Derivation of Hamilton's eqs. from stationarity principle 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?

 A: You should read up on functional derivatives.


*

*This is exactly the definition of the functional derivative as given on Wikipedia:
If $\delta I = \int (\cdots)\, \delta q(t) \,\mathrm dt$, then the $(\cdots)$-part is by definition the functional derivative $\frac{\delta I}{\delta q(t)}$.
Compare this to the definition of the gradient:
$$ F(\vec x + \delta \vec x) - F(\vec x) = \nabla F(\vec x) \cdot \delta \vec x + \mathcal O(|\delta \vec x|^2) . $$
In the functional case, the vector $\vec x$ is replaced by the function $q(t)$ and the dot product by the integral.

*After the integration by parts, you have
$$ \delta I = \int \left[ \delta p\, (\dot q - \frac{\partial H}{\partial p} ) -
 \delta q\, (\dot p + \frac{\partial H}{\partial q}) \right] \mathrm dt , $$
because $\delta q$ is defined to be zero at times $t_1$ and $t_2$.

*Your formula is wrong, it is not $\frac{\mathrm d I}{\mathrm d F}$ but
$$ \frac{\delta I}{\delta F(t)} = \int \mathrm d\tau \frac{\delta L'(\tau)}{\delta F(t)} = \int \mathrm d\tau q(\tau) \delta(t-\tau) = q(t) . $$
The object $\frac{\delta L'(\tau)}{\delta F(t)}$ is something slightly new again, it obeys the rule $\frac{\delta q(\tau)}{\delta q(t)} = \delta(t-\tau)$, compare with $\frac{\partial x_i}{\partial x_j} = \delta_{ij}$ in the more familiar case. Try to understand the fact that, if $I[q] = \int L[q(t)] \,\mathrm dt$,
$$ \frac{\delta I}{\delta q(t)} = \int \frac{\delta L[q(\tau)]}{\delta q(t)} \mathrm d\tau .$$
