A Question on Hamilton's Principle In some literatures, the Hamilton's principle for conservative systems is introduced by this equation:
$$\delta \int_{t_1}^{t_2}(T-V) ~\mathrm{d}t~=~0$$
In some others, this principle is introduces as follow:
$$\int_{t_1}^{t_2}\delta(T-V) ~\mathrm{d}t~=~0$$
What is the difference between two equations? Are those expressions the same?
 A: They're the same, because integration is linear:
$$\int _{t_1}^{t_2} \left( f(t) - g(t) \right) \,dt = \int_{t_1}^{t_2}f(t)\,dt - \int_{t_1}^{t_2}g(t)\,dt$$
Addendum:
Consider two paths. Let the system trajectory in the first path be denoted $X_1(t)$ and the trajectory for the second path be denoted $X_2(t)$. Let the Lagrangian be denoted $L$. Then the action for the first trajectory is
$$S_1 = \int_{t_1}^{t_2}L(X_1(t))\,dt$$
and the action for the second trajectory is
$$S_2 = \int_{t_1}^{t_2}L(X_2(t))\,dt.$$
The difference in the action between the two paths is \begin{eqnarray}
\Delta S \equiv S_1 - S_2 &=& \int_{t_1}^{t_2}L(X_1(t))\,dt - \int_{t_1}^{t_2}L(X_2(t))\,dt \\
&=& \int_{t_1}^{t_2} \left[ L(X_1(t))-L(X_2(t)) \right]\,dt \\
&=& \int_{t_1}^{t_2}\Delta L(t)\,dt \end{eqnarray}
In the last equation we wrote $\Delta L$ to denote $L(X_1) - L(X_2)$. This is not quite what we're going for: we want to get $\delta L$ inside the integral, where $\delta L$ is the variation of the Lagrangian. The variation is the derivative of a function with respect to something. When you think about the definition of a derivative, you have something like
$$\frac{df}{dx}(x_0)\equiv \lim_{h\rightarrow 0}\frac{f(x_0+h)-f(x_0)}{h}.$$
So really, taking the $\delta$ inside the integral requires the linearity property we already demonstrated and the ability to move the limit inside the integral. This is ok in almost every case you will ever encounter. To prove why you can move the limit inside the integral you have to do some analysis which I really don't want to recall and type here. You can find that sort of thing in analysis books.
