= Solution
Substitution of $e^{i(kx-\omega t)}$ in the <Klein-Gordon equation> gives $\boxed{\omega^2=1+k^2}$. On the positive-frequency branch, the <phase velocity> is $\boxed{c_p=\sqrt{1+k^2}/k}$ for $k\ne0$, and the <group velocity> is $\boxed{c_g=k/\sqrt{1+k^2}}$. The group speed is less than one, although the phase speed exceeds one in magnitude.
With <Fourier transform> convention $\widehat f(k)=\int f(x)e^{-ikx}\,dx$, the initial velocity has transform $2/(1+k^2)$. Each Fourier mode solves a harmonic-oscillator equation, giving
$$
\boxed{\phi(x,t)=\frac1\pi\int_{-\infty}^{\infty}\frac{e^{ikx}\sin(t\sqrt{1+k^2})}{(1+k^2)^{3/2}}\,dk.}
$$
For $x=Vt$, symmetry makes this the imaginary part of $\pi^{-1}\int(1+k^2)^{-3/2}e^{it(\sqrt{1+k^2}-Vk)}\,dk$. When $0<V<1$, the phase has a unique stationary point $k_*=V/\sqrt{1-V^2}$, with phase value $\sqrt{1-V^2}$ and positive second derivative $(1-V^2)^{3/2}$. The <stationary phase> method gives
$$
\boxed{\phi(Vt,t)=\sqrt{\frac2{\pi t}}(1-V^2)^{3/4}\sin\left(t\sqrt{1-V^2}+\frac\pi4\right)+O(t^{-3/2}).}
$$
For $V>1$ there is no real stationary point. More precisely, <finite propagation speed> makes the solution at $x>t$ depend only on the initial data on $[x-t,x+t]\subset(0,\infty)$. On that half-line the initial velocity is $e^{-x}$, and $te^{-x}$ exactly solves the equation with zero initial displacement. Uniqueness on the domain of dependence therefore gives
$$
\boxed{\phi(Vt,t)=te^{-Vt}\quad(V>1).}
$$
This <exponential initial-velocity tail for the Klein-Gordon equation> is exponentially small, rather than identically zero: the initial data already have a nonzero tail outside the origin.
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