Given that the stochastic process follows,
$$ \frac{dS_t}{S_t} = \mu dt + \sigma dW_t $$
How do i calculate the expected value of,
$$ \int_0^T S_te^{r(T-t)} dt $$
in terms of T.
What I tried was the below,
$$ \begin{align} \int_0^T S_te^{r(T-t)} dt &= \left[ -\frac{1}{r}S_te^{r(T-t)}\right]_0^T + \frac{1}{r}\int_0^T e^{r(T-t)} dS_t\\ &=\frac{1}{r}\left[ S_Te^{rT}-S_T \right] + \frac{\mu}{r} \int_0^T S_te^{r(T-t)} dt + \frac{\sigma}{r} \int_0^T S_te^{r(T-t)} dW_t\\ \Rightarrow \left( 1-\frac{\mu}{r}\right) \int_0^T S_te^{r(T-t)} dt &= \frac{1}{r}\left[ S_Te^{rT}-S_T \right] + \frac{\sigma}{r} \int_0^T S_te^{r(T-t)} dW_t\\ \int_0^T S_te^{r(T-t)} dt &= \frac{1}{r-\mu}\left[ S_Te^{rT}-S_T \right] + \frac{\sigma}{r-\mu} \int_0^T S_te^{r(T-t)} dW_t\\ \Rightarrow \mathbb{E}\left[\int_0^T S_te^{r(T-t)} dt\right] &= \frac{1}{r-\mu}\left[ e^{rT}-1 \right]\mathbb{E}[S_T] \end{align} $$
Can anyone verify if what I did was correct?
Assuming $T$ is just a number and not a stopping time, $$ E\int_0^T S_t e^{r(T-t)}dt $$ is just just a double integral of a positive function, so we can exchange the order of integration $$ = \int_0^T E [S_t] e^{r(T-t)} dt $$ $$ = \int_0^T S_0 e^{\mu t} e^{r(T-t)}dt $$ and I presume you can finish from here as nothing left is random.