# Differentiating Stochastic Integral

I was wondering how to write the following integral in differential form: $$\int^t_0 f(s,t)dW_s$$ where $W_s$ is a standard Brownian Motion. In my understanding, if $f(s,t)$ can be written as $f(s)g(t)$, we can take $g(t)$ out of the integral and apply Ito's Lemma. However, if we are unable to separate the functional form of $f(s,t)$, how can I differentiate the integral? Thanks for your help!

We look at the integral at time $t+dt$:

$$\begin{split} \int_0^{t+dt}f(s,t+dt)dW_s &= \int_0^{t+dt}\left[f(s,t)+\partial_t f(s,t)dt \right]dW_s \\ &= \int_0^{t}\left[f(s,t)+\partial_t f(s,t)dt \right]dW_s+\left[f(t,t)+\partial_t f(t,t)dt \right]dW_t \end{split}$$

In the first line we expanded the integrand around $t$ in the second argument of $f$. In the second line we approximated the integral from $t$ to $t+dt$ by its integrand evaluated at $t$. Now we may use the rules of stochastic calculus to drop the term $dt\, dW_t$ and write

$$\begin{split} \int_0^{t+dt}f(s,t+dt)dW_s &= \int_0^{t}f(s,t)dW_s + dt\cdot \int_0^{t}\partial_tf(s,t)dW_s + f(t,t)dW_t. \end{split}$$

Thus, if we define

$$\begin{split} X^{(0)}_t=\int_0^{t}f(s,t)dW_s, \quad X^{(n)}_t=\int_0^{t}f^{(0,n)}dW_s, \end{split}$$ where $f^{(0,n)}$ is the $n$'th derivative of $f$ in the second argument, $X^{(0)}_t$ satisfies the Ito SDE $$\begin{split} dX^{(0)}_t=X^{(1)}_t dt+f(t,t)dW_t. \end{split}$$

One remark: In practice this SDE is not very helpful on its own, since $X^{(1)}$ is of the same form as $X^{(0)}$ and its SDE depends on $X^{(2)}$, and so on. This generates a closure problem, requiring us to always know one of the integrals beforehand. However, if $f^{(0,n)}\equiv 0$ for some positive integer $n$, we get a closed system of SDEs