I need some help solving the following
$$E_t \int_t^\infty{M_s^{(\gamma-1)/\gamma}ds}$$
Where $$dM_s/M_s = -rdt - \theta dZ_s$$ and $dZ_s$ is a Brownian motion, and $t<s$.
I know that I can move the expectation inside the integral, and also I know the solution of the Geometric Brownian Motion here is $M_t = M_0 \exp{(-r+1/2 \theta^2)t - \theta Z_t}$ but I think the fact that the integral and the expectation start from $t$ rather than from $0$ confuses me somewhat.
The integral limit should not be an issue. Note that \begin{align*} E_t\left(\int_t^{\infty} M_s^{(1-\gamma)/\gamma} ds \right) = \int_t^{\infty} E_t\left(M_s^{(1-\gamma)/\gamma} \right) ds. \end{align*} You can now continue for the remaining.