Let $\gamma$ be a continuous function from $\mathbb{R}^+$ to $\mathbb{R}^+$ and consider the real valued inhomogeneous OrnsteinUhlenbeck process satisfying
$$
d X_t = \gamma_t X_t d t + d W_t, \quad X_0 = x \in \mathbb{R}.
$$
Assume that $\gamma_t \geq \nu > 0$. It seems reasonable to believe that $X_t$ has an invariant probability measure which should be gaussian.
Indeed, a quick calculation gives
$$
X_t = x \exp \left (  \int_0^t \gamma_s d s\right )
+ \int_0^t \exp \left (  \int_s^t \gamma_u d u\right ) d W_s.
$$
In order to identify the variance of the limiting gaussian variable, we shall need to calculate
$$
\lim_{t \to \infty} \exp \left ( 2 \int_0^t \gamma_s d s \right ) \int_0^t \exp \left ( 2\int_0^s \gamma_u d u\right ) d s.
$$
Question Is this limit easily computable?
I am expecting something which looks like (I may be wrong)
$$
\frac{1}{2 \Gamma}
$$
where
$$
\Gamma \triangleq \lim_{t \to \infty} \frac{1}{t} \int_0^t \gamma_s d s.
$$

$\begingroup$ I think the $+2$ and $2$ in the two exponents where you take the limit $t\rightarrow\infty$ should be interchanged, see the answer below. $\endgroup$ – Carlo Beenakker Mar 25 '17 at 13:56

$\begingroup$ Indeed, they have to be interchanged. thank you. it is updated. $\endgroup$ – megaproba Mar 26 '17 at 1:11
If we define $$\Omega(t)= \exp \left ( 2 \int_0^t \gamma_s d s \right ) \int_0^t \exp \left ( 2\int_0^s \gamma_u d u\right ) d s,$$ then this function satisfies the differential equation $$\frac{d}{dt}\Omega(t)=12\gamma_t\,\Omega(t).$$ The limit $t\rightarrow\infty$ of $\Omega(t)$ exists if $\lim_{t\rightarrow\infty}\gamma_t=\Gamma>0$ exists, and is then equal to $$\lim_{t\rightarrow\infty}\Omega(t)=\frac{1}{2\Gamma}.$$