Let $W(t)$ be a Brownian motion, and say we want to find $(t_{2}\geq t_{1})$:
$\mathbb{E}[W(t_{1})e^{W(t_{2})}]$
Could we just rewrite this as finding the result to:
$\int_{-\infty }^{\infty }\sqrt{t_{1}}xe^{\sqrt{t_{2}}x}f(x)dx$
(where is the density of a standard normal random variable)?
No, that won't work because $W(t_1)$ and $W(t_2)$ are not perfectly correlated. Instead, you would want to use the independent increments property to write \begin{align*} \mathbb{E}[W(t_1)e^{W(t_2)}] &= \mathbb{E}[W(t_1)e^{W(t_2)-W(t_1)+W(t_1)}] \\ &= \mathbb{E}[W(t_1)e^{W(t_1)}]\mathbb{E}[e^{W(t_2)-W(t_1)}]. \end{align*} You can evaluate the second expectation as the moment generating function of a $N(0,t_2-t_1)$ random variable. The first expectation you can evaluate as the integral $\int \sqrt{t_1}xe^{\sqrt{t_1}x}f(x) dx$ where $f$ is the pdf of a $N(0,1)$ random variable.