Radon Transform--Square

$$R(p, \tau)=\int_{-\infty}^\infty \int_{-\infty}^\infty f(x,y) \delta[y-(\tau+px)]\,dy\,dx,$$ (1)

where

$$f(x, y)\equiv \cases{ 1 & for $x, y\in [-a, a]$\cr 0 & otherwise\cr}$$ (2)

and

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

is the Delta Function.

$$R(p, \tau) = {1\over 2\pi} \int_{-a}^a \int_{-a}^a \int_{-\infty}^\infty e^{-ik[y-(\tau+px)]}\,dk\,dy\,dx$$

$$= {1\over 2\pi} \int_{-\infty}^\infty e^{ik\tau} \left[{\int_{-a}^a e^{-ky}\,dy \int_{-a}^a e^{ikpx}\,dx}\right]\,dk$$

$$= {1\over 2\pi} e^{ik\tau} {1\over -ik} [e^{-iky}]^a_{-a} {1\over ikp} [e^{ikpx}]^a_{-a}\,dk$$

$$= {1\over 2\pi} \int_{-\infty}^\infty e^{ik\tau} {1\over k^2p} [-2i\sin(ka)][2i\sin(kpa)]\,dk$$

$$= {2\over\pi p} \int_{-\infty}^\infty {\sin(ka)\sin(kpa)e^{ik\tau}\over k^2}\,dk$$

$$= {4\over\pi p} \int_0^\infty {\sin(ka)\sin(kpa)\cos(k\tau)\over k^2}\,dk$$

$$= {2\over\pi p} \int_0^\infty {\sin[k(\tau+a)]-\sin[k(\tau-a)]\over k^2} \sin(kpa)\,dk$$

$$= {2\over\pi p} \left\{{\int_0^\infty {\sin[k(\tau+a)]\sin(kpa)\over k^2}\, dk -\int_0^\infty {\sin[k(\tau-a)]\sin(kpa)\over k^2}\,dk}\right\}.$$ (4)

From Gradshteyn and Ryzhik (1979, equation 3.741.3),

$$\int_0^\infty {\sin(ax)\sin(bx)\over x^2}\,dx = {\textstyle{... ...n}\nolimits (ab) \mathop{\rm min}(\vert a\vert, \vert b\vert),$$ (5)

so

$$R(p, \tau) = {1\over p}\left\{{\mathop{\rm sgn}\nolimits [(\tau+a)pa]\mathop{\rm min}(\vert\tau+a\vert, \vert pa\vert)}\right.$$

$$\phantom{=} - \left.{\mathop{\rm sgn}\nolimits [(\tau-a)pa]\mathop{\rm min}(\vert\tau-a\vert, \vert pa\vert)}\right\}.$$ (6)

References

Gradshteyn, I. S. and Ryzhik, I. M. Tables of Integrals, Series, and Products, 5th ed. San Diego, CA: Academic Press, 1979.