degree $p$ on $\R^n$ defined as follows. If $x\in \R^n,$ then $x=r\theta,$ where $r=|x|_2$ and $\theta= x/|x|_2.$ We put $$f\cdot r^p(x) = f\left(\theta\right) r^p.$$ It was proved in [@K1 Lemma 3.16] that, for any $p\in (-n,0)$ and infinitely smooth function $f$ on $S^{n-1},$ the Fourier transform of $f\cdot r^{-p}$ is equal to another infinitely smooth function $h$ on $S^{n-1}$ extended to an even homogeneous of degree $-n+p$ function $h\cdot r^{-n+p}$ on the whole of $\R^n.$ The following Parseval formula on the sphere was proved in [@K1 Corollary 3.22].

Let $f,g$ be even infinitely smooth functions on $S^{n-1},$ and $p\in (0,n).$ Then $$\label{parseval}
\int_{S^{n-1}} (f\cdot r^{-p})^\wedge(\theta) (g\cdot r^{-n+p})^\wedge(\theta) =
(2\pi)^n \int_{S^{n-1}}  f(\theta)g(\theta) \ d\theta.$$

We need a simple observation that will, however, provide the basis for applications of the Fourier transform to complex bodies.

\[const\] Suppose that $f\in C_c(S^{2n-1})$ is an even infinitely smooth function. Then for every $0<p<2n$ and $\xi\in S^{2n-1}$ the Fourier transform of the distribution $f\cdot r^{-p}$ is a constant function on $S^{2n-1}\cap H_\xi^\bot.$

By [@K1 Lemma 3.16], the Fourier transform of $f\cdot r^{-p}$ is a continuous function outside of the origin in $\R^{2n}.$ The function $f$ is invariant with respect to all $R_\theta$, so by the connection 