What is the stochastic integral of $\frac{dW_t}{W_t}$ Does anyone know the solution to the Ito integral with the scaling factor on $dW_t$ being $\frac{1}{w_t}$?  In other words what is:
$\int \frac{dW_t}{W_t}$  ?
It looks dangerously close to what mathemattical finance people do when they look at geometric Brownian motion, but they start with the assumption that the price, $S_t$ follows an exponential trajectory of the form:
$S_t=S_0e^{((\mu -\frac{1}{2}\sigma^2)t+\sigma W_t)}$
so, when they take the natural log the $e^{(-)}$ falls away.
My brain may just be failing, but I think this does not help find $\int \frac{dW_t}{W_t}$.
In fact, I am curious if there is a 'standard' (cookbook) way of getting the integral of any polynomial Brownian motin, i.e., $\int (W_t)^kdW_t$, but $k=-1$ is fine for now...
 A: On the particular case, $k=-1$, one shouldn't be surprised by this being problematic. Certainly, the hyperbola, $\frac{1}{x}$, is undefined at the origin. Therefore, it does not satisfy the usual conditions that guarantee the existence and uniqueness of a solution to the SDE, $\mathrm df(W_t) = \frac{1}{W_t}\mathrm dW_t,\ W_0=0$. This problem does not arise in finance since the stock price, $S$, cannot be zero.
On the cases of $k=0,1,2,\ldots$, a useful first step is the use of Ito's lemma to justify:

$$
\int\limits_{0}^{t} W_s^{k}\ \mathrm dW_s = \frac{1}{k+1}\left(W_t^{k+1} {}+{} {\bf 1}_{k>0}(k)\ {k+1\choose 2}\int\limits_{0}^{t}W_s^{k-1}\ \mathrm ds\right),
$$

where the r.h.s. now casts the problem in terms of Riemann integrals. While, in general, determining the distributions of $\ ``\int\limits_{0}^{t}W_s^{k-1}\ \mathrm ds"$ can be involved, less ambitious goals, such as computing $\mathbb E[\int\limits_{0}^{t}W_s^{k}\ \mathrm dW_s]$, can be accomplished using this alternate form.
