Let $(X_t, Y_t)$ be a two-dimensional $(\mathscr{F}_t)$-Brownian motion started from 0. We set, for every $t \geq 0$ $$\mathscr{A}_t = \int_0^t X_s dY_s - \int_0^t Y_s dX_s$$ (Lévy's area). My goal is to show $E[e^{i\lambda\mathscr{A}_r}] = \frac{1}{\cosh(\lambda r)}$ ($\lambda \in \mathbb{R}$).
I already know the following, which I believe should be helpful. (They were previous parts of the same question from Le Gall's book, Brownian Motion, Martingales, and Stochastic Calculus.)
First, I know that $E[e^{i\lambda\mathscr{A}_t}] = E[\cos(\lambda \mathscr{A}_t)]$.
Second, I know that $\langle \mathscr{A}, \mathscr{A} \rangle_t = \int_0^t (X_s^2 + Y_s^2) ds$.
Third, I know that, if $$Z_t = \cos(\lambda \mathscr{A}_t)$$ $$W_t = -\frac{f'(t)}{2} (X_t^2 + Y_t^2) + f(t)$$ and if $f(t) = -\log\cosh(\lambda(r-t))$, then $Z_t \exp(W_t)$ is a martingale. I also know what the canonical decompositions of these processes are and that $\langle Z, W \rangle_t = 0$. (See this question.)
The fact that $Z_t \exp(W_t)$ allows me to compute $E[Z_t \exp(W_t)] = E[Z_0 \exp(W_0)] = \frac{1}{\cosh(\lambda r)}$. This should be relevant but I don't know how to relate this back to $E[\cos(\lambda \mathscr{A}_t)] = E[Z_t]$.
Any thoughts?
You're practically done. The statement $E[Z_t \exp(W_t)] = \frac{1}{\cosh(\lambda r)}$ is true for any $t$; in particular it's true when $t=r$. But $f(r) = f'(r) = 0$ so $W_r = 0$ and hence you have $E[Z_r] = \frac{1}{\cosh(\lambda r)}$ as desired.