Given a time horizon N, I want to know the time-$t$ Black-Scholes fair price of $$\int_0^T S_u du$$ where $S_u$ denotes the time-$u$ stock price. I have used the formula I have been given as follows: $$\begin{align*} &e^{-r(T-t)}E^Q\left(\int_0^T S_u du\mid \mathcal{F}_t \right)\\=&e^{-r(T-t)}\int_0^t S_u du+e^{-r(T-t)}E^Q\left(\int_t^T S_u du\mid \mathcal{F}_t\right)\\=&e^{-r(T-t)}\int_0^t S_u du+S_te^{-r(T-t)}E^Q\left(\int_t^T \frac{S_u}{S_t} du\mid \mathcal{F}_t\right)\\=&e^{-r(T-t)}\int_0^t S_u du+S_te^{-r(T-t)}E^Q\left(\int_t^T e^{(r-\sigma^2/2)(u-t)}e^{\sigma(W_u-W_t)} du\mid \mathcal{F}_t\right)\\=&e^{-r(T-t)}\int_0^t S_u du+S_te^{-r(T-t)}E^Q\left(\int_t^T e^{(r-\sigma^2/2)(u-t)}e^{\sigma(W_u-W_t)} du\right) \end{align*}$$ We remove the conditioning since $W_u-W_t$ is independent of the sigma algebra. Now we swap the integrals and use the mgf of a normal distribution. $$\begin{align*}=&e^{-r(T-t)}\int_0^t S_u du+S_te^{-r(T-t)}\left(\int_t^T e^{(r-\sigma^2/2)(u-t)}e^{\sigma^2(u-t)^2/2} du\right) \end{align*}$$
I was wondering if firstly, this is correct and secondly, if this was as simple as we could go for the pricing or if we could simplify this expression.
EDIT: Thanks to @LucaMac for pointing out $E(e^{\sigma(W_u-W_t)})=e^{\sigma^2(u-t)/2}$