TITLE: Asymptotic behaviour of transformed Schwartz functions
QUESTION [3 upvotes]: Given two Schwartz functions $f, g$, I consider the following simultaneous transform of them:
$$\Lambda_{f,g}(\lambda) := \int^{\infty}_{-\infty} dx f(x) g(x) e^{i\sinh(x+\lambda)}.$$
Of course, the product of Schwartz functions is still Schwartz so I could consider just a single function here, however I keep it this was for my attempt at the problem.
What I would like to show is a Riemann-Lebesgue type property:
$$\lim_{\lambda \to \pm \infty} \Lambda_{f,g}(\lambda) = 0.$$
Due to the presence of the $\sinh$ term in the exponential, the usual arguments for proving Riemann-Lebesgue don't seem to be useable here as I cannot seem to find a primitive for $e^{i\sinh(x-\lambda)}$.
My attempt to show this was to apply a hyperbolic trig identity, then split the integral into two using a delta distribution:
$$\int^{\infty}_{-\infty} dx f(x) g(x) e^{i\sinh(x+\lambda)} = \int^{\infty}_{-\infty}\int^{\infty}_{-\infty} dx dy f(x)e^{i\cosh(\lambda)\sinh(x)} g(y) e^{i\sinh(\lambda)\cosh(y)} \delta(x-y)$$
Then I apply a substitution:
$$= \int^{\infty}_{-\infty} dx dy \frac{f(x)}{\cosh(x)} \cosh(x)e^{i\cosh(\lambda)\sinh(x)} \frac{g(y)}{\sinh(y)}\sinh(y) e^{i\sinh(\lambda)\cosh(y)} \delta(x-y) \\= 2\int^{\infty}_{-\infty}\int^{\infty}_{0}dx dy\frac{f(x)}{\cosh(\sinh^{-1}(x))}e^{i\cosh(\lambda)x} \frac{g(y)}{\sinh(\cosh^{-1}(y))}e^{i\sinh(\lambda)y} \delta(\cosh^{-1}(x) -\sinh^{-1}(y))$$
Now the "altered" test functions $\frac{f(x)}{\cosh(\sinh^{-1}(x))}$ and $\frac{g(y)}{\sinh(\cosh^{-1}(y))}$ are still Schwartz as the denominator contributions are only polynomial.
From here, I was hoping to apply a Riemann-Lebesgue type argument to the integral over x to show the final integrals vanishes asymptotically in $\lambda$, but I'm not convinced due to the presence of the delta distribution.
Does this argument work? Any other possible directions of investigation to show this, other than what I have attempted, is also welcome.

REPLY [2 votes]: Here is a sketch, following the proof of the Riemann–Lebesgue lemma, per your suggestion. It suffices to show that for an indicator function $f = \mathbb{1}_{[a, b]}$, the integral
$$
\int_{\mathbb{R}} f(x) e^{i \sinh(x+\lambda)} \mathrm{d}x =
\int_{a}^{b} e^{i \sinh(x+\lambda)} \mathrm{d}x
$$
converges to zero as $\lambda \to \infty$ (the case $\lambda \to -\infty$ being similar). The rest follows by additivity of limits and density of step functions (see this page).
While we cannot find a primitive for $e^{i \sinh(x+\lambda)}$, it is not too difficult to show that the above integral converges to zero. We can assume $\lambda$ is much larger than $a, b$. In that case
$$
\sinh(x + \lambda) = \frac{1}{2} \Big( e^{x+\lambda} + e^{-(x+\lambda)} \Big) \approx \frac{1}{2} e^{\lambda} e^{x}
$$
and thus $e^{i \sinh(x+\lambda)} \approx e^{i e^{\lambda} e^{x}/2} = e^{i \mu e^{x}}$ where $\mu = \frac{1}{2}e^{\lambda}$. We are left with an oscillatory integral of the form
$$I(\mu) \colon= \int_{a}^{b} e^{i \mu \phi(x)} \mathrm{d} x, \qquad \phi(x) = e^{x}.$$
This integral is seen to vanish in the limit $\mu \to \infty$ by the van der Corput lemma,
because $\phi'$ is monotone with $|\phi'(x)| \geq c$ bounded below on the interval, and we are done.

The van der Corput lemma is easily proven by integration by parts:
$$
I(\mu) = \frac{1}{i \mu \phi'(x)} e^{i \mu \phi(x)}\bigg|_{a}^{b} -
\int_{a}^{b} \bigg( \frac{d}{dx} \frac{1}{i \mu \phi'(x)} \bigg) e^{i \mu \phi(x)} \mathrm{d}x.
$$
Since $|\phi'| \geq c$, the first term is bounded by $\frac{2}{c} \mu^{-1}$. For the second term, we have
$$
\bigg| \int_{a}^{b} \bigg( \frac{d}{dx} \frac{1}{i \mu \phi'(x)} \bigg) e^{i \mu \phi(x)} \mathrm{d}x \bigg| \leq \frac{1}{\mu}
\int_{a}^{b}\bigg| \frac{d}{dx} \frac{1}{\phi'(x)} \bigg| \mathrm{d}x
$$
Now since $\phi'$ is monotone, so is $1/\phi'$ and thus $\frac{d}{dx} \frac{1}{\phi'}$ has constant sign, which allows us to move the absolute value back outside the integral:
$$
\bigg| \int_{a}^{b} \bigg( \frac{d}{dx} \frac{1}{i \mu \phi'(x)} \bigg) e^{i \mu \phi(x)} \mathrm{d}x \bigg| \leq \frac{1}{\mu}
\bigg| \int_{a}^{b}\frac{d}{dx} \frac{1}{\phi'(x)} \mathrm{d}x \bigg| =
\frac{1}{\mu} \bigg| \frac{1}{\phi'(b)} - \frac{1}{\phi'(a)} \bigg|
$$
We conclude that $|I(\mu)| \leq \frac{C}{\mu}$ for some constant $C$.
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