Expectation of integral of involving geometric brownian motion

Question

Compute

$$\mathbb{E_P} \left( \exp{(\alpha W_t)} \int_0^t \exp{(\gamma W_u)} \,du \right)$$

where $\alpha$ and $\gamma$ are real numbers and $W_t$ is a Brownian Motion.

Answer 1

Hint

$$\mathbb{E} \left( \exp(\alpha \cdot W_t) \cdot \int_0^t \exp (\gamma \cdot W_u) \, du \right) = \mathbb{E} \left( \int_0^t e^{\alpha \cdot (W_t-W_u)+(\gamma+\alpha) \cdot W_u} \, du \right)$$

Apply Fubini's theorem to interchange expectation with integration and note that for fixed $u \in [0,t]$, $W_t-W_u$ is independent of $W_u$, i.e.

$$\mathbb{E}(e^{\alpha \cdot (W_t-W_u)+(\gamma+\alpha) \cdot W_u}) = \mathbb{E}(e^{\alpha \cdot (W_t-W_u)}) \cdot \mathbb{E}(e^{(\gamma+\alpha) \cdot W_u})$$

Since the exponential moments of a normal distributed random variable are well-known, this allows you to compute the expression you are looking for.