Let $dX_t = \mu_t^Xdt + \sigma_t^X dZ_t$. Compute the quantity $E_0[\int_0^{\infty}e^{-\rho t} log X_t dt]$.
Here's what I have so far.
$log X_t = log X_0 + \int_0^{t} dlog X_s ds$. Also, by Ito's lemma $dlog X_t = (\mu_t^X -0.5(\sigma_t^X)^2)dt + \sigma_t^XdZ_t$. By plugging it into the expectation, I have the following $E_0[\int_0^{\infty}e^{-\rho t} log X_t dt] = \int_0^{\infty} e^{-\rho t} log X_0 dt + \int_0^{\infty} e^{-\rho t} E_0\{\int_0^{t} (\mu_s^X -0.5(\sigma_s^X)^2)ds]dt\}$ This further reduces to $E_0[\int_0^{\infty}e^{-\rho t} log X_t dt] = \frac{1}{\rho} log X_0 + \int_0^{\infty} e^{-\rho t} E_0\{\int_0^{t} (\mu_s^X -0.5(\sigma_s^X)^2)ds]dt\}$.
It is not clear how to handle the second term in the R.H.S. Is the following correct? $\int_0^{\infty} e^{-\rho t} E_0\{\int_0^{t} (\mu_s^X -0.5(\sigma_s^X)^2)ds]dt\} = \frac{1}{\rho}\int_0^{\infty} E_0\{\int_0^{t} (\mu_s^X -0.5(\sigma_s^X)^2)ds]\}$
If yes, then how to reduce it further? Really appreciate your help.