 Solution

Set $\theta =(\mu -r)/\sigma$. The positive densityhas mean one, and the Girsanov theorem makes $W_t^Q=W_t+\theta t$ a Brownian motion under the equivalent measure $Q$. HenceThe Black-Scholes equation makes $e^{-rt}V(t,S_t)$ a local martingale. Bounded $V$ on the finite horizon makes it a true martingale, so its terminal condition gives the risk-neutral valuationThe geometric Brownian motion solution under $Q$ isThe Brownian increment is independent of ${\mathcal{F}}_t$ and equals $\sqrt{T-t}Z$ in distribution, where $Z$ has the standard normal distribution. ThereforeThe expectation on the right is only an integral against the standard normal law; the physical drift $\mu$ has disappeared.