 Solution

Fix an integer $N\ge 0$ and writeFor 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 givesTo control the remainder, choose $\delta >0$ such that $|r_N(x)|\le ϵx^{\alpha +N\beta }$ for $0<x<\delta$. Its integral there is at most $ϵ\Gamma (\alpha +N\beta +1){\lambda }^{-\alpha -N\beta -1}$. On $[\delta ,b]$ the remainder is bounded and its integral is $O(e^{-\lambda \delta })$. Since $ϵ$ is arbitrary, we have the full Watson lemma expansionMore 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.