= Solution
For $x>0$, the upper and lower boundary values on the <branch cut> are
$$
F(-x+i0)=-\frac{e^{-x-i\sqrt x}}x,\qquad F(-x-i0)=-\frac{e^{-x+i\sqrt x}}x.
$$
Write $x=A_R$ for the outer endpoint distance on the cut. The upper path runs from $x=A_R$ to $x=\epsilon$, while the lower path runs back from $\epsilon$ to $A_R$; in both cases $dz=-dx$. Therefore
$$
\int_{C_3}F\,dz=-\int_\epsilon^{A_R}\frac{e^{-x-i\sqrt x}}x\,dx,\qquad
\int_{C_5}F\,dz=\int_\epsilon^{A_R}\frac{e^{-x+i\sqrt x}}x\,dx,
$$
and
$$
\boxed{\int_{C_3}F\,dz+\int_{C_5}F\,dz=2i\int_\epsilon^{A_R}e^{-x}\frac{\sin\sqrt x}{x}\,dx.}
$$
The source's limits zero and infinity require $\epsilon\to0$ and $R\to\infty$; they are not an exact identity on a finite <contour>. For a concrete counterexample take $\epsilon=1/4$ and $A_R=1$: the finite real integral is at most $\int_{1/4}^1x^{-1/2}\,dx=1$, whereas the full-range integral is $\pi\operatorname{erf}(1/2)\simeq1.6352$, as evaluated in the next part. Near zero the combined integrand is $O(x^{-1/2})$, so the combined <improper integral> converges, although the individual cut integrals have logarithmic divergences. At infinity the factor $e^{-x}$ ensures convergence. Consequently
$$
\boxed{\lim_{R\to\infty,\,\epsilon\to0}\left(\int_{C_3}F\,dz+\int_{C_5}F\,dz\right)=2i\int_0^\infty e^{-x}\frac{\sin\sqrt x}{x}\,dx.}
$$
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