I'm struggling to say anything about the following random variable:
$$\Bigg(\int_0^\infty e^{aB_t-bt}dt\Bigg)^\eta \;\;\;(1)$$
where $B_t$ is a Brownian motion, $b>0$ and $\eta\in\mathbb{R}$. I found a result in Yor (Exponential functionals of Brownian Motion and related processes, 2001) which indicates that
$$\int_0^\infty e^{aB_t-bt}dt$$
where $B_t$ is a Brownian motion and $b>0$ is distributed as the reciprocal of a gamma variable (up to a multiplicative constant).
I'd like to go further, at least computing the mean of (1) if possible. Thanks for your insights.
I don't know if this is what you are looking for, but assuming that you can find the shape and scale parameter of the aforementioned gamma distributed random variable, i.e $$ X = \int_0^\infty e^{aB_t-bt}dt \stackrel{\mathcal{D}}{=} Y^{-1}, $$ where $Y\sim \Gamma(\alpha,\beta)$ for some shape parameter $\alpha>0$ and scale parameter $\beta>0$ then $X$ is inverse gamma distributed with shape parameter $\alpha$ and scale parameter $\beta$. Thus
\begin{align*} E(X^\eta) &= \int_0^\infty x^\eta \frac{ \beta^\alpha}{\Gamma(\alpha)} x^{-\alpha-1} \exp\left( -\frac{\beta}{x} \right) dx \\ &= \frac{ \beta^\alpha}{\Gamma(\alpha)} \frac{\Gamma(\alpha-\eta) }{\beta^{\alpha-\eta}} \int_0^\infty \frac{ \beta^{\alpha-\eta}}{\Gamma(\alpha-\eta)} x^{-(\alpha-\eta)-1} \exp\left( -\frac{\beta}{x} \right) dx \\ &= \frac{ \beta^\alpha}{\Gamma(\alpha)} \frac{\Gamma(\alpha-\eta) }{\beta^{\alpha-\eta}} , \end{align*} if $\alpha> \eta$.