= Solution
Let $t$ label the reception event in the source <reference frame>. Its distance from the source and the relevant <retarded time> are
$$
r(t)=\sqrt{L^2+v^2t^2},\qquad t_e=t-\frac{r(t)}c.
$$
The received phase is $2\pi f_0t_e$. Because the observer's <proper time> obeys $d\tau=dt/\gamma$, its measured <frequency> is
$$
\boxed{f_{\rm in}(t)=f_0\frac{dt_e}{d\tau}
=\gamma f_0\left(1-\frac{v^2t}{c\sqrt{L^2+v^2t^2}}\right).}
$$
Equivalently, write $\beta=v/c$ and $s(t)=vt/r(t)$. The incoming <Doppler factor> is $D_{\rm in}=\gamma(1-\beta s)$, obtained also by contracting the <photon> <four-momentum> with the observer's <four-velocity>. The <frequency> decreases monotonically, with
$$
\boxed{f_{\rm in}(-\infty)=\gamma(1+\beta)f_0,\quad
f_{\rm in}(0)=\gamma f_0,\quad
f_{\rm in}(+\infty)=\gamma(1-\beta)f_0.}
$$
In particular, the instant of closest approach is not the zero-shift point: a moving receiver looking perpendicular to its motion sees a factor $\gamma$, not $1/\gamma$. The latter factor applies to a different moving-emitter geometry. The graph uses source-frame reception time, with observer <proper time> $\tau=t/\gamma$.
\Image[/past-exam-of-the-symmetry-and-relativity-course-of-the-university-of-oxford/2019-passing-observer-doppler.png]
{title=Frequency measured by an observer passing a stationary monochromatic source}
{description=Each curve gives the received frequency divided by the emitted frequency against source-frame time ct/L. The vertical line marks closest approach; the black points have height gamma. Approaching and receding asymptotes are labelled for the fastest observer.}
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