I want to solve the following:
$\int\limits_{0}^{T} \exp[S(t)-t/2] \mathrm dW(t)$
where $\mathrm dS=µS \mathrm dt+\sigma S \mathrm dW$ is the Brownian motion.
The Ito's formula I need to use reads:
$\mathrm df=(\frac{\partial f}{\partial S}\mu + \frac{\partial f}{t}+\frac{1}{2}\sigma^2 \frac{\partial^2 f}{\partial S^2}) \mathrm dt + \sigma \frac{\partial f}{\partial S} \mathrm dW$
I am choosing $µ = 0$ and $\sigma = 1$; yet, I am having trouble in deciding and guessing what function $f(S,t)$ is.
Shall I look at the single $\exp[S(t)-t/2]$ and integrate this from $0$ to $T$ as I would do in the standard way?
Thanks!
Since you already know that $S$ solves the SDE $\,d S_t = \mu S_t \,d t + \sigma S_t\,d W_t$, and you want to express the stochastic integral $$\int_0^T e^{S_t - t/2} \,d W_t $$ using other integrals, you may want to reverse engineer $f(t,x)$ using Ito's formula as follows.
Ito's formula gives you $$ \,d f(t,S_t) = \left( \frac{\partial f}{\partial t}(t,S_t) + \mu S_t \frac{\partial f}{\partial x}(t,S_t) + \frac{1}{2}\sigma^2 S_t^2\frac{\partial^2 f}{\partial x^2}(t,S_t)\right)\,d t + \sigma S_t \frac{\partial f}{\partial x}(t,S_t)\,d W_t$$ or,
$$ \int_0^T\sigma S_t \frac{\partial f}{\partial x}(t,S_t)\,d W_t = f(t,S_t) - f(0,S_0) - \int_0^T\left( \frac{\partial f}{\partial t}(t,S_t) + \mu S_t \frac{\partial f}{\partial x}(t,S_t) + \frac{1}{2}\sigma^2 S_t^2\frac{\partial^2 f}{\partial x^2}(t,S_t)\right)\,d t.$$
So you will be done if you can find $f$ such that $$ \sigma x \frac{\partial f}{\partial x}(t,x) = e^{x-t/2}$$ which will lead you to the integral $$ f(t,x) - f(t,0) = \frac{e^{-t/2}}{\sigma}\int_0^x \frac{e^y}{y}\,d y. $$ Unfortunately, it seems that this integral does not have a closed form expression (see here and here).
Hope this helps.