= Solution
Fix an integer $N\geq0$ and write
$$
f(x)=\sum_{j=0}^Na_jx^{\alpha+j\beta}+r_N(x),\qquad
r_N(x)=o(x^{\alpha+N\beta}).
$$
For each power, substitute $y=\lambda x$ and extend the upper limit to infinity; the discarded tail is exponentially small. Since $\alpha+j\beta>-1$, the <Gamma integral> gives
$$
\int_0^\infty x^{\alpha+j\beta}e^{-\lambda x}\,dx
=\Gamma(\alpha+j\beta+1)\lambda^{-\alpha-j\beta-1}.
$$
To control the remainder, choose $\delta>0$ such that $|r_N(x)|\leq\epsilon x^{\alpha+N\beta}$ for $0<x<\delta$. Its integral there is at most $\epsilon\Gamma(\alpha+N\beta+1)\lambda^{-\alpha-N\beta-1}$. On $[\delta,b]$ the remainder is bounded and its integral is $O(e^{-\lambda\delta})$. Since $\epsilon$ is arbitrary, we have the full <Watson lemma> expansion
$$
\boxed{I(\lambda)\sim\sum_{j=0}^\infty
\frac{a_j\Gamma(\alpha+j\beta+1)}{\lambda^{\alpha+j\beta+1}}.}
$$
More precisely, after the term $j=N$, the remainder is $o(\lambda^{-\alpha-N\beta-1})$. This proof uses neither differentiability of $f$ nor convergence of its asymptotic series.
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