For $a > 0$ is the It$\hat{o}$ integral,
$$\int_{0}^{t} e^{e^{B(s)}}\mathbb{1}_{s<\tau_{a}}dB(s)$$ a martingale on the interval $[0,T]$ for $T < \infty$?
Here, $\tau_{a}$ is the first time such that B(t) = a.
My approach -
I know that in order for a It$\hat{o}$ integral to be martingale, we can check if $$\int_{0}^{T} \mathbb{E}[e^{e^{B(s)}}\mathbb{1}_{s<\tau_{a}}]ds < \infty$$ holds or not (because in this case, the drift coefficient is $0$).
$$\int_{0}^{T} \mathbb{E}[e^{e^{B(s)}}\mathbb{1}_{s<\tau_{a}}]ds = \int_{0}^{\tau_{a}} \mathbb{E}[e^{e^{B(s)}}]ds = ? $$
This is where I am slightly stuck. I know how to calculate $e^{B(s)}$ but in order to calculate $e^{e^{B(s)}}$, I will have to use integration.
I have two questions -
1) Am I on the right track?
2) Is there any alternate (maybe easier) approach to solve this?
Any help will be appreciated.
First of all, two remarks on your attempt: In order to prove that an Itô integral is a martingale, you need to check square integrability, i.e. $\int_0^t f(s) \, dB_s$ is a martingale if
$$\int_0^t \mathbb{E} (|f(s)|^2) \, ds < \infty.$$
In your example, it doesn't make much of a difference whether there is a square or not, but in general it does. Secondly,
$$\int_0^T \mathbb{E}(e^{e^{B_s}} 1_{\{s<\tau_a\}}) \, ds = \int_0^{\tau_a} \mathbb{E}(e^{e^{B_s}}) \, ds$$
cannot hold true. Why? Well, the left-hand side is a deterministic object, i.e. just a real number. In contrast, on the right-hand side the limit of the upper integral is a random variable (namely, $\tau_a$), and hence the integral will be a random variable.
Here is one possible apprach:
Since the Brownian motion has continuous sample paths (with probability $1$), we have $B_s(\omega) \leq a$ for all $s \leq \tau_a(\omega)$. Because of the monotonicity of the exponential function, this implies $$e^{B_s(\omega)} \leq e^a, \qquad s \leq \tau_a(\omega).$$ Using once more the monotonicity, we get $$e^{2e^{B_s(\omega)}} \leq e^{2e^{a}}, \qquad s \leq \tau_a(\omega).$$ This gives $$\mathbb{E}(e^{2e^{B_s}} 1_{\{s<\tau_a\}}) \leq e^{2e^{a}} \mathbb{P}(s \leq \tau_a) \leq e^{2e^{a}}.$$ Hence, $$\int_0^T\mathbb{E}\big( \big|e^{e^{B_s}} 1_{\{s<\tau_a\}} \big|^2 \big)\, ds = \int_0^T\mathbb{E}\big( e^{2e^{B_s}} 1_{\{s<\tau_a\}} \big)\, ds \leq e^{2e^{a}}< \infty,$$which implies that the Itô integral $\int_0^t e^{e^{B_s}} 1_{\{s<\tau_a\}} \, dB_s$ is a martingale.