Yes, it's completely right what TrickyDicky wrote. Take a good standard textbook of quantum theory, where scattering is described correctly. Then you'll see that you need wave packets to define the cross section. In the generalized (!) momentum eigenstates, aka plane waves, you get
$$S_{fi}=\delta_{fi} - (2 \pi)^4 \mathrm{i} T_{fi} \delta^{(4)}(p_f-p_i).$$
The transition probability is the modulus squared of the amplitude, which naively is ##S_{fi}##. Of course, this doesn't make sense, but you first have to fold it with the incoming wave packets take the square and then take the limit to vanishing momentum spread. The final result is that there's only one energy-momentum conserving ##\delta^{(4)}## distribution in the transition probability density, which can then be meaningfully integrated out.
Admittedly, here the whole issue is complicated also by the problem of how to define asymptotic free states, where you have to use an appropriate switching procedure. If I remember right, a good source for the discussion of this is Messiah's classical quantum mechanics text on this issue. For the relativistic case, it's nicely dealt with in Peskin&Schroeder.
Of course, there are shortcuts in the literature like in Landau/Lifshitz vol. 4, using a "box regularization", where the momenta become descrete, and there's no problem in squaring the S-matrix elements to get transition probabability rates. After deviding over the finite four-volume and then taking the limit of the four-volume to ##\infty## you get the same result as with the proper procedure with the wave packets. However, the latter is much more physical, although a bit more complicated mathematics wise. When discussing the physics, one should use the wave-packet approach.