The solution to Langevin SDE approaches to a normal distribution

Question

I am reading L. Evans SDE lecture notes. The Langevin's Equation $$\dot{X}=-bX\it{dt}+\sigma\it{dW}$$ where X is a stochastic process, W a brownian motion, initial value $X(0)=X_0$, and constants $b$ and $\sigma$. The solution to it is $$X(t)=e^{-bt}X_0+\sigma\int_0^te^{-b(t-s)}\it{dW}.$$ However he loosely claims that $X(t)\to \mathcal{N}(0,\sigma^2/2b)$, when $t\to\infty$. I tried the characteristic function but fail to show that the convergence of $\sigma\int_0^te^{-b(t-s)}\it{dW}$. Thanks for any hint to the problem.

Answer 1

Indeed by Itô-isometry we get that the variance of $X_{t}$ is

$$\sigma^{2}\int_{0}^{t}e^{-2b(t-s)}ds=\frac{\sigma^{2}}{2b}(1-e^{-2bt})\to \frac{\sigma^{2}}{2b}.$$