Vrftsjtryk

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
$$
lim_{sto infty}F_s(x) = 0
$$
(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
$$
F_s(x)
= left[ frac{-e^{-t}big(scos(sx)+sin(sx)big)}{1+s^2} right]_0^x
= frac{1-e^{-s}big(scos(sx)+sin(sx)big)}{1+s^2}>0.
$$