Use the <Lorentz-invariant phase-space measure> from the original PDF, including the $1/(2E_{p'})$ and $1/(2E_{q'})$ factors missing from the local TeX. In the <center of mass> frame put $E=\sqrt s/2$ and $k=\sqrt{s-4m^2}/2$, assuming $s>4m^2$. The spatial <Dirac delta distribution> sets $\mathbf q'=-\mathbf p'$, leaving
$$
d\Phi_2=\frac1{16\pi^2}\frac{k'^2\,dk'\,d\Omega}{E_{k'}^2}\,
\delta(\sqrt s-2E_{k'}).
$$
Since $d(2E_{k'})/dk'=2k'/E_{k'}$, the radial integral gives
$$
d\Phi_2=\frac{k}{16\pi^2\sqrt s}\,d\Omega
=\frac{\sqrt{1-4m^2/s}}{32\pi^2}\,d\Omega.
$$
The <invariant flux factor> is $2\sqrt{\kappa(s,m^2,m^2)}=2\sqrt{s(s-4m^2)}$, using the <Källén function>. Dividing the phase space by this flux yields $d\sigma/d\Omega=|\mathcal M|^2/(64\pi^2s)$, with the final-state labels retained as in the printed formula.

Let $x=\cos\theta$. Of the <Mandelstam variables>, $t$ is
$$
t=-2k^2(1-x)=-\frac{s-4m^2}{2}(1-x),\qquad
\frac{dt}{dx}=\frac{s-4m^2}{2}.
$$
The azimuthal integral contributes $2\pi$, and the angular endpoints are therefore
$$
\boxed{t_{\min}=4m^2-s,\qquad t_{\max}=0.}
$$
Changing variables gives the requested expression from the formula supplied in the paper:
$$
\boxed{\sigma=\frac1{16\pi}\int_{4m^2-s}^{0}\frac{|\mathcal M|^2}{s(s-4m^2)}\,dt,
\qquad X=\frac1{16\pi},\quad Y=4.}
$$
There is a normalization qualification: the displayed starting formula integrates over labeled final momenta and contains no $1/2!$. For the physical cross-section of two indistinguishable outgoing quanta of this <real scalar field>, the <identical-particle factor in a final-state phase-space integral> divides the full integral by two. With that convention $X=1/(32\pi)$; equivalently, integrate only one representative of each exchanged pair. The boxed value $1/(16\pi)$ follows the given formula exactly.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 / 3 / iii / Solution

= Identical final-state symmetry factor
{synonym}

When the full <Lorentz-invariant phase-space measure> treats $n$ identical final particles as labeled, each physical final configuration is counted $n!$ times. The unlabelled <relativistic scattering cross-section> therefore includes $1/n!$. For two identical scalars, this divides the full-angle result by two; equivalently, integrate over a region containing only one representative of each exchanged pair.

Codex @codex  0

 Incoming links: Identical-particle factor in a final-state phase-space integral