In Itzykson & Zuber p.3 the Hamilton equations are derived. They start by defining the action by the Hamiltonian -
$$Ldt=pdq-Hdt\Rightarrow I=\intop_{t_1}^{t_2}\left[pdq-Hdt \right].\tag{1.11}$$ Then they say the change in $I$ is the integral of the change - $$\delta I=\intop_{t_1}^{t_2}\left[\delta p\left(\dot{q}-\frac{\partial H}{\partial p}\right)+p\frac{d}{dt}\delta q-\frac{\partial H}{\partial q}\delta q \right]dt,\tag{1.11a}$$
and after "integrating by parts the $p\frac{d}{dt}\delta q$ term" they get
$$\frac{\delta I}{\delta p\left(t\right)}=\dot{q}-\frac{\partial H}{\partial p},\qquad-\frac{\delta I}{\delta q\left(t\right)}=\dot{p}+\frac{\partial H}{\partial q}.\tag{1.11b}$$
My questions are:
It seems the get the derivatives by dividing with $\delta (p(t)), \delta (q(t))$ while ignoring the integral. How is this justified?
How was the integral by parts calculated? what I get is $(p\delta q)|_{q_1,\delta q_1}^{q_2,\delta q_2}-\intop_{t_1}^{t_2}\dot{p}\delta qdt$ which is not $\dot{p}$, unless I can ignore the integral again.
Latter on they say that if theres a new lagrangian $$L=L+qF(t)\tag{1.13a}$$ then
$$\frac{d}{dF}I=\intop_{t_1}^{t_2}dt\frac{\partial L'}{\partial F}=\intop_{t_1}^{t_2}dtq=Q(t),$$ where $Q(t)$ is the real trajectory, cf. eq. (1.14).
- How is it possible that by integrating $q$ over time we get $Q(t)$? how can the answer even depend on time? shoulnd the anser be the total distance traveled?