# Riemann - Lebesgue lemma for generalized Fourier integrals

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?

• I don't think $\psi(t)$ is not constant means $\psi'(t)\neq 0$ Apr 26, 2021 at 14:26
• @LuOH3 I think that if $\psi(t)$ has a stationary point, you can split the integral in a small area around it and estimate its behavior there, still the rest will converge to zero
– mosx
Apr 26, 2021 at 16:43
• I am not sure whether it is correct when there are countable split points. Apr 27, 2021 at 11:01
• I think another problem is $c$ is not a constant, it is a function of $t$, so the change of variable is not valid Apr 27, 2021 at 11:02
• @LuOH3 If we split the integral in many very small integrals, then isn't it approximated by that in each integral?
– mosx
Apr 27, 2021 at 11:23

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.