Let $f:(0,\infty)\to\mathbb{R}_+$ be a non-negative non-increasing function such that $\int_0^1 xf(x)\,dx<\infty$. Define $A(b)=\int_0^\infty e^{-x}\sin(bx)f(x)\,dx$. Is it possible that $\limsup_{b\to\infty} A(b)>0$ and $\liminf_{b\to\infty} A(b)=0$ simultaneously?
Note: Using periodicity of $\sin$ function, it can be shown that $A(b)\ge 0$ for any $b$.
Since $|e^{-x} f(x)| < f(a)e^{-x}$ for $x \geqslant a > 0$, it follows that $e^{-x}f(x)$ is integrable over $[a,\infty)$ and by the Riemann-Lebesgue lemma
$$\lim_{b \to \infty}\int_a^\infty e^{-x} \sin(bx)f(x) \, dx = 0$$
If the limit of the integral over $[0,\infty)$ fails to converge to $0$, then the problem arises from the behavior near $x = 0$.
Since $xf(x)$ is integrable over $[0,1]$ and $f$ is monotone we can only conclude that $f(x) = o(x^{-2})$ as $x \to 0$. It is possible that $e^{-x}f(x)$ is not integrable over $[0,1]$ and the Riemann-Lebesgue lemma cannot be invoked. That leaves the possibility that $\limsup_{b\to \infty}A(b) \neq \liminf_{b\to \infty}A(b) \neq 0$.
However, I suspect that $\limsup_{b \to \infty} A(b) > \liminf_{b \to \infty} A(b) = 0$ may not be possible.
For example with $f(x) = 1/x$ we have
$$A(b) = \int_0^\infty e^{-x} \frac{\sin(bx)}{x} \, dx = \arctan (b),$$
and $\limsup_{b \to \infty}A(b) = \liminf _{b \to \infty}A(b) = \frac{\pi}{2}$.