Stochastic integration: Computing $\mathbb{E}[\exp(- \lambda \int_0^{t∧T_\epsilon}\frac{ds}{B_s^2}) ]$

Recently I started taking Stochastic calculus class, and I am struggling still with computation of stochastic integrals and I would appreciate any help with the following example:

We define: $$Y_t = (B_{t∧T_\epsilon})^\alpha e^{- \lambda \int_0^{t∧T_\epsilon}\frac{ds}{B_s^2}}$$ We defined $$T_\epsilon = \inf\{t \geq 1, B_t = \epsilon\}$$,$$B=(B_t)_t$$ is a one-dimensional Brownian motion starting from 1, i.e. $$B_0 = 1$$, and $$\lambda > 0$$ and $$\alpha \ne 0$$ are real numbers.

I have to compute $$\mathbb{E}[\exp(- \lambda \int_0^{T_\epsilon}\frac{ds}{B_s^2}) ]$$ We are allowed to assume that Y is a martingale and that $$\alpha(\alpha - 1) = 2 \lambda.$$

I don't know how to approach this problem or how to start at all. I would appreciate any help with it.

So far I have shown that $$T_\epsilon$$ is unique stopping time but more than that I couldn't do.

Since $$T_\epsilon$$ is a stopping time and $$Y_t$$ is a martingale, then by Optional Stopping Theorem we have: $$\mathbb{E}[Y_T]=\mathbb{E}[Y_0]$$ for any stopping time $$T$$.

We can easly comptute $$\mathbb{E}[Y_0],$$ precisely:

$$\mathbb{E}[Y_0]= (B_0)^{\alpha}e^{-\lambda \int_{0}^{0}\frac{1}{B_s^2}}=(1)^\alpha e^0=1$$ Since as initially stated: $$B_0 = 1$$ Now for $$T=T_\epsilon$$ a stopping time we have: $$\mathbb{E}[Y_T]=\mathbb{E}[(B_{T_\epsilon})^\alpha e^{- \lambda \int_0^{T_\epsilon}\frac{ds}{B_s^2}}] = \mathbb{E}[\epsilon^\alpha e^{- \lambda \int_0^{T_\epsilon}\frac{ds}{B_s^2}}] = \epsilon^\alpha \mathbb{E}[e^{- \lambda \int_0^{T_\epsilon}\frac{ds}{B_s^2}}]$$ And this implies that $$\mathbb{E}[e^{- \lambda \int_0^{T_\epsilon}\frac{ds}{B_s^2}}] = \frac{1}{\epsilon^\alpha}$$.

My only question left is: Does $$Y_t$$ fulfils all the requirement for Optional Stopping Theorem? I would assume that $$T_\epsilon$$ is finite a.s., and $$(B_{t∧T_\epsilon})$$ is a bounded martingale, hence close, but what about the other part?

The trick to being able to use the Optional Stopping Theorem here is to realize that, in the equation $$\alpha (\alpha - 1) = 2\lambda$$, we can always choose $$\alpha < 0$$. This implies $$Y_t = (B_{t \wedge T_{\epsilon}})^{\alpha}e^{-\lambda \int_0^{t \wedge T_{\epsilon}} \frac{ds}{B_s^2}} \le \epsilon^{\alpha}$$ for all $$t$$, and hence $$Y_t$$ is a uniformly integrable martingale.
$$T_{\epsilon}$$ is, as you said, finite almost surely by the recurrence properties of Brownian Motion.
Therefore, your use of the Optional Stopping Theorem is justified, and we do have $$\mathbb{E}[e^{-\lambda \int_0^{T_{\epsilon}} \frac{ds}{B_s^2}}] = \epsilon^{-\alpha}$$.