Consider the following stochastic integral of a deterministic function $f(t,s)$ with respect to the Wiener process $W_s$:
$$\int_0^\infty f(t,s) d W_s$$
My questions are:
Is such an integral suitably well-defined that it defines a stochastic process $Y_t$?
If so, is there a simple expression for $dY_t$?
I'm aware that the Ito integral with $t$ as the upper limit in the integration defines a stochastic process, but it is unclear what happens in this more general case (we can recover the usual case by $f(t,s)=f(s)(1-\Theta(s-t))$, where $\Theta(x)$ is the Heaviside step function).
This post here (Stochastic process as an Ito integral with time-dependent integrand) seems to imply that (1) may be true, but doesn't answer (2).
The process $$ Y_t=\int_0^\infty f(t,s)\,d W_s $$ is well defined when the usual condition $P[\int_0^\infty f^2(t,s) ds<\infty]=1$ holds which in your deterministic case boils down to $\int_0^\infty f^2(t,s) ds<\infty$. When $f(t,s)$ is differentiable in $t$ and $\int_0^\infty \partial_t f^2(t,s) ds<\infty$ then $$ dY_t=\left(\int_0^\infty \partial_t f(t,s)\,dW_s\right)\,dt\,. $$