What is the thermal average of time-dependent Hamiltonian?

$DeclareMathOperator{tr}{tr}$Let $H(t)$ be a time-dependent Hamiltonian. Then what does it mean to calculate the thermal average?

My thought process.
Usually, if $H$ is time-independent, this would just mean
$$
langle Arangle= frac{1}{Z} tr(e^{-beta H}A)
$$
So it seems that the reasonable answer would to just replace $Hmapsto H(t)$ so that the thermal average is now time-dependent too. However, this doesn’t seem to be the correct formulation in linear response theory. Indeed, in the context of linear response where $H(t) = H_0 +f(t)A$ where $f$ is real function depending on time, we see that
$$
Z(t) equiv tr e^{-beta H(t)}=Z_0langle T_tau exp left(-f(t)int_0^beta A(tau)dtau right)rangle_0, qquad A(tau)=e^{H_0tau}Ae^{-H_0tau}
$$
Yet, in textbooks (e.g., Coleman. Many-body Physics), we see that $f(t)$ is put inside the imaginary time integral and replaced with $f(tau)$, which results in the RHS being independent of time $t$, i.e.,
$$
Z_0langle T_tau exp left(-int_0^beta f(tau)A(tau)dtau right)rangle_0
$$
The above is equal to
$$
Zequivtr left( T_tau exp left(-int_0^beta H(tau)dtau right)right)
$$
This would imply that the termal average of time-dependent Hamiltonian would be something like
$$
langle Arangle=frac{1}{Z} trleft( Acdot T_tau exp left(-int_0^beta H(tau)dtau right)right)
$$
I understand that if $H(t)= H$ was time-independent (e.g., $H=H_0 +fA$ where $f$ is a constant), then
$$
T_tau exp left(-int_0^beta H(tau)dtau right) = e^{-beta H}
$$
Hence, the formula above reduces to what we expect in the time-independent case. However, I’m confused about the motivation of the time-dependent formulation.