= Accretive complex quadratic reciprocal Fourier transform
{title2=$\mathcal F[(x^TAx)^{-1}](\xi)=\dfrac{2\pi^2}{d(A)\sqrt{\xi^TA^{-1}\xi}}$}
For a complex symmetric three-dimensional $A=G+iB$ with real symmetric $B$ and positive-definite $G$, the reciprocal is a regular <tempered distribution> since $|x^TAx|\geq x^TGx$. Holomorphic continuation of the real quadratic transform proves the displayed expression. The <determinant> branch is the <analytic determinant square root for accretive symmetric matrices>. The other root has positive real part because $\operatorname{Re}(\xi^TA^{-1}\xi)=\xi^TG^{-1/2}(1+C^2)^{-1}G^{-1/2}\xi>0$ for nonzero real $\xi$. Both the original and transformed singularities are locally integrable, so the continued identity holds as <distributions> at the origin too.
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