Riemann - Lebesgue lemma for generalized Fourier integrals

Question

Consider a generalized Fourier integral of the form $\int_a^bf(t)e^{ix\psi(t)}\ dt,\ x\rightarrow\infty $. Show that $\int_a^bf(t)e^{ix\psi(t)}\ dt \rightarrow0$ provided $|f(t)|$ is integrable, $\psi(t)$ is continuously differentiable, and $\psi(t)$ is not constant on any subinterval of $a\leq t\leq b.$

My attempt:
Using the mean value theorem: $\psi(t)=(t-a)\psi'(c)+\psi(a)$.
$\int_a^bf(t)e^{ix\psi(t)}\ dt=e^{ix[\psi(a)-a\psi'(c)]}\int_a^bf(t)e^{ix\psi'(c)t}\ dt$. Setting $s=\psi'(c)t$, $\int_a^bf(t)e^{ix\psi'(c)t}\ dt=\frac{1}{\psi'(c)}\int_{a\psi'(c)}^{b\psi'(c)}f(\frac{s}{\psi'(c)})e^{ixs}\ ds=0$ by Riemann - Lebesgue lemma.

Is this approach okay or not? Is there a better one?

Answer 1

The use of the MVT given in the OP to justify the result is not correct since $c$ depends on the endpoint $x$.

Under the assumption that $\phi$ is strictly increasing and that $\phi'\neq0$ in $(a, b)$, it is possible to prove the result using the inverse function theorem. For simplicity, assume $\phi'>0$ on $(a, b)$. Then

\begin{align} \int^b_a f(t) e^{ix\phi(t)}\,dt&=\int^b_a f(\phi^{-1}(\phi(t))\frac{\phi'(t)}{\phi'(\phi^{-1}(\phi(t)))}e^{ix\phi(t)}\,dt\\ &=\int^{\phi(b)}_{\phi(a)}f(\phi^{-1}(u))\frac{1}{\phi'(\phi^{-1}(u))}e^{ixu}\,du\\ &=\int^{\phi(b)}_{\phi(a)} F(u)\,e^{ixu}\,du \end{align} where $F:=\frac{f\circ \phi^{-1}}{\phi'\circ\phi^{-1}}$ defined on the interval $(\phi(a),\phi(b))$. Notice that \begin{align} \int^{\phi(b)}_{\phi(a)}|F(u)|\,du&=\int^{\phi(b)}_{\phi(a)}|f(\phi^{-1}(u))|\frac{1}{\phi'(\phi^{-1}(u))}\,du\\ &=\int^{\phi(b)}_{\phi(a)}|f(\phi^{-1}(u))|(\phi^{-1})'(u)\,du\\ &=\int^b_a|f(t)|\,dt<\infty \end{align}

That is, $F\in L_1(\phi(a),\phi(b))$. Extending $F$ to $\mathbb{R}$ by setting $F=0$ outside the interval $(\phi(a),\phi(b))$ yields a function $F\in L_1(\mathbb{R})$. The conclusion the follows from the Riemann-Lebesgue theorem for $L_1(\mathbb{R})$ functions.

If there are a countable number of critical points for $\phi$, that is points where $\phi'=0$ within the interval; $(a, b)$, one can split the interval in subintervals $(a_n,a_{n+1})$ where $\phi\neq0$ ($\phi'(a_n)=0$ for all $n$) and apply the result to $\int^{a_{n+1}}_{a_n}f(t)e^{-ix\phi(t)}\,dt$. When the number of critical points is infinite, dominated convergence will justify that the validity of this splitting.