Radon Transform--Cylinder

Let the 2-D cylinder function be defined by

$$f(x,y)\equiv\cases{ 1 & for $r<R$\cr 0 & for $r>R$.\cr}$$ (1)

Then the Radon transform is given by

$$R(p, \tau)=\int_{-\infty}^\infty \int_{-\infty}^\infty f(x,y) \delta[y-(\tau+px)]\,dy\,dx,$$ (2)

where

$$\delta(x)={1\over 2\pi} \int_{-\infty}^\infty e^{-ikx}$$ (3)

is the Delta Function.

$R(p, \tau) = {1\over 2\pi} \int_0^{2\pi}\int_0^R \int_{-\infty}^\infty e^{-ik(r\sin\theta-pr\cos\theta)}r\,dr\,d\theta\,dk$
$= {1\over 2\pi} \int_{-\infty}^\infty e^{ik\tau}\int_0^{2\pi}\int_0^R e^{-ikr(\sin\theta-p\cos\theta)}r\,dr\,d\theta\,dk.$
	(4)

Now write

$$\sin\theta-p\cos\theta=\sqrt{1+p^2}\,\cos(\theta+\phi)\equiv \sqrt{1+p^2}\,\cos\theta',$$ (5)

with $\phi$ a phase shift. Then

$$R(p, \tau) = {1\over 2\pi}\int_{-\infty}^\infty e^{ik\tau}\int_0^R\left({\int_0^{2\pi} e^{-ik\sqrt{1+p^2}\,r\cos\theta'}\,d\theta'}\right)\,r\,dr\,dk$$

$$= {1\over 2\pi} \int_{-\infty}^\infty e^{ik\tau} \int_0^R 2\pi J_0(k\sqrt{1+p^2}\, r)r\,dr\,dk$$

$$= \int_{-\infty}^\infty e^{ik\tau} \int_0^R J_0(k\sqrt{1+p^2}\,r)r\,dr\,dk.$$ (6)

Then use

$$\int_0^z t^{n+1}J_n(t)\,dt=z^{n+1}J_{n+1}(z),$$ (7)

which, with $n=0$, becomes

$$\int_0^z tJ_0(t)\,dt=zJ_1(z).$$ (8)

Define

$$t \equiv k\sqrt{1+p^2}\,r$$ (9)

$$dt = k\sqrt{1+p^2}\,dr$$ (10)

$$r\,dr = {t\,dt\over k^2(1+p^2)},$$ (11)

so the inner integral is

$$\int_0^{R\sqrt{1+p^2}} J_0(t) {t\,dt\over k^2(1+p^2)} = {1\over k^2(1+p^2)} kR\sqrt{1+p^2} J_1(kR\sqrt{1+p^2}\,)$$ (12)

$$= {J_1(kR\sqrt{1+p^2}\,)\over k\sqrt{1+p^2}} R,$$ (13)

and the Radon transform becomes

$$R(p, \tau) = {R\over \sqrt{1+p^2}} \int_{-\infty}^\infty {e^{ik\tau}J_1(kR\sqrt{1+p^2}\,)\over k}\,dk$$

$$= {2R\over \sqrt{1+p^2}} \int_0^\infty {\cos(k\tau)J_1(kR\sqrt{1+p^2}\,)\over k}\,dk$$

$$= \left\{\begin{array}{ll} {2\over 1+p^2} \sqrt{R^2(1+p^2)-\tau^2} ... ...tau^2<R^2(1+p^2)$}\\ 0 & \mbox{for $\tau^2\geq R^2(1+p^2)$.}\end{array}\right.$$ (14)

Converting to $R'$ using $p=\cot\alpha$,

$$R'(r, \alpha) = {2\over\sqrt{1+\cot^2\alpha}} \sqrt{(1+\cot^2\alpha)R^2-r^2\csc^2\alpha}$$

$$= {2\over \csc\alpha}\sqrt{\csc^2\alpha R^2-r^2\csc^2\alpha}$$

$$= 2\sqrt{R^2-r^2},$$ (15)

which could have been derived more simply by

$$R'(r, \alpha) = \int_{-\sqrt{R^2-r^2}}^{\sqrt{R^2-r^2}} dy.$$ (16)