Calculus of Variations commutes with Integrals I have a question about the variational calculus.
Assume a function $q(t,x)$ gives rise for another function
$$f(x) := \int dt q(t,x).$$
My question is why the variation $\delta$ commutes with the integral, therefore why holds
(*) $$\delta f(x) := \int dt \delta q(t,x)~?$$
I know that intuitively it seems to works since variations are linear and we can interpret in integral as a sum passing to a limit.
Is there a way to verify rigorously that (*) holds? I'm working with wiki's definitions describing the "variation operator" $\delta$.
Futhermore which "tools" are recomendable to handle/investivate similar properies in variational calculus when one "struggles" with it? Does it suffice only to work with the "defining equation" $$\begin{align} \delta F[\rho ;\phi ]= \int \frac{\delta F}{\delta\rho}(x) \phi(x) \; dx &= \lim_{\varepsilon\to 0}\frac{F[\rho+\varepsilon \phi]-F[\rho]}{\varepsilon} \\&= \left [ \frac{d}{d\epsilon}F[\rho+\epsilon \phi]\right ]_{\epsilon=0}~?\end{align} $$
 A: I think the notation $\delta f(x)$ while possibly intuitive is extremely ambiguous and not well defined. The formula at the end of your post: (with minor notational change)

\begin{equation}
\delta F_{\phi}(\rho) := \dfrac{d}{d \varepsilon} \bigg|_{\varepsilon = 0} F(\rho + \varepsilon \phi)
\end{equation}

is well defined, and unambiguous (as long as you specify the domain and target space of the function $F$). The way to "read" the symbol $\delta F_{\phi}(\rho)$ is "the directional derivative of the function $F$ at the point $\rho$, along the direction $\phi$". If you look at any advanced calculus textbook such as Loomis and Sternberg's Advanced Calculus, you'll see that this is precisely how directional derivatives are defined (some books require $\phi$ to be a unit vector... but that's not needed). In the subject of Calculus of variations, this is often called "the first variation of $F$ at $\rho$, along $\phi$" (or simply, the first variation of $F$). Regardless of what you want to call it, the formula above is well defined, and thus we can apply it to your question.
If we define $f(x) = \displaystyle\int_a^b q(t,x) \, dt$, then we can compute the first variation of $f$ at the point $x$, along $\phi$ as follows:
\begin{align}
\delta f_{\phi}(x) &:= \dfrac{d}{d \varepsilon} \bigg|_{\varepsilon = 0} f(x + \varepsilon \phi) \\
&:= \dfrac{d}{d \varepsilon} \bigg|_{\varepsilon = 0} \int_a^b q(t,x + \varepsilon \phi) \, dt \\
&= \int_a^b \dfrac{\partial}{\partial \varepsilon} \bigg|_{\varepsilon = 0} q(t,x + \varepsilon \phi) \, dt
\end{align}
($:=$ means "by definition")
In the last equality I made use of the Leibniz Integral rule for differentiating under the integral. The quantity inside the integral can be expressed using the multi-variable chain rule as $\dfrac{ \partial q}{ \partial x} (t,x) \cdot \phi$. But if you want to express it as a variation, using $\delta$, then to be proper, you would have to do the following: for each $t \in [a,b]$, define $Q_t(x) = q(t,x)$. Then, the quantity inside the integral is precisely $\delta (Q_t)_{\phi}(x)$ (the first variation of the function $Q_t$ at $x$, along $\phi$). So, what we have shown is
\begin{equation}
\delta f_{\phi}(x) = \int_a^b \delta(Q_t)_{\phi}(x) \, dt.
\end{equation}
And now, if you abuse notation by suppressing the direction of variation $\phi$, and if you're too lazy to define a new function $Q_t$, so that domains etc match up, then you get the claimed formula:
\begin{equation}
\delta f(x) = \int_a^b \delta q(t,x) \, dt.
\end{equation}
A: This is the definition of functional variation for integral functionals. You vary the function that is the integrand, and then see how that changes the value of the resulting integral.
A: I think I understand your question - the first part of it at least.  When you ask:

My question is why the variation δ commutes with the integral

I think you mean to ask why is it that when we apply the variational derivative:
$$\delta f(x) := \delta \int dt q(t,x)$$
we seem to jump right to:
$$\delta f(x) := \int dt \delta q(t,x)$$
The answer to this is that even though $\delta$ is not a traditional derivative like $dx$ from first semester calculus, it still holds many of the same basic properties.  Commuting with an integral (or being able to be brought inside an integral) is one of those properties.  Remember that the fundamental theorem of calculus states (among other things) that:
$$ \frac{d}{dx} \int f(x) dx = \int \frac{d}{dx} f(x) dx$$
The $\delta$ of variational calculus ends up having this same property.  I gather that you just want an explanation, not a proof. 
A: A useful definition of $\delta$ is that it is a first order change in a quantity for corresponding first order change in the quantity that it depends on.
Proof:
We wish to calculate the first order change in $f(y) = \int q(t,y(t))\,dt$, if the curve $y(t)$ is changed to $Y(t)$.
$$Y(t) = y(t) + \delta y(t)$$
It is useful to introduce an infinitesimal $\epsilon$ to keep track of the order of terms.
$$\delta y(t) = \epsilon \phi(t)$$
The corresponding change in the function $q(t,y)$ at a given time $t$ can be calculated using taylor series :
$$\Delta q = \epsilon\phi\frac{\partial q}{\partial y}+ \frac{1}{2!}\epsilon^2\phi^2\frac{\partial^2q}{\partial y^2} + O(\epsilon^3)$$
$$\Delta f = \int \epsilon\phi\frac{\partial q}{\partial y}+ \frac{1}{2!}\epsilon^2\phi^2\frac{\partial^2q}{\partial y^2} + O(\epsilon^3)\, dt$$
By the definition of $\delta$ : $\delta q$ is the first term of $\Delta q$, and $\delta q$ contains only the first term of the integral since other terms are of higher order. Hence
$$\delta f  = \int \delta q\, dt$$
