Does a minimiser exist for this Gaussian-like functional? Let $f(x)$ be a strictly increasing function such that $\lim\limits_{x\to\pm\infty}f(x)=\pm\infty$ and $\lim\limits_{x\to+\infty}f'(x)e^{f(x)-x}=+\infty$. If $f(a)=0$ for some prescribed $a\in\Bbb R$, does a minimum exist for $$\int_{\Bbb R}xe^{f(x)-x^2}\,dx$$ and if so, what would be the minimiser $f^*$? The presence of the exponential terms comes from a log-normal distribution integral.
The given conditions mean that $f$ has asymptotic order $x^\alpha$ for some $\alpha\in[1,2)$ as $x\to+\infty$. The goal is to maximise the negative area on the negative reals (whilst minimising the positive area) so perhaps we may need to define $f\in C^1$ piecewise, as there is no constraint on the rate of increase on the negative side.
 A: $\newcommand\R{\mathbb R}$For any $a\in\R$, there is no minimizer of
\begin{equation*}
    I(f):=\int_\R xe^{f(x)-x^2}\,dx \tag{-1}\label{-1}
\end{equation*}
over all $f\in F_a$, where $F_a$ is the set of all strictly increasing functions $f\colon\R\to\R$ such that $\lim_{x\to\pm\infty}f(x)=\pm\infty$,  $\lim_{x\to\infty}f'(x)e^{f(x)-x}=\infty$, and  $f(a)=0$. Also,
\begin{equation*}
    \inf_{f\in F_a}I(f)=0 \tag{0}\label{0}
\end{equation*}
for any $a\in\R$.
Indeed, take any $a\in\R$ and any $f\in F_a$.
Since $f$ is strictly increasing, we have
\begin{equation*}
     xe^{f(x)-x^2}>xe^{f(0)-x^2} \tag{1}\label{1}
\end{equation*}
for all nonzero real $x$.
Consider now the following two possible cases:
Case 1: $a\le0$. Then $f(x)<f(a)=0$ and $x<0$ for real $x<a$ and hence
\begin{equation*}
\begin{aligned}
    \int_{-\infty}^a xe^{f(x)-x^2}\,dx&>\int_{-\infty}^a xe^{-x^2}\,dx.  
\end{aligned}
\tag{2}\label{2}
\end{equation*}
Also, by \eqref{1},
\begin{equation*}
\begin{aligned}
    \int_a^\infty xe^{f(x)-x^2}\,dx&>\int_a^\infty xe^{f(0)-x^2}\,dx \\ 
    &=e^{f(0)}\int_a^\infty xe^{-x^2}\,dx \\ 
    &\ge e^{f(a)}\int_a^\infty xe^{-x^2}\,dx \\ 
    &=\int_a^\infty xe^{-x^2}\,dx, 
\end{aligned}
\tag{3}\label{3}
\end{equation*}
because $f$ is increasing, $a\le0$, $\int_a^\infty xe^{-x^2}\,dx\ge0$,
and $f(a)=0$.
In view of \eqref{-1}, \eqref{2}, \eqref{3}, and the equality $\int_{-\infty}^\infty xe^{-x^2}\,dx=0$,
\begin{equation*}
    I(f)>0 \tag{4}\label{4}
\end{equation*}
in Case 1.
Case 2: $a>0$. Then, again by \eqref{1},
\begin{equation*}
\begin{aligned}
    \int_{-\infty}^a xe^{f(x)-x^2}\,dx&>\int_{-\infty}^a xe^{f(0)-x^2}\,dx \\ 
    &=e^{f(0)}\int_{-\infty}^a xe^{-x^2}\,dx \\ 
    &>e^{f(a)}\int_{-\infty}^a xe^{-x^2}\,dx \\ 
    &=\int_{-\infty}^a xe^{-x^2}\,dx, 
\end{aligned}
\tag{5}\label{5}
\end{equation*}
because $f$ is increasing, $a>0$, $\int_{-\infty}^a xe^{-x^2}\,dx<0$,
and $f(a)=0$.
Also, here $f(x)>f(a)=0$ and $x>0$ for real $x>a$ and hence
\begin{equation*}
\begin{aligned}
    \int_a^\infty xe^{f(x)-x^2}\,dx&>\int_a^\infty xe^{-x^2}\,dx.  
\end{aligned}
\tag{6}\label{6}
\end{equation*}
In view of \eqref{-1}, \eqref{5}, \eqref{6}, inequality \eqref{4} holds
in Case 2 as well.
Take now any $k$ and $A$ in $(0,\infty)$, and let
\begin{equation*}
    f_{a,k,A}(x):=k(x-a)\,1(x\le a+A)+(kA+2(x-A))\,1(x>a+A)
\end{equation*}
for real $x$. Then $f_{a,k,A}\in F_a$ and, by dominated convergence,
\begin{equation*}
    \lim_{k\downarrow0}\lim_{A\to\infty}I(f_{a,k,A})
    =\lim_{k\downarrow0}\int_\R xe^{k(x-a)-x^2}\,dx
        =\int_\R xe^{-x^2}\,dx=0. 
\end{equation*}
So, $\inf_{f\in F_a}I(f)\le0$. Now \eqref{0} follows by \eqref{4}, which also shows that there is no minimizer of $I(f)$ over $f\in F_a$. $\quad\Box$.
