Sorry about all the "of"s in the title... here's my problem:
I want to compute the expected value of $$ \exp\bigg\{ C \int_0^t |W_s|ds\bigg\} $$ where $W$ is a Brownian motion and $C$ is a positive constant.
My trouble comes from the absolute value (if the absolute value weren't there it would be the exponential of a Gaussian process). What do you think?
The distribution for the $$Y = \int_0^1 |W_s|ds$$ is worked on in On the Distribution of the Integral of the Absolute Value of the Brownian Motion Takacs (1993). It isn't very pretty: $$P(Y \leq x) = H(x)$$ where $$\frac{dH(x)}{dx} = \frac{\sqrt{3}}{x \sqrt{\pi}} \sum_{j=1}^\infty C_j e^{-v_j}v_j^{2/3} U(1/6, 4/3, v_j)$$ where $$C_j = \frac{1 + 3\int_0^{a_j'} Ai(-u) du}{3 a_j'Ai(-a_j')}$$ where $a_j'$ are the zeros of the Airy integral (Ai), and $U$ is a confluent hypergeometric function, and $$v_j = \frac{2(a_j')^3}{27x^2}$$ So from all there, you can compute $$E[e^{\int_0^1 |W_s| ds}] = \int_0^\infty e^{Cx}\frac{dH(x)}{dx}dx$$
The paper also does work out expression for the $k$th moment $E[Y^k]$ of $H(x)$. Perhaps you can combine those to calculate $\sum_{k=0}^\infty \frac{E[Y^k]}{k!}$.