Suppose you have a function $f$ in $L^1$. Now let $$ F_s(x) = \int_0^x f(t)\sin( s \cdot t)\,\mathrm dt $$ According to the Riemann-Lebesgue lemma we can tell that $$ \lim_{s\to \infty}F_s(x) = 0 $$ (pointwise).
Can I use that fact in order to tell that for each fixed $x$ there is an infinite set $S_x$ of values of $s(x)$ such that $F_{s(x)}(x)=0$? And is there a (more or less) explicit way to exhibit that set $S_x$?
EDIT: Fixed notation
It seems that your conjecture is false. If $f$ is a positive, decreasing function then $F_s(x)>0$ for all $x>0$: \begin{align*} F_s(x) &= \int_0^x f(t) \sin(st) \mathrm dt = \int_{t=0}^x \left( f(x)-\int_{u=t}^x \mathrm df(u) \right) \sin(st) \mathrm dt \\ &= f(x)\int_{t=0}^x \sin(st) \mathrm dt + \int_{u=0}^x \left( \int_{t=0}^u \sin(st) \mathrm dt \right) \big(-\mathrm df(u)\big) \\ &= f(x)\frac{1-\cos(sx)}{s} + \int_{u=0}^x \left( \frac{1-\cos(su)}{s}\right) \big(-\mathrm df(u)\big) >0. \end{align*}
A concrete example: if $f(t)=e^{-t}$ then $$ F_s(x) = \left[ \frac{-e^{-t}\big(s\cos(sx)+\sin(sx)\big)}{1+s^2} \right]_0^x = \frac{1-e^{-s}\big(s\cos(sx)+\sin(sx)\big)}{1+s^2}>0. $$