= Solution
Define
$$
H(x)=\frac{e^{-x}}{\sqrt x}\qquad(x>0).
$$
This is a <Lebesgue integrable function>: on $(0,1]$ it is bounded by $x^{-1/2}$, whose <integral> is two, while on $[1,\infty)$ it is bounded by $e^{-x}$, whose <integral> is $e^{-1}$. Since $|\sin(xy)|\leq x|y|$,
$$
|h_y(x)|\leq|y|H(x).
$$
Thus \b[$h_y$ is <Lebesgue integrable> for every $y\geq0$], including $h_0=0$; the same argument actually works for all real $y$.
The <mean value theorem> for the <sine> gives $|\sin u-\sin v|\leq|u-v|$. Hence
$$
|h_y(x)-h_z(x)|\leq|y-z|H(x)
$$
and the <triangle inequality for an integral> gives
$$
\boxed{|F(y)-F(z)|\leq\left(\int_0^\infty H(x)\,dx\right)|y-z|.}
$$
Thus $F$ is a <globally Lipschitz function>, in particular \b[continuous on $[0,\infty)$], including the endpoint zero.
For the <derivative>, fix $y>0$ and take nonzero $t$ sufficiently small that $y+t>0$. The difference quotient has the <integrand>
$$
D_t(x)=\frac{e^{-x}}{x\sqrt x}\frac{\sin(x(y+t))-\sin(xy)}{t}.
$$
For every $x>0$, $D_t(x)\to H(x)\cos(xy)$ as $t\to0$. The same <mean value theorem> gives the uniform-in-$t$ bound $|D_t(x)|\leq H(x)$.
The <dominated convergence theorem> states that if <measurable functions> $u_n$ converge <almost everywhere> to $u$ and $|u_n|\leq G$ for a single <Lebesgue integrable function> $G$, then $u$ is <Lebesgue integrable> and $\int|u_n-u|\to0$, hence $\int u_n\to\int u$. Apply it to the difference quotients along every sequence $t_n\to0$, using the common dominator $H$. The resulting <limit> is independent of the sequence, so
$$
\boxed{F'(y)=\int_0^\infty e^{-x}\frac{\cos(xy)}{\sqrt x}\,dx\qquad(y>0).}
$$
The right-hand <integral> is absolutely convergent because its <integrand> has absolute value at most $H$. This is <dominated differentiation under the integral sign>, justified by an explicit integrable bound rather than a formal interchange.
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