Euler-Lagrange equation: integral over positive real line of a perturbed functional My goal is to minimize the functional
$I[f] = \int_{0}^{\infty}{L(x,f(x),f'(x)) e^{-x} dx  }$
However, the solution of the Euler-Lagrange equation is usually stated
as minimizing a functional of the form
$\int_{a}^{b}{  G(x,f(x),f'(x)) dx}$
My question is the following: for $f$ to be a minimizer of $I[f]$,
is it still a necessary condition that it solve the Euler-Lagrange
equation?
I should also mention that
$I[f] = \lim_{d\rightarrow \infty}\int_{0}^{d}{L(x,f(x),f'(x)) e^{-x} dx  }$
since $L[\cdot]$ is uniformly bounded by a constant.
 A: Yes $e^{-x}$ has an influence on the E.L. equation due to the Leibniz rule for derivatives, even if  the "extra factor", i.e. $e^{-x}$ does not depend on $f$, which is the argument of the functional. Very sketchy:
$$\frac{1}{\epsilon}\left(I[f+\epsilon\varphi]-I[f]\right)=\frac{1}{\epsilon}\int_0^\infty e^{-x}\left(L(f+\epsilon\varphi,f'+\epsilon\varphi',x)-L(f,f',x)\right)dx\rightarrow
\int_0^\infty e^{-x}\left(\frac{\partial L}{\partial f}\varphi+\frac{\partial L}{\partial f'}\frac{d\varphi}{dx}\right)dx $$ 
for $\epsilon\rightarrow 0$ and for all variations $\varphi$'s of $f$.
Then
$$\lim_{\epsilon\rightarrow 0}\frac{1}{\epsilon}\left(I[f+\epsilon\varphi]-I[f]\right)=
\int_0^\infty e^{-x}\left(\frac{\partial L}{\partial f}\varphi+\frac{d}{dx}\left[\frac{\partial L}{\partial f'}\varphi\right]-\frac{d}{dx}\left(\frac{\partial L}{\partial f'}\right)\varphi\right)dx. $$
Now
$$\int_0^\infty e^{-x}\frac{d}{dx}\left[\frac{\partial L}{\partial f'}\varphi\right]=
\int_0^\infty \frac{d}{dx}\left[e^{-x}\frac{\partial L}{\partial f'}\varphi\right]-
\int_0^\infty \frac{de^{-x}}{dx}\frac{\partial L}{\partial f'}\varphi;$$
in summary we arrive at
$$\delta I:=\lim_{\epsilon\rightarrow 0}\frac{1}{\epsilon}\left(I[f+\epsilon\varphi]-I[f]\right)=
\int_0^\infty e^{-x}\varphi\underbrace{\left(\frac{\partial L}{\partial f}-\frac{d}{dx}\left(\frac{\partial L}{\partial f'}\right)+\frac{\partial L}{\partial f'}\right)}_{\text{these terms lead to the E.L. equations}}dx+ \underbrace{\int_0^\infty \frac{d}{dx}\left[e^{-x}\frac{\partial L}{\partial f'}\varphi\right]}_{\text{this is a boundary term(=0 choosing good funct.spaces)}}.$$
The E.L. equations for the modified functional have a special structure given as above; if $f$ satisfies the "classical" E.L. equations 
$$ \frac{\partial L}{\partial f}-\frac{d}{dx}\left(\frac{\partial L}{\partial f'}\right)=0,$$
then it is a solution of the modified E.L. for the modified functional $I$ (in which $e^{-x}$ appears) only if
$$\frac{\partial L}{\partial f'}=0. $$
