 Solution

In $I(\lambda )$ put $t=u/\sqrt{\lambda }$. Expanding the cubic exponential near the endpoint $t=0$ givesThe substitution $v=u^2$ evaluates the coefficient using the Gamma function. Thus, numbering the first term by $n=0$,For any fixed truncation the exponential's Taylor remainder is bounded by the next power times $e^{-u^2}$, and replacing the finite upper limit by infinity adds an exponentially small tail. This justifies the asymptotic expansion for $\alpha >-1/2$.

For large $n$, the ratio of magnitudes of consecutive terms, by Stirling's formula, isThe least term consequently has index $\boxed{n\sim 8\lambda /27}$. An indexing convention starting at one changes this by one, without changing the estimate.

For the second integral, both endpoints minimize $t^2-t^3$. The endpoint $t=0$ gives the same expansion with positive coefficients. Its first two contributions areTo check their ordering, near $t=1$ write $t=1-v/\lambda$. Then $\lambda (t^2-t^3)=v+O(v^2/\lambda )$ and $t^{2\alpha }=1+O(v/\lambda )$, producing a contribution $1/\lambda +O({\lambda }^{-2})$. Because $\alpha +1<1$, this is smaller than the second displayed term. It is the third term, ahead of the next $t=0$ contribution of order ${\lambda }^{-\alpha -3/2}$. Splitting the integral into small endpoint neighborhoods and an interior interval, where the phase is strictly positive, makes the endpoint estimates rigorous.