I have the transform functions (forward and backward projections) such as:
$$FP\{f(x,y)\} = \int_{-\infty}^{\infty}f(r\cos(\theta) - z\sin(\theta), r\sin(\theta) + z\cos(\theta))dz$$ $$BP\{g_{\theta}(r)\} = \int_{0}^{\pi}g_{\theta}(x\cos(\theta) + y\sin(\theta))d\theta$$
What i want to solve is that first forward projection, which is: $FP\{\delta(x,y)\} = ?$, then i want to solve the backward projection of the result, which actually is: $BP\{FP\{\delta(x,y)\}\}=?$.
I know that the answers must be $BP\{FP\{\delta(x,y)\}\}= \frac{1}{\sqrt{x^2 + y^2}}$.
This problem occurs during study of the Radon transform.
Let's –boringly– just insert the functional in question and see where it takes us:
$$\newcommand{\FP}[1]{\mathrm{FP}\{#1\}(r,\theta)} $$
\begin{align} \FP{\delta(x,y)} &= \int_{-\infty}^{\infty}\delta(r\cos(\theta) - z\sin(\theta), r\sin(\theta) + z\cos(\theta))dz\\ &=\#\left\{z|r\cos(\theta) - z\sin(\theta) = 0 \wedge r\sin(\theta) + z\cos(\theta)=0\right\}\\ &= \#\left\{z|r\cos(\theta) = z\sin(\theta) \wedge r\sin(\theta) = - z\cos(\theta)\right\}\tag{*}\\[2em] &\underline{\text{1. case: $\cos\theta\ne0,\,r\ne0,\,n\in\mathbb Z$}}\\ &= \#\left\{z| r= z\tan(\theta) \wedge r\tan(\theta) = - z\right\}\\ &= \#\left\{z\Big| \frac rz=\tan(\theta) \wedge \tan(\theta) = - \frac zr\right\}\\ &\text{$\tan\theta$ takes any arbitrary but fixed value $\in\mathbb R$:}\\ &= \#\left\{z\Big| \frac rz = - \frac zr\right\}\\ &= \#\left\{z| r^2 = - z^2\right\}\\ &= 0 \text{ (because $r^2>0\ge-z^2$)}\\[2em] &\underline{\text{2. case: $r=0$}}\\ (*) &= \#\left\{z| 0 = z\sin(\theta) \wedge 0 = - z\cos(\theta)\right\}\\ &= \#\left\{z|0 = z\sin(\theta) \wedge 0 = z\cos(\theta)\right\}\\ &=1 \text{ (namely, $z=0$, since $\sin\theta\ne\cos\theta$)}\\[2em] &\underline{\text{3. case: $\cos\theta=0,\,r\ne0,\,n\in\mathbb Z$}}\\ (*) &= \#\left\{z| r0 = z1 \wedge r1 = - z0\right\}\\ &= \#\left\{z|z=0 \wedge r = 0\right\}\\ &= 0 \end{align}
Thus, in total:
$$\FP{\delta}(r,\theta) = \begin{cases} 1 &\text{when} &r=0 \\ 0 &\text{else.} \end{cases}$$