= Solution
The initial <Fourier transform> is $\widehat\zeta(k,0)=2b\zeta_0/(1+b^2k^2)$. Initial rest gives zero <time derivative>, so each wave mode evolves by $\cos(\sqrt{g|k|}t)$. <Fourier inversion> therefore gives
$$
\boxed{\zeta(x,t)=\frac{2b\zeta_0}{\pi}\int_0^\infty\frac{\cos(kx)\cos(\sqrt{gk}t)}{1+b^2k^2}\,dk.}
$$
On $x=Vt$, the product of cosines produces phases $\varphi_\pm(k)=Vk\pm\sqrt{gk}$ with prefactor $b\zeta_0/\pi$. Only the minus phase has a <stationary point>:
$$
k_*=\frac g{4V^2},\qquad\varphi_-(k_*)=-\frac g{4V},\qquad\varphi_-''(k_*)=\frac{2V^3}{g}>0.
$$
Expanding quadratically there and evaluating the Gaussian <oscillatory integral> gives the stationary-phase factor $\sqrt{2\pi/(t\varphi_-''(k_*))}$ and phase shift $+\pi/4$. The other phase has no <stationary point>. Using $q=\sqrt k$ makes both endpoint phases smooth with nonzero endpoint derivative and amplitude vanishing at zero, so endpoint contributions are smaller. Thus
$$
\boxed{\zeta(Vt,t)=\frac{b\zeta_0}{1+(bg/(4V^2))^2}\sqrt{\frac g{\pi V^3t}}\cos\left(\frac{gt}{4V}-\frac\pi4\right)+o(t^{-1/2}).}
$$
The selected wavenumber has <group velocity> $V$, explaining why it determines the long-time signal on this ray.
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