Question about a function in the argument of a Delta function (distribution)

Question

referring to

https://physics.stackexchange.com/questions/160503/what-is-the-wave-propagated-away-from-an-impulsively-excited-spherical-shell

my question concerns the last part of the answer which I quote in the following:

" $$\frac{R}{2z_o} \int_{-R}^{R} \delta\left(z_s-\frac{R^2+z_o^2 -c^2t^2}{2z_o}\right) d z_s$$

Now $\int_{-R}^{R} \delta\left(z_s-\frac{R^2+z_o^2 -c^2t^2}{2z_o}\right) d z_s$ is either one or zero. We know from the way we picked the delta function that the integral is one when $|z_o−R|<ct<z_o+R$ and the integral is zero otherwise. Therefore

$$f(z_o,t)=\begin{cases} \frac{R}{2 z_o}, & |z_o-R|<ct<z_o+R \\ 0, & \textrm{otherwise.} \end{cases}$$ "

( where $f(z_0,t)$ is the result of the integral. )

My question is: Is there a way to 'analytically' deduce this from the argument of the delta distribution?

The existing answer's method does not seem totally 'mathematical' otherwise I think the answer may be correct. Refer to the link for the complete question and answer. The following figure also may help.

Answer 1

Yes. All of the mass is concentrated on $z_s = \frac{R^2+z_0^2-c^2 t^2}{2z_0}$. Since this quantity is fixed, there is only one such point $z_s$ in the integration range.

Such a value $z_s$ must lie in $[-R,R]$, hence $-2Rz_0 < R^2 +z_0^2 -c^2 t^2 < 2R z_0$ thus you can see that the inequality follows from this.