Solving Stochastic Integral with Ito's lemma

I want to solve the following:

$$\int\limits_{0}^{T} exp[S(t)-t/2] dW(t)$$

where $$dS=µSdt+\sigma S dW$$ is the Brownian motion.

The Ito's formula I need to use reads:

$$df=(\frac{\partial f}{\partial S}\mu + \frac{\partial f}{t}+\frac{1}{2}\sigma^2 \frac{\partial^2 f}{\partial S^2})dt + \sigma \frac{\partial f}{\partial S} dW$$

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!

• What do you mean by "solving" the integral? – Sayantan Apr 11 at 14:43
• @Sayantan , I mean finding a solution to it, using ito's lemma. For example, representing it in terms of other stochastic integrals. I wouldn't know how to guess $f(S,t)$ – Waie Apr 11 at 17:58

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).