I am having trouble figuring out how to integrate $$ \int_0^t e^{as} dW_s $$ where $a \in \mathbb{R}$ and $W_t$ is a Brownian motion.
I (think) that when you have an integral w.r.t. Brownian motion you can express it as $$ \int_0^t e^{as} dW_s = \lim_{n\xrightarrow{} \infty} \sum_{i = 0}^{n-1} e^{at} (W_{t_{i+1}} - W_{t_i}).$$
(1) I'm not sure if that expression is even relevant for solving my problem.
(2) How can I simplify this if it is relevant?
In the end I will need to multiply the result by $e^{-at}$ and end up with just $W_t$ to show that an expression satisfies an SDE. So I am hoping there is a simple answer here, but I'm just not sure about the applying the proper "rules" of stochastic calculus since they are different from ordinary calculus.
Also, if there are any particularly good resources out there for introductory stochastic calculus I'd be interested in consulting those for more background on the topic. Please link if possible! (I have scoured the internet for good resources but there's so much out there and some of it is more advanced than I can understand right now.)
Using Ito Lemma:
$$ \begin{align} d(e^{at}W_t) &= ae^{at}W_tdt+e^{at}dW_t\\ \Rightarrow e^{at}W_t - W_0 &= a\int_0^te^{as}W_sds+\int_0^te^{as}dW_s\\ \Rightarrow\int_0^te^{as}dW_s&=e^{at}W_t-a\int_0^te^{as}W_sds \end{align}$$