Laplace transform of integrated geometric Brownian motion

Is there any closed form of the Laplace transform of an integrated geometric Brownian motion ?

A geometric Brownian motion $X=(X_t)_{t \geq 0}$ satisifies $dX_t = \sigma X_t \, dW_t$ where $W=(W_t)_{t \geq 0}$ denotes a Brownian motion and the associated integrated Brownian motion is $\int_0^t X_s \, ds$. The Laplace transform of an integrated gometric Brownian motion is thus

$$\mathcal{L}(\lambda) = \mathbb{E}\left[e^{-\lambda \int_0^t X_s ds } \right]$$

I would be very surprised (in a good way) if there was a closed form solution of this. Even though it is straight-forward to calculate $$E\left[\int_0^t X(s) ds\right]$$ the distribution of $$\int_0^t X(s) ds$$ is unknown. And without knowing the distribution, I believe it will be difficult to calculate the expectation. The only thing I can figure out is a lower bound $$E\left[e^{- \lambda \int_0^t X(s) ds} \right] \geq e^{-\frac{2\lambda}{\sigma^2}( e^{\frac{\sigma^2 t}{2}}-1)}$$ which is of little use.
• $\forall s \geq 0, \ \mathbb{E}(X(s)) = x_0$ since it is a martingale. Thus by Jensen $$\mathbb{E}(e^{-\lambda \int_0^t X(s) ds}) \geq e^{-\lambda \mathbb{E}(\int_0^t X(s)ds)} \geq e^{-\lambda x_0 t}$$ – vanna Jul 18 '12 at 12:12