Let $R := (R_1, R_2)$ be a two-dimensional diffusion process defined by the following SDE:
$$\mathrm{d}R_{1,t} = -\lambda_1 R_{1,t} \, \mathrm{d}t + \lambda_1 \sigma(R_{1,t}, R_{2,t}) \, \mathrm{d}W_t$$ $$\mathrm{d}R_{2,t} = \lambda_2\left(\sigma^2(R_{1,t}, R_{2,t}) - R_{2,t}\right) \, \mathrm{d}t$$
where $W$ is a standard brownian motion, $\lambda_1, \lambda_2 >0$, $\sigma: (R_1, R_2) \mapsto \beta_0 + \beta_1 R_1 + \beta_2 \sqrt{R_2}$ ; $\beta_0 > 0, \beta_1 < 0, \beta_2 \in (0, 1)$
for $T >0$ and $t\in (0, T]$, i know that the law of $R_{1,t}$ is : $$R_{1,t} \sim \mathcal{N}\left(R_{1,0}e^{-\lambda_1 t}, \lambda_1^2 \int_0^t e^{-2\lambda_1 (t - s)}\mathbb{E}\left(\sigma^2(R_{1,s}, R_{2,s})\right) \, \mathrm{d} s \right)$$
I'm interested in knowing the law of $R_{2,t}$, or at least its laplace transform : $$\mathbb{E}\left(e^{-\lambda R_{2,t}}\right)$$
The goal is to compute $\mathbb{E}\sqrt{R_{2,t}}$, using the identity : $$\mathbb{E}\sqrt{R_{2,t}} = \frac{\sqrt{\pi}}{2} \int_0^{+\infty} \frac{1 - \mathbb{E}\left(e^{-\lambda R_{2,t}}\right)}{\sqrt{\lambda^3}} \mathrm{d} \lambda$$
