What about $p(Z_{si})$?

The roots of  $f(E_s)$ remains as roots even if  $p(Z_{si})$  is allowed to vary widely, and as long as the integral is take from root to root,   $p(Z_{si})$ can be ignored,  The boundary from the last  $E_s$ to $E_s\rightarrow\infty$  is still not change as  $e^{f(E_s)=0}=1$  and  $e^{f(E_s\rightarrow\infty)}\rightarrow0$ irrespective of  $p(Z_{si})$.

In the case where  $E_s=0$  is not valid then we might consider the integral along the y-axis up till the first valid  $E_s$ and bounded by $Z_s1$ and the x-axis.  $p(E_{si}=0)$  will change $Z$ as  the y-intercept changes.

The value of the end point of this integral does not change.  But the starting point,

$Z=\beta (1-e^{f(E_s=0)})$+unchanged tail integral

and

$f(E_s=0)=p(E_s=0)\prod\limits_{i}^n{E_{si}}$.

Only one point on  $p(E_s)$  is needed.  But  $E_s=0$  should be valid although  $Z_s=0$.

Hypothetically Yours,