Is it possible to integrate this asymptotic expansion?

Question

Suppose that some smooth function $f \in C^\infty\bigl(\mathbb R^n \times (0,+\infty)\bigr)$ possesses an asymptotic development $$ f(x,t) \sim t^{-\alpha} e^{ith(x)} \sum\limits_{k=0}^{+\infty} a_k(x)t^{-k}, \quad t \to + \infty, \;\text{uniformly w.r.t. $x$} $$ Here $h$ is some real-valued smooth function and $\alpha>0$. Suppose further that functions $$ F_k(t) = \int\limits_{\Omega} e^{it h(x)}a_k(x) \, dx, \quad \Omega \Subset \mathbb R^n, $$ possess their own asymptotic developments $$ F_k(t) \sim t^{-\alpha} e^{i t \lambda } \sum\limits_{l=0}^\infty b_{k,l} t^{-l}, \quad t \to + \infty, \;\; \lambda \in \mathbb R, \;\; k \in \mathbb N \cup \{0\}. $$ Is it possible to say then that function $F(t) = \int_\Omega f(x,t) \, dx$ has some natural asymptotic development? In the case $\alpha = 0$ it is true but in the case $\alpha > 0$ I don't see which asymptotic sequence I should take.

Answer 1

We have that \begin{align*} & \left| {F( t) - \sum\limits_{k = 0}^{N - 1} {F_k ( t)t^{ - k - \alpha } } } \right| = \left| {\int_\Omega {\left( {f( x,t) - \sum\limits_{k = 0}^{N - 1} {a_k (x)\mathrm{e}^{\mathrm{i}th( x )} t^{ - k - \alpha } } } \right)\mathrm{d}x} } \right| \\ & \le \int_\Omega {\left| {f( x,t) - \sum\limits_{k = 0}^{N - 1} {a_k( x )\mathrm{e}^{\mathrm{i}th( x)} t^{ - k - \alpha } } } \right|\mathrm{d}x} \le \int_\Omega {C_N \left| {\mathrm{e}^{\mathrm{i}th( x )} } \right|t^{ - N - \alpha } \mathrm{d}x} = \operatorname{Vol}( \Omega)C_N t^{ - N - \alpha } , \end{align*} for large enough $t$ and bounded $\Omega$. So we have $$ F( t ) = \sum\limits_{k = 0}^{N - 1} {F_k ( t )t^{ - k - \alpha } } + \mathcal{O}( t^{ - N - \alpha }), $$ for large $t$ and any fixed $N$. Substitute truncated asymptotic series for the $F_k(t)$s and collect the error terms to show that $F(t)$ has an asymptotic expansion.