I want to calculate the following expected value: $$ \mathbb{E} [e^{-\int_0^{t}B_s \ ds} \cdot e^{B_t}], $$ where $B_t$ is a Brownian motion, so $B_t \sim \mathcal{N}(0, t)$.
Moreover I know that $\int_0^{t}B_s \ ds \sim \mathcal{N}(0, \frac{t^3}{3})$.
The problem is that $\int_0^{t}B_s \ dt$ and $B_t$ are dependent random variables.
What I can do in this situation to calculate this expected value?
Hints:
Step 1: Show that $X_t := B_t - \int_0^t B_s \, ds$ is Gaussian. To this end, recall that $(B_t)_{t \geq 0}$ is a Gaussian process which implies, in particular, that $$B_t - \sum_{j=1}^n B_{t_j} (t_{j+1}-t_j)$$ is Gaussian for any $t>0$ and $t_1 < \ldots <t_n$. For $t_j := t \frac{j}{n}$, we find that $$ X_t = \lim_{n \to \infty} \left( B_t - \sum_{j=1}^n B_{tj/n} \frac{1}{n} \right)$$ is Gaussian as a pointwise limit of Gaussian random variables.
Step 2: Determine the mean and variance of $X_t$.
Step 3: Use the fact that $\mathbb{E}e^Y = \exp(\sigma^2/2)$ for any $Y \sim N(0,\sigma^2)$.