It is well known how to calculate functional derivative if a functional depends of the function and it's derivatives (Euler-Lagrange rule): $\mathcal{L}=\int F(x,\dot{x},t)dt$. There is also straight-forward generalization for a functional that depend on higher order function derivatives.
From now on I will use $x$ instead of $t$ and will denote functions by small english letters (e.g. $f$, $\phi$. etc.).
In my problem, a functional depends on the function's antiderivative:
$$
\mathcal{L}=\int dx \int_{-1}^x f(x')dx'.
$$
How can I calculate ${\delta \mathcal{L}}/{\delta f}$ ? I tried to do it by definition:
$$
\int \frac{\delta \mathcal{L}}{\delta f} \phi(x) dx = \left[\frac{dF[f+\epsilon\phi]}{d\epsilon}\right]_{\epsilon=0}.
$$
However, after simplifying the right hand side I found out that the result cannot be factorized to $\int (…)\phi(x)dx$.
Denoting $F=\int_{-1}^xf(x')dx'$, I was also trying to use the chain rule, but did not succeed.
In my actual problem I need to minimize $\mathcal{L}=\int dx\int \exp(-f(x'))dx'$.
Thanks,
Mikhail
Edit: Sorry, I should have been more specific from the beginning. The actual problem is find the functional gradient w.r.t. $f(x)$ of the following functional:
$$
\mathcal{L} = \int_{-1}^1 dx \exp\left[\int_{-1}^x f(x')dx'\right]\beta(x).
$$
Note, that the second integral is entirely inside of the exponent. $\beta(x)$ does not depend on f(x).
Best Answer
OP first considers a functional $S\equiv {\cal L}$ of the form $$\begin{align}S[f]~=&~\int_a^b\!\mathrm{d}x \int_a^x\!\mathrm{d}x^{\prime} ~g\circ f(x^{\prime}) \cr ~=&~\iint_{[a,b]^2}\!\mathrm{d}x ~\mathrm{d}x^{\prime} ~\theta(x-x^{\prime})~g\circ f(x^{\prime}) \cr ~=&~ (b-x^{\prime}) \int_a^b\!\mathrm{d}x^{\prime}~g\circ f(x^{\prime}), \end{align} \tag{1a}$$ where $\theta$ is the Heaviside step function and $g$ is a fixed function. (In OP's examples $g(y)=y$ and $g(y)=e^{-y}$ and $a=-1$.)
The functional/variational derivative is then $$\forall x^{\prime}~\in~[a,b]:~~\frac{\delta S[f]}{\delta f(x^{\prime})} ~=~ (b-x^{\prime}) ~g^{\prime}\circ f(x^{\prime}) .\tag{1b}$$
In an edit OP considers a functional $S\equiv {\cal L}$ of the form $$S[f]~=~\int_a^b\!\mathrm{d}x ~g(x,F(x)), \qquad F(x)~:=~\int_a^x\!\mathrm{d}x^{\prime}~f(x^{\prime}) ~=~ \int_a^b\!\mathrm{d}x^{\prime}~\theta(x-x^{\prime})~f(x^{\prime}). \tag{2a}$$ Let $g_F=\frac{\partial g}{\partial F}$ denote the partial derivative wrt. the second argument of the $g$-function. The functional/variational derivative is then $$\forall x^{\prime}~\in~[a,b]:~~\frac{\delta S[f]}{\delta f(x^{\prime})} ~=~ \int_a^b\!\mathrm{d}x~\theta(x-x^{\prime})~g_F(x,F(x)) ~=~ \int_{x^{\prime}}^b\!\mathrm{d}x~g_F(x,F(x)).\tag{2b}$$