<s>[INST] Evaluating an integral Gaussian-profile initial condition has the solution,
$$\phi (r,t)=\frac{R^{3}}{2}\frac{A}{\sqrt{\pi }}\int_{0}^{\infty
}ke^{-R^{2}k^{2}/4}\frac{\sin (kr)}{r}\cos (\omega t)\ dk,$$
where A is an arbitrary initial amplitude and the dispersion relation is $\omega= (k^2 + \omega^2_{mass})^\frac{1}{2}$, where the mass frequency $ \omega_{mass}= \sqrt{2}$.
Details : see the article equation 4.
Now how do we get the equation:
$$\phi (0,t)=\frac{A_{0}}{\left( 1+\frac{2t^{2}}{R^{4}}\right) ^{\frac{3}{4}}}\cos \left( \sqrt{2}t+\frac{3}{2}\tan ^{-1}\left[ \frac{\sqrt{2}t}{R^{2}}\right] \right) .$$
 [/INST] They take the limit $r\to0$ to obtain
$$
\phi (0,t)=\frac{R^{3}}{2}\frac{A}{\sqrt{\pi }}\int_{0}^{\infty
}k^2e^{-R^{2}k^{2}/4}\cos (\sqrt{k^2+2} t)\ dk.
$$
Then, since the exponent $e^{-R^{2}k^{2}/4}$ goes very quickly to zero as $k$ goes to infinity, the authors say that  the integral is dominated by small values of $k$, $k\sim 2R^{-1}$. For whose values the root  in $\cos (\sqrt{k^2+2} t)$ is approximated as $\sqrt{k^2+2}=\sqrt2\sqrt{k^2/2+1}\approx \sqrt2(1+k^2/4)\ $.  Now if to add an imaginary part,
$$
\cos(\sqrt2(t+tk^2/4))+i\sin (\sqrt2(t+tk^2/4))= e^{i\sqrt2(t+tk^2/4)},
$$
the integral turns into
$$
\frac{R^3 e^{i \sqrt{2} t} A}{2\sqrt{\pi }} \int_0^{\infty } k^2
   e^{-\frac{1}{4} k^2 R^2} e^{\frac{1}{4} i \sqrt{2} k^2
   t} \, dk=
\frac{R^3 e^{i \sqrt{2} t} A}{2\sqrt{\pi }} \int_0^{\infty } k^2
   e^{-\frac{1}{4}(  R^2-i \sqrt{2}  t)k^2} \, dk=
$$
$$
\frac{AR^3 e^{i \sqrt{2} t}}{\left(R^2-i \sqrt{2}
   t\right){}^{3/2}}.
$$
Taking the real part gives the answer:
$$
\phi (0,t)\approx\frac{A}{\left( 1+\frac{2t^{2}}{R^{4}}\right) ^{\frac{3}{4}}}\cos \left( \sqrt{2}t+\frac{3}{2}\tan ^{-1}\left[ \frac{\sqrt{2}t}{R^{2}}\right] \right).
$$
</s>