Recently I'm starting to learn Ito calculus from the book, Stochastic Process for Physicists by Kurt Jacobs. However, I have been puzzled by the following basic statement in pp.46 and pp.47 of the book for a long time.
Suppose we have the following integral:
\begin{equation} I = \int_0^{T} \left[\int_0^{T} f(s)dW(s) \right] g(t)dW(t) \end{equation}
The book states that the integral $I$ does not have an analytical expression. This is the point that confuses me a lot. In my understanding, if you write this integral as a limit of finite summations, we have:
\begin{equation} I = \lim_{N\to \infty} \sum_{i=1}^{N} \sum_{j=1}^{N} f_i g_j \Delta W_i \Delta W_j \end{equation}
Where $f_i = f(iT/N)$, $g_j = f(jT/N)$, $\Delta W_i = \Delta W(iT/N)$, $\Delta W_j = \Delta W(jT/N)$. You can separate the summation above as a product of two summations to get:
\begin{aligned} I &= \lim_{N\to \infty} \left(\sum_{i=1}^{N} f_i \Delta W_i \right) \left(\sum_{j=1}^{N} g_j \Delta W_j \right) \\ &= \lim_{N\to \infty} \left(\sum_{i=1}^{N} f_i \Delta W_i \right) \lim_{N\to \infty} \left(\sum_{j=1}^{N} g_j \Delta W_j \right) \\ &= F(T)G(T) \end{aligned}
Where $F(T)=\int_{0}^{T} f(t)dW(t)$ and $G(T)=\int_{0}^{T} g(s)dW(s)$. With this representation, we have expressed $I$ as the product of two random variables, $F(T)G(T)$. Since $F(T)$ and $G(T)$ are mutually dependent as they both reply on the same stochastic process $dW$, we can know $G(T)$ given $F(T)$. With that, we should be able to derive the distribution of $F(T)G(T)$.
So why is the integral $I$ claimed not to have an analytical expression? Could there be something wrong or missing in my derivation, or did I misunderstand the phrase "analytical expression"? If I misunderstood, what does it mean for an Ito integral to have "analytical expression"? I will greatly appreciate your guidance!