Srednicki Ch. 11 (p.84) provides an argument for introducing by hand a symmetry factor $S$ in the final integral for the total cross-section.
$$ \sigma = \frac1{S} \int d\sigma. \tag{11.36} $$
The argument claims that the final state is an unordered list of outgoing momenta and since the integral treats it otherwise, we should account for the miscalculation by introducing the symmetry factor by hand.
I don't get it. Could you please illustrate what he means?
Besides, from kindergarten mathematics, I know that $\int df = f$ for any integrable function $f$. Therefore, the argument sounds apologetic and wrong to me.
So, why is the symmetry factor really introduced?
Example:
Peskin & Schroeder Exercise 4.2 : Decay of a scalar particle
$$ \mathcal L = \frac12(\partial_\mu \Phi)^2 - \frac12 M^2\Phi^2 + \frac12(\partial_\mu \phi)^2 - \frac12 m^2\phi^2 - \mu \Phi \phi^2$$
The decay $\Phi \to \phi\phi$ is characterized by the amplitude $\mathcal M = -\color{red}2\mu$. Note the symmetry factor of $\color{red}2$ here in the amplitude already accounts for the multiple identical contractions. The differential decay rate would now be given as
$$ \frac{d\Gamma}{d\Omega} = \frac{\mu^2}{16\pi^2M}\sqrt{1-(\frac{2m}M)^2}\,.$$
Now, to get $\Gamma$ from this, why does everyone divide by a factor of $2$ after integrating over the solid angle?