Functional derivative: How to obtain $\delta F=\int \frac{\delta F}{\delta f}\delta f dx$? I have been reading about functionals and functional derivatives, but having some trouble to reach the common expression 
$\delta F=\int \frac{\delta F}{\delta f}\delta f dx$
As I understand, $\delta f(x)$ can be represented as $\epsilon \eta (x)$.
Then, the variation of the functional F can be written:
$\delta F = F[f+\delta f] - F[f] = F[f+\epsilon \eta] - F[f]$.
Now, as F is a function of the arbitrary number $\epsilon$, it can be expanded as power series of it:
$F[f+\epsilon \eta]=F[f]+\left( \frac{dF[f]+\epsilon \eta}{d \epsilon}  \right)_{\epsilon=0} \epsilon +...$
Until this point, everything is clear. But the text I am reading follows with the definition:
$\left(\frac{dF[f+\epsilon \eta]}{d \epsilon}\right)_{\epsilon=0}=:\int dx_1 \frac{\delta F[f]}{\delta f(x_1)} \eta(x_1)$
And, honestly, I can not understand it. I mean, where did $\epsilon$ go? Where do the integral symbol comes from?
Thanks.
 A: To start, I am going to use a somewhat different notation here to simplify things a bit and thus I'll build up from the beginning so excuse the length of my post.
Let $\mathcal{B}$ be a Banach space and $J:\mathcal{B} \to \mathbb{R}$ be a linear functional given by
$$
J[f]=\int^{b}_{a} \Lambda(f(x),f'(x),x) \, dx
$$
 Now let $\eta \in C^{1}([a,b])$ be an arbitrary function such that 
$$
\lim_{x \to a} \eta(x) =0
$$
and 
$$
\lim_{x \to b} \eta(x)=0
$$
and is identically zero outside of the interval $[a,b]$. If the functional $J[f]$ attains a local minimum at $f^{*}$ then
$$
J[f^{*}+\epsilon \eta] \geq J[f^{*}]
$$
for all $\epsilon > 0$.
Now, let $\Phi(\epsilon)$ be given by
$$
\Phi(\epsilon)=J[f^{*}+\epsilon \eta]=J[f^{*}+\delta f^{*}]
$$
then 
$$
\Phi'(0)= \frac{d \Phi}{d \epsilon}\bigg\rvert_{\epsilon=0} =\int^{b}_{a} \frac{d\Lambda}{d\epsilon}\bigg\rvert_{\epsilon=0} \,dx=0
$$
Now we need to take the total derivative for the function $\Lambda(f(x),f'(x),x)$ and we let $f(x)=f^{*}(x)+\epsilon \eta$ and $f'(x)=f^{*}(x)+\epsilon \eta'(x)$. Now note that $f(x)$ and $f'(x)$ are functions of $\epsilon$ but $x$ is not. The total derivative of $\Lambda$ is given by
$$
\frac{d \Lambda}{d \epsilon}=\frac{\partial\Lambda}{\partial f}\frac{df}{d\epsilon}+\frac{\partial \Lambda}{\partial f'}\frac{df'}{d\epsilon}
$$
and then we substitute in $\frac{df}{d\epsilon}=\eta$ and $\frac{df'}{d\epsilon}=\eta'$ to give us
$$
\frac{d \Lambda}{d \epsilon}=\frac{\partial\Lambda}{\partial f}\eta+\frac{\partial \Lambda}{\partial f'}\eta'$$
which gives us
$$
\int^{b}_{a} \frac{d\Lambda}{d\epsilon}\bigg\rvert_{\epsilon=0} \,dx=\int^{b}_{a}\left(\frac{\partial\Lambda}{\partial f}\eta+\frac{\partial \Lambda}{\partial f'}\eta'\right) \, dx
$$
Now here is where it gets a tadbit tricky. We are going to integrate by parts to get
$$
\int^{b}_{a} \frac{d \Lambda}{d \epsilon} \bigg\rvert_{\epsilon=0} \, dx=\int^{b}_{a} \left(\frac{\partial\Lambda}{\partial f}\eta-\eta\frac{d}{dx}\frac{\partial \Lambda}{\partial f'} \right)\, dx+ \frac{\partial\Lambda}{\partial f'}\eta\bigg\rvert^{b}_{a}$$
Now recall that $\eta(a)=\eta(b)=0$ so the last term vanishes and we finally arrive at 
$$
\int^{b}_{a} \frac{d \Lambda}{d \epsilon} \bigg\rvert_{\epsilon=0} \, dx=\int^{b}_{a} \eta \left(\frac{\partial\Lambda}{\partial f}-\frac{d}{dx}\frac{\partial \Lambda}{\partial f'} \right)\, dx
$$
where
$$
\frac{\partial\Lambda}{\partial f}-\frac{d}{dx}\frac{\partial \Lambda}{\partial f'}=0$$
which is (as you probably know), the Euler-Lagrange Equation. Now, the right hand side of this equation is the functional derivative of $J$, which we denote by
$$
\frac{\delta J}{\delta f(x)}=\frac{\partial\Lambda}{\partial f}-\frac{d}{dx}\frac{\partial \Lambda}{\partial f'}$$
which should answer your first question about how to derive the functional derivative expression. 
As for your second question, I think the confusion is caused by bad/confusing notation. Do you have a link or can you paste a picture with more context?
