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,