I want to find the non-decreasing, positive-everywhere (except at $x=0$), twice differentiable function $g$ which minimizes the following integral of a functional of a function: $$ \int_0^\infty g'(x) e^{-x} \ dx $$ subject to the following constraints: $g(0)=0, g(1)=1, \lim\limits_{x\to\infty} g(x) = +\infty, \lim\limits_{x\to\infty} g'(x) = +\infty$.
Any ideas for that? It seems that the standard Euler-Lagrange equation cannot be used because it would require that $x e^{-x} = 0$ which is of course impossible (let alone the three-constraint structure of this instance).
If convenient, I might potentially be ready to suppose that $ \int_0^\infty g_0(x) e^{-x} \ dx $ converges for the minimizer functional $g_0$ that satisfies the above constraints (e.g., really, any polynomial) but that might not yield optimal results in the full generality of the above problem, so I don't want to assume that out-of-the-box.
Let $$g(x)=\int _0^x f(t)dt.$$
The constraints are now: $0=0,\;\int_0^1 f(x)dx=1$ and $\int_0^\infty f(x)=+\infty$.
There could be many $f(x)$'s that minimize the integral, but here I am showing one possible $f(x)$ that minimizes it into $-\infty$.
Let:
$$ f(x)= \begin{cases} 1& (x < 1)\\ -k& (1\le x\le 2)\\ e^{x/2}& (x>2) \end{cases} $$
where $k$ is some constant for which we can take an arbitrarily large positive number.
This function obviously satisfies the first and the second constraint. Also, if we set $k$ to diverge slower than $e^{x/2}$, $f(x)$ satisfies the third constraint as well.
$$\int_0^\infty f(x)e^{-x}dx=\int_0^1 e^{-x}dx + \int_1^2 -ke^{-x}dx + \int_2^\infty e^{-x/2}dx=-(\frac{1}{e}-\frac{1}{e^2})k+1-\frac{1}{2e}$$
$k\rightarrow+\infty$ then this diverges to $-\infty$.