In Weinberg QFT V1, we got the general path integral of time-ordered operators product, equation (9.1.38), $$\langle{q',t'|\text{T}\left\{\mathcal{O}_A\left(P(t_A),Q(t_A)\right)\mathcal{O}_B\left(P(t_B),Q(t_B)...\right)\right\}\rangle}=$$ $$\int{\prod_{\tau,a}dq_{a}(\tau)}\prod_{\tau,b}\frac{dp_{b}(\tau)}{2\pi}\mathcal{O}_{A}\left(p(t_A),q(t_A)\right)\mathcal{O}_B\left(p(t_B),q(t_B)\right)...$$ $$\times\exp\left[i\int_{t}^{t'}d\tau\,\left\{\sum_{a}\dot{q}_a(\tau)p_{a}(\tau)-H(q(\tau),p(\tau))\right\}\right]\tag{9.1.38}$$ $a$ and $b$ are particles' indices. He says that $q_a(\tau)$ and $p_a{\tau}$ appearing in the equation are merely variables of integration, and are not constrained to obey the classical Hamiltonian eom. (see 9.1.39 and 9.1.40), take 9.1.39 for instance, $$\dot{q}_{a}(\tau)-\frac{\partial H (q,p)}{\partial p_a(\tau)} = 0\tag{9.1.39}$$ However, if we take $\mathcal{O}_{A}\left(p(t_A),q(t_A)\right)$ to be LHS of above equation, we would have, $$\mathcal{O}_{A}\left(p(t_A),q(t_A)\right)\exp(iI[q,p]) = -i\frac{\delta}{\delta p_a(t_A)}\exp(iI[q,p])\tag{p.383} $$ Where $I$ is the exponent of (9.1.38).
Weinberg then claims on p. 383 that
As long as $t_A$ doesn't approach $t$ or $t'$ (which $q_a(t)$ and $q_a(t')$ are fixed), then the integrations 9.1.38 over $q_a(t_A)$ and $p_a(t_A)$ are unconstrainted, which I can understand.
There is reasonable assumptions that the integrations are convergent.
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- and 2. imply the integral with such variational derivatives must vanish.
I don't understand how 3. is valid at all since I know that $p_a(\tau)$ and $q_a(\tau)$ don't obey classical eom, so the variational derivative wouldn't trivially vanish at all, how come Weinberg can use 1. and 2. to imply 3.?
