How to extract small-frequency behaviour for Fourier-type integrals?

Question

Considering an integral of the form $$ G(\omega) = \int_a^\infty dx \ \sin(\omega x) f(x) $$ where $a, \omega \in \mathbb{R}$. I see commonly in the literature how to extract the behaviour of $G(\omega)$ for very large $\omega \to \infty$. This relies on the small-$x$ behaviour of the function $f(x)$.

Is there a way to extract the behaviour of $G(\omega)$ for small $\omega$? My guess would be that you need to know the large-$x$ behaviour of $f(x)$.

EDIT: To be concrete, pick $a=0$ and $f(x) = e^{-x}$ then its easy to $G(\omega) = \frac{\omega}{\omega^2 +1}$. It's clear that a $0 < \omega \ll 1$ expansion yields $G(\omega) \simeq \omega + \mathcal{O}(\omega^3)$. Is there a technique you can use which allows you to learn about this behaviour (without explicitly evaluating the integral)?

Answer 1

We may exploit the periodicity of $\sin \omega x$ to extract the small frequency behaviour.

$$ \begin{align} \int^\infty_a f(x)\sin\omega x \,dx &=\int^\infty_0 f(x+a)\sin(\omega x+\omega a) \,dx \\ &=\frac1\omega\int^\infty_0 f\left(\frac{x}{\omega}+a\right)\sin(x+\omega a) \,dx \\ &=\frac1\omega\sum^\infty_{n=0}\int^{2(n+1)\pi}_{2n\pi}f\left(\frac{x}{\omega}+a\right)\sin(x+\omega a)dx\\ &=\frac1\omega\sum^\infty_{n=0}\int^{2\pi}_{0}f\left(\frac{x+2n\pi}{\omega}+a\right)\sin(x+\omega a)dx\\ &=\frac1\omega\int^{2\pi}_{0}\sin(x+\omega a)\left[\sum^\infty_{n=0}f\left(\frac{x+2n\pi}{\omega}+a\right)\right]dx\\ \end{align} $$

Typically, $f(x)\to 0$ as $x\to\infty$. Hence, as $\omega\to 0$ and $n\to\infty$, terms in the summation are decaying quickly. Therefore, keeping only the first few terms of the summation may give a good approximation.

On the other hand, we want to imitate Watson's Lemma to give an asymptotic expansion for analytic $f$. In particular, if $f(z)$ is analytic at $z=\infty$, then $f$ can be expanded as $$f(z+a)=\sum_{k=1}^\infty a_k z^{-k}$$ (It is assumed that $f(\infty)=0$.)

Thus, $$\begin{align} \int^\infty_a f(x)\sin\omega x \,dx &=\frac1\omega\int^{2\pi}_{0}\sin(x+\omega a)\left[\sum^\infty_{n=0}\sum^\infty_{k=1}a_k\left(\frac{\omega}{x+2n\pi}\right)^k\right]dx \\ &=\sin(\omega a)\sum^\infty_{k=1}a_k\left[\int^{2\pi}_0 \cos x\sum^\infty_{n=0}\frac1{(x+2n\pi)^k}\right]\omega^{k-1} \\ &~~~~+\cos(\omega a)\sum^\infty_{k=1}a_k\left[\int^{2\pi}_0 \sin x\sum^\infty_{n=0}\frac1{(x+2n\pi)^k}\right]\omega^{k-1} \end{align} $$

Surprisingly, we found a series of constants $$A_k=\frac1{(2\pi)^k}\int^{2\pi}_0 \cos x\, \zeta\left(k,\frac{x}{2\pi}\right)dx =\frac1{(2\pi)^{k-1}}\int^{1}_0 \cos (2\pi x)\zeta\left(k,x\right)dx$$ $$B_k=\frac1{(2\pi)^k}\int^{2\pi}_0 \sin x\, \zeta\left(k,\frac{x}{2\pi}\right)dx =\frac1{(2\pi)^{k-1}}\int^{1}_0 \sin (2\pi x)\zeta\left(k,x\right)dx$$ where $\zeta$ is the Hurwitz zeta function.

Hence,

$$\int^\infty_a f(x)\sin\omega x \,dx =\sin(\omega a)\sum^\infty_{k=1}a_k A_k\,\omega^{k-1} +\cos(\omega a)\sum^\infty_{k=1}a_k B_k\,\omega^{k-1}$$

For a full expansion, you may expand the sine and cosine into Maclaurin series, at the cost of losing neatness.