolution>], in fact pointwise for this <continuous> representative.
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