Consider the random variable:
$$ g(x) = \int_{0}^{\infty} \delta e^{-\delta t} \phi(x, \mu t + \sigma Z_t, \nu^2 t)dt $$ where $\phi(x, \mu, \nu^2)$ is the density of a normal variable with mean $\mu$ and variance $\nu^2$, $\delta >0$, and $Z_t$ is a Brownian Motion.
When $\sigma = 0$, this is a deterministic problem and we have $g(x) = C e^{-\zeta x}$ for $x > 0$, where $\zeta$ solves the equation $\mu \zeta + \frac{1}{2}\zeta^2\nu^2 - \delta = 0$.
When $\sigma > 0$, I think we still have $\lim_{x \to \infty} \frac{1}{x}\log g(x) = -\zeta x$ but I'm not sure how to prove it.