In literature when calculating IRT SE I found multiple times Fisher Information being mentioned.

Being curious I started to try to play around with the fisher information in order to obtain the typical Information reported as P(theta)Q(theta)a^2.

My understanding of the process failed me when I started to check why the variance of the score is defined as follows

score = s = d/dtheta ln( f(x,theta) )

Var(s) = E[s^2]

Given that the variance is

Var(s) = E[s^2] - E[s]^2

I started looking in why E[s]^2 is zero. As long as f(x,theta) is a density function I can write

E[s]^2 = [\integral{ d/dtheta ln( f(x,theta) ) * f(x,theta) dx }]^2

= [\integral{ d/dtheta( f(x,theta) ) * f(x,theta)/f(x,theta) dx }]^2

= [\integral{ d/dtheta( f(x,theta) )dx }]^2

= [d/dtheta( \integral{ f(x,theta)dx } ) ]^2

= [d/dtheta( 1 ) ]^2

= 0

But as soon as we use the IRF (Item Response Function), that gives us the probability of getting score x given theta, all the computations done above are not working anymore. The reason being that the integral of the IRF is not finite, hence

[d/dtheta( 1 ) ]^2

not valid.

I have demonstrated that

E[d/dtheta ln( f(x,theta) ) ^ 2] = -1 *E[d/dtheta(d/dtheta ln( f(x,theta) ))]

but that holds when integrals are one for f(x, theta) and simplifications can be done.

Any input on my approach and (not) understanding of the problem?