Assume $\sigma(t)$ is a deterministic function of time such that $\int_{0}^{t} \sigma^{2} (s) ds < \infty$ for $\forall t \geq0$. let the process $X$ be defined by $X(t) = \int_{0}^{t} \sigma(s) dW(s)$ how to show that for a fixed $t$ the characteristic function $X(t)$ is given by
$$\mathbb{E} e^{i uX(t)} = \exp \left( - \frac{u^{2}}{2} \int_{0}^{t} \sigma^{2} (s) ds \right)$$
where $u\in \mathbb{R}$.
The solution seems to require writing $e^{iuX(t)}$ as an Itô process and then derive an ODE for $\mathbb{E} e^{iuX(t)}$. However, how can the Itô process be written and how should the ODE be derived?
Let $$f(t,x)=e^{iux}$$According to Ito's lemma(https://en.wikipedia.org/wiki/Itô%27s_lemma), $$de^{iuX_t}=df(t,X_t)=-\frac{1}{2}u^2\sigma_t^2e^{iuX_t}dt+iu\sigma_te^{iuX_t}dW_t$$
so $$e^{iuX_t}=\int_{0}^{t}-\frac{1}{2}u^2\sigma_s^2e^{iuX_s}ds+\int_{0}^{t}iu\sigma_se^{iuX_s}dW_s$$so$$ Ee^{iuX_t}=E\int_{0}^{t}-\frac{1}{2}u^2\sigma_s^2e^{iuX_s}ds+E\int_{0}^{t}iu\sigma_se^{iuX_s}dW_s$$
Because of independent increment of Brownian process, the second term equals 0.
For the Dominated Convergence Theorem(hint:$|e^{ix}|=1, x\in R$), the first term can exchange E and $\int$.
Then let $$m(t)=Ee^{iuX_t}$$,we get $$m(t)=\int_{0}^{t}-\frac{1}{2}u^2\sigma_s^2m(s)ds$$ so $$dm(t)=-\frac{1}{2}u^2\sigma_t^2m(t)dt$$, this is the ODE. Then we can get what we want easily.
Ps:My English in math is not very good, forgive me.
You can see my answer by pen, and you can understand it without understanding the Chinese:):[1]: https://i.stack.imgur.com/5CkIr.jpg