= Solution
In $I(\lambda)$ put $t=u/\sqrt\lambda$. Expanding the cubic exponential near the endpoint $t=0$ gives
$$
I(\lambda)\sim\lambda^{-\alpha-1/2}\sum_{n=0}^\infty\frac{(-1)^n}{n!\lambda^{n/2}}\int_0^\infty u^{2\alpha+3n}e^{-u^2}\,du.
$$
The substitution $v=u^2$ evaluates the coefficient using the <Gamma function>. Thus, numbering the first term by $n=0$,
$$
\boxed{I(\lambda)\sim\sum_{n=0}^\infty\frac{(-1)^n\Gamma(\alpha+(3n+1)/2)}{2n!\,\lambda^{\alpha+(n+1)/2}}.}
$$
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>, is
$$
\frac{\Gamma(\alpha+(3n+4)/2)}{(n+1)\sqrt\lambda\,\Gamma(\alpha+(3n+1)/2)}
\sim\left(\frac32\right)^{3/2}\sqrt{\frac n\lambda}.
$$
The least term consequently has index $\boxed{n\sim8\lambda/27}$. 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 are
$$
\boxed{J(\lambda)=\frac{\Gamma(\alpha+1/2)}{2\lambda^{\alpha+1/2}}+
\frac{\Gamma(\alpha+2)}{2\lambda^{\alpha+1}}+O(\lambda^{-1}),\quad-1/2<\alpha<0.}
$$
To 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.
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