Let $s \in [0,1]$ and define diffusion processes, $$dS(s)_t = \mu(s) dt + \sigma(s) dW_t$$
The question is if the following make sense,
$$ \int_0^1 dS(s)_t ds = \int_0^1 \mu(s) ds dt + \int_0^1 \sigma(s) ds dW_t $$ or, $$ d \int_0^1 S(s)_t ds = \int_0^1 \mu(s) ds dt + \int_0^1 \sigma(s) ds dW_t $$ Assuming of course that $\int_0^1 \mu(s) ds$ and $\int_0^1 \sigma(s) ds$ exist.
I know it works when $s$ is in a finite set, but I cannot prove it in this case, as Ito's lemma is for vectors. Thanks
Formally, $dS(t)=\mu(t) dt +\sigma(t) dW$ is shorthand for: $$S(t) = S_0+\int_0^t \mu(t') dt'+\int_0^t \sigma(t') dW(t')$$ In this case, if we let $dS_s(t)=\mu(s)dt+\sigma(s)dW(t)$ $$S_s(t)=S_0+\mu(s)t+\sigma(s)W(t)$$ So then: $$\int_0^1S_s(t) ds=S_0+t\int_0^1 \mu(s) ds+W(t)\int_0^1 \sigma(s) ds$$ Letting $\int_0^1 \mu(s) ds=C_1$, $\int_0^1 \sigma(s) ds=C_2$ and $\int_0^1S_s(t) ds=R(t)$ we have that: $$R(t)=S_0+C_1t+C_2W(t)$$ Or in differential form: $$dR(t)=C_1dt+C_2dW(t)$$ Which is certainly an Ito process.
Edit: I want to ask, what do you mean by "Ito's formula is only for vectors"?