= Solution
A photon follows a <null geodesic>, so the parameter called proper time in the question must instead be an <affine parameter>: proper time is identically zero on a null worldline. Denote this parameter by $\tau$ for <continuity> with the displayed formulas.
The <geodesic> <Lagrangian> in the equatorial plane is $\tfrac12(-F\dot t^2+F^{-1}\dot r^2+r^2\dot\phi^2)$. The cyclic coordinates give conserved quantities $E=F\dot t$ and $h=r^2\dot\phi$, proportional to the energy measured at infinity and angular momentum. The null constraint then gives
$$
\boxed{\dot r^2=E^2-\frac{h^2F}{r^2}.}
$$
The overall scale depends on the <affine parameter>, while $b=h/E$ is the invariant impact parameter. For $u=1/r$, we have $\dot r=-h\,du/d\phi$, giving
$$
\boxed{(u')^2=b^{-2}-u^2+r_su^3.}
$$
At zeroth order choose the direction of the orbit so that $u_0=\sin\phi/b$. Then $r\sin\phi=b$, a straight line at distance $b$ from the origin. Differentiating the orbit equation gives $u''+u=\tfrac32r_su^2$. To first order,
$$
u=\frac{\sin\phi}{b}
+\frac{r_s}{4b^2}(3+\cos2\phi)
$$
solves this equation; the homogeneous first-order correction can be absorbed into the choice of asymptotic direction and impact parameter. It also satisfies the first integral to this order: the terms $2u_0'u_1'+2u_0u_1-u_0^3$ cancel.
Near the incoming asymptote its zero is at $\phi=-r_s/b+O(r_s^2/b^2)$, and near the outgoing asymptote at $\phi=\pi+r_s/b+O(r_s^2/b^2)$. Thus the excess angular change over a straight trajectory is
$$
\boxed{\Delta\phi=\frac{2r_s}{b}=\frac{4GM}{c^2b}}
$$
to first order. Solar light-deflection measurements and <gravitational lensing> provide observational evidence for this effect. The perturbation requires $r_s/b\ll1$.
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