Let $(W_s)_{s \geq 0}$ be a Wiener process and $\tau$ be a random variable with an exponential distribution with parameter $\lambda$. Suppose that $W$ and $\tau$ are independent. In this question, we see that the distribution of the stopped Wiener process $W_\tau$ corresponds to a Laplace distribution with scale parameter $\frac{\sigma}{\sqrt{2\lambda}}$ where $\sigma$ is the instantaneous variance of the Wiener process.
Suppose now, that the variance $\sigma^2$ also follows a stochastic process such that we have: $$dS = \sigma SdW_s \\ d\sigma^2 = \alpha\sigma^2dt + \xi\sigma^2dW_\sigma$$ where $\alpha$ and $\xi$ are independent of $S$ and $dW_s$ and $dW_\sigma$ are independent Wiener processes.
My aim is to derive the distribution of $\log S_\tau$.
First, if $\bar{V}_T$ denotes the mean variance over some time interval $[0,T]$ defined by $$\bar{V}_T = \frac{1}{T} \int\limits_0^T\sigma^2(t)dt,$$ it is easy to show (Lemma of Îto) that $$\log S_T = \log S_0 - \bar{V}_TT/2 + \int\limits_0^T\sqrt{\bar{V}_T}dW_3 $$ where $W_3$ is a Wiener process. Making use of the answer provided here I get for the characteristic function of the stopped Process $\log S_\tau$:
$$\varphi_{S_{\tau}^{b, s}}\left(u\right) = E\left( e^{iu S_\tau} \right) \\ = E\left( E\left( e^{iuS_\tau} \mid\tau \right)\right) \\ = E\left( e^{iu\bar{V}_\tau\tau/2-\tau\left(\bar{V}_\tau\right)u^2/2} \right) \\ =\lambda \int_0^\infty e^{-\lambda + iu\bar{V}_\tau\tau/2-\tau\left(\bar{V}_\tau\right)u^2/2} \, d\tau. $$ Here I am stuck as I do not see how to integrate $\bar{V}_\tau$ over $\tau$..is there any way to represent the distribution of the stopped Wiener Process $\log S_\tau$ as a function of $\bar{V}_\tau$?