= Solution
The <Itô formula> for $B_t^2$ yields $\int_0^tB_u\,dB_u=(B_t^2-t)/2$. Substituting into the expression for $Z_t$ gives the pathwise identity
$$
Z_t=\exp\!\left[-\frac\beta2B_t^2+\frac{\beta t}{2}
-\frac{\beta^2}{2}\int_0^tB_u^2\,du\right].
$$
Consequently
$$
Z_t\exp\!\left[\left(\frac\beta2-\alpha\right)B_t^2-\frac{\beta t}{2}\right]
=\exp\!\left[-\alpha B_t^2-\frac{\beta^2}{2}\int_0^tB_u^2\,du\right].
$$
Taking <expected values> proves
$$
\boxed{L(t,\alpha,\beta)
=\mathbb E\!\left[Z_t\exp\!\left(\left(\frac\beta2-\alpha\right)B_t^2-\frac{\beta t}{2}\right)\right].}
$$
The random variable on the preceding line is bounded by one for $\alpha\geq0$, so its integrability is immediate.
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