= Solution
This is <endpoint asymptotics after odd Gaussian cancellation>. All interior algebraic terms cancel because $\sin t\,e^{-\lambda t^2}$ is odd. Split the interval into $[-1,1]$ and the remaining tail:
$$
\int_{-2}^1\sin t\,e^{-\lambda t^2}\,dt
=-\int_1^2\sin t\,e^{-\lambda t^2}\,dt.
$$
The right-hand integral has its minimum phase at its left endpoint, with $\phi'(1)=2$ and nonzero amplitude $\sin1$. Endpoint <Laplace method> therefore give
$$
\boxed{I(\lambda)\sim-\frac{\sin1}{2\lambda}e^{-\lambda}.}
$$
The answer is exponentially small, rather than zero; applying only the interior Gaussian approximation misses the asymmetric boundary tail.
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