Let $V: \mathbb R \rightarrow \mathbb R$ be such that $V(x) \to \infty$ when $|x| \to \infty$ and $e^{-V}$ is integrable. Can
$$ f(x) := e^{-V(x)}\int_0^x e^{V(y)} \, \mathrm d y $$
be in $L^1(\mathbb R)$? If it can, are there conditions on $V$ under which $f(x)$ is guaranteed not to be in $L^1(\mathbb R)$?
An example with closed-form integrals, but $V(0)=-\infty$. $$ V(x) = x^4+\log(|x|^3) $$ works. We have $$ e^{V(y)} = |y|^3 e^{y^4};\quad e^{-V(x)} = \frac{e^{-x^4}}{|x|^3} $$ For $x>0$, $$ \int_0^x V(y)\;dy = \int_0^x y^3 e^{y^4}\;dy = \frac{e^{y^4}}{4}\bigg\vert_0^x = \frac{e^{x^4}-1}{4} > 0 \\ f(x) = \frac{e^{-x^4}}{x^3}\;\frac{e^{x^4}-1}{4} = \frac{1-e^{-x^4}}{4x^3} > 0 $$ and similarly for $x<0$, $$ f(x) = \frac{-1+e^{-x^4}}{4x^3} > 0 $$ And finally $$ \int_{-\infty}^\infty |f(x)|\;dx = 2 \int_0^\infty \frac{1-e^{-x^4}}{4x^3}\;dx < +\infty $$ so that $f \in L_1(\mathbb R)$.
remark
I expect $V(x) = x^4$ similarly yields $f \in L^1(\mathbb R)$. But the integral has no nice closed form, meaning that it will require more estimates.