Assume that $u(c_s)$ is some concave function. The goal is to evaluate the following recursive stochastic process $V_t$ given by $$V_t = E_t[\int_t^{\infty} (u(c_s) - \beta V_s)ds] $$ for some constant $\beta$ and some adaptible stochastic process $c_s$. The expectation is conditional with respect to filtration at time-$t$. The solution is $V_t = E_t[\int_t^{\infty}e^{-\beta(s-t)}u(c_s)ds]$.
What is the easiest way to prove this? Following are my steps \begin{eqnarray} V_t = E_t[\int_t^{\infty} (u(c_s) - \beta V_s)ds] = E_t[\int_t^{\infty} u(c_s)ds] - \beta E_t[\int_t^{\infty}V_sds] \end{eqnarray} $$V_t + \beta E_t[\int_{t}^{\infty}V_s ds] = E_t[\int_t^{\infty} u(c_s)ds]$$
Then what? Thanks!
How about proving that the provided expression satisfies the recursive formula?
\begin{align} \mathbb E_t\left[\int_t^\infty \left(u(c_s)-\beta\mathbb E_s\left[\int_s^\infty e^{-\beta(w-s)}u(c_w)dw\right]\right)ds\right]&=\mathbb E_t\int_t^\infty \left(u(c_s) - \int_s^\infty \beta e^{-\beta(w-s)} \mathbb E_s\left[u(c_w)\right]dw\right)ds \end{align}
Using the fact that $$\int_s^\infty \beta e^{-\beta (w-s)} \mathrm dw = 1$$ then the previous expression is:
\begin{align} \int_t^\infty \int_s^\infty\beta e^{-\beta (w-s)} \mathbb E_t\mathbb E_s[u(c_s)-u(c_w)]dwds&=\int_t^\infty \int_s^\infty\beta e^{-\beta (w-s)} \mathbb E_t[u(c_s)-u(c_w)]dwds \end{align}
Can you finish the proof using Fubini?