The value of a stochastic integral, in this case integrating a Wiener process with respect to itself $\int_0^T W(t)\;dW(t)$ is dependent on the chosen position of the endpoint of the subintervals.
For example the Ito integral (leftpoint chosen) gives: $\int_0^T W(t)\;dW(t)=\frac{W(T)^2}{2}-\frac{T}{2}$ The extra term $-T/2$ is the so called Ito correction term.
My question concerns this very term: If $W(t)$ wasn't stochastic but deterministic the classical calculus would be used and this extra term wouldn't be there. You can e.g. show via simulation that it disappears when you set $\sigma=0$ and thereby render Brownian motion deterministic.
My question
When the deterministic integral is interpreted as the limiting case ($\sigma \rightarrow 0$) of the stochastic integral where the extra term $-T/2$ vanishes when $\sigma$ reaches $0$ (and thereby becomes deterministic), how can it be that the extra term $-T/2$ is only dependent on $T$ and not on $\sigma$?