# Stochastic process as integral over window function

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:

1. Is such an integral suitably well-defined that it defines a stochastic process $$Y_t$$?

2. 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). Apologies in advance if this question has already been answered elsewhere.

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 $$dY_t=\left(\int_0^\infty \partial_t f(t,s)\,dW_s\right)\,dt\,.$$