A smooth radial phantom and its sinogram

For $u>0$, repeated differentiation gives ${\eta }^{(k)}(u)=P_k(1\mathrm{/}u)e^{-1\mathrm{/}u}$ for a polynomial $P_k$. For every integer $m\ge 0$, $v^m{e}^{-v}\to 0$ as $v\to \mathrm{\infty }$: if $v>0$, the exponential series gives $e^v\ge v^{m+1}\mathrm{/}(m+1)!$. Hence each right derivative tends to $0$ at $u=0$, and the quotient ${\eta }^{(k)}(u)\mathrm{/}u$ also tends to $0$ because it is again a polynomial in $1\mathrm{/}u$ times $e^{-1\mathrm{/}u}$. Induction on $k$ shows the zero extension is smooth on $\mathbb{R}$.

The composition $f(x)=\eta (1-\mathrm{\mid }x{\mathrm{\mid }}^2)$ is smooth and radial; it vanishes whenever $\mathrm{\mid }x\mathrm{\mid }\ge 1$ and is positive at $x=0$. Thus it is nonzero and has compact support in the closed unit disk.

For every $\theta \in S^1$, orthogonality gives $\mathrm{\mid }s\theta +t{\theta }^{\perp }{\mathrm{\mid }}^2=s^2+t^2$, so [F1] yields $Rf(\theta ,s)={\int }_{\mathbb{R}}\eta (1-s^2-t^2)dt$. If $\mathrm{\mid }s\mathrm{\mid }\ge 1$, then $1-s^2-t^2\le 0$ for every $t$, so the line integral is $0$; if $\mathrm{\mid }s\mathrm{\mid }<1$, the integrand vanishes outside $\mathrm{\mid }t\mathrm{\mid }<\sqrt{1-s^2}$ and equals the exponential displayed in the Example inside that interval, with value $0$ at its endpoints because $\eta (0)=0$. It depends on $\theta$ only through the norm identity, so the sinogram is radial-data independent of orientation. The continuous compactly supported integrands are Riemann integrable by [F2].