Under the B-S model: $$\mathrm{d}S_t = (r-q) S_t \mathrm{~d}t + \sigma S_t \mathrm{~d}B_t,\quad t \geq 0.$$ I want to get the distribution of \begin{equation}\label{eq:G(s,t)} G(s,t) = \left(\frac{1}{t-s} \int_{s}^{t}\ln S_{u} \mathrm{~d} u\right),\, t>0, \end{equation} which is known as part of geometric average Asian option.
To get what I want, I am confused about the integral interval of it.
Following is what I did: \begin{equation} \mathrm{d} \ln S_u = (r - q - \frac{1}{2} \sigma^2)\mathrm{d}u + \sigma \mathrm{d} B_u, \end{equation}
integral on $[0,x]$ (I am not sure whether this should be integral on [0,x] or [s,t] to get $G(s,t)$), then \begin{equation}\label{eq:ln S_x} \ln S_x = \ln S_0 + (r - q - \frac{1}{2}\sigma^2) x + \sigma B_{x}. \end{equation}
integral $\ln S_x$ on $[s,t]$, then： \begin{equation} \begin{aligned} \int_{s}^{t} \ln S_x \mathrm{~d}x &= \int_{s}^{t} \ln S_0 \mathrm{~d}x + \int_{s}^{t} (r - q - \frac{1}{2}\sigma^2) x \mathrm{~d} x + \sigma \int_{s}^{t} B_x \mathrm{~d} x\\ &= (t - s)\ln S_0 + \frac{r - q - \frac{1}{2} \sigma^2}{2} (t^2 - s^2) + \sigma \int_{s}^{t} B_x \mathrm{~d} x.\\ \end{aligned} \end{equation}
If I integral on $[s,t]$ in step 1. ,then I would get: \begin{equation}\label{eq:aaa} \ln S_{t}-\ln S_{s}=\left(r-q-\frac{1}{2} \sigma^{2}\right)(t-s)+\sigma\left(B_{t}-B_{s}\right) \end{equation}
But how can I integral again like step 2. ? It seems no variable here in above equation.