\left(\int_0^\infty r\hat{\phi}(r\xi) dr \right) d\xi = 
\frac1{2\pi} \int_{S^{2n-1}} f(\xi) \left(\int_{H_\xi^\bot} \hat{\phi}(x) dx\right) d\xi$$ $$= 2\pi \int_{S^{2n-1}} f(\xi) \left(\int_{H_\xi} \phi(x) dx \right) d\xi$$ $$= 2\pi \int_{S^{2n-1}} f(\xi) \left(\int_{S^{2n-1}\cap H_\xi} \left(\int_0^\infty r^{2n-3} \phi(r\theta) dr \right) d\theta \right) d\xi$$ $$= 2\pi \int_{S^{2n-1}} f(\xi)\  {\cal{R}}_c\left(\int_0^\infty r^{2n-3}\phi(r\cdot)dr\right)(\xi) d\xi$$ $$= 2\pi \int_{S^{2n-1}} \left(\int_0^\infty r^{2n-3}\phi(r\theta)dr\right) {\cal{R}}_cf(\theta) d\theta$$ $$= 2\pi \int_{\R^{2n}} |x|_2^{-2} {\cal{R}}_cf(x/|x|_2) \phi(x) dx.$$ We get that for every even $R_\theta$-invariant test function $\phi,$ $$\langle \left(f\cdot r^{-2n+2}\right)^\wedge, \phi \rangle = 2\pi \langle |x|_2^{-2}  {\cal{R}}_cf(x/|x|_2) , \phi \rangle,$$ so even $R_\theta$-invariant distributions $ \left(f\cdot r^{-2n+2}\right)^\wedge$ and $2\pi |x|_2^{-2}  {\cal{R}}_cf(x/|x|_2)$ are equal.

Lemma \[r-s\] implies the following Fourier transform formula for the volume of sections of star bodies. Note that the real version of this formula was proved in [@K2], and that the complex formula below was proved in [@KKZ] for infinitely smooth bodies by a different method; here we remove the smoothness condition.

\[volume-ft\] Let $K$ be an origin symmetric complex star body in $\R^{2n}, n\ge 2.$ For every $\xi\in S^{2n-1},$ we have $$|K\cap H_\xi| = \frac{1}{4\pi(n-1)} \left(\|x\|_K^{-2n+2}\right)^\wedge(\xi).$$

By (\[v-r\]) and Lemma \[r-s\] applied to the function $f(\theta)= \|\theta\|_K^{-2n+2},$ $$|K\cap H_\xi| = \frac1{2n-2} {\cal{R}}_c\left(\|\cdot\|_K^{-2n+2}\right)(\xi)= \frac{1}{4\pi(n-1)} \left(\|x\|_K^{-2n+2}\right)^\wedge(\xi).$$

We use Theorem \[volume-ft\] to prove the complex version of the Minkowski-Funk 