feynman-diagramsquantum-field-theoryrenormalizationsingularities
In section 10.5 of Peskin and Schroeder, Peskin and Schroeder says the value of the last diagram in (10.51) is just a constant, which we can freely adjust to absorb any divergent terms that are independent of the external momenta.
My question is: Doesn't this mean that we still have an infinity in our scattering cross section? Doesn't it need to be cancelled by something else?
The formula given by Srednicki $$d \sigma_{\textrm{CM}} = \frac{1}{64 \pi^2 s } \frac{|\bf{p}'|}{|\bf{p}|}|\mathcal{M}|^2 d \Omega_{\textrm{CM}}$$ is the general result for the CM $2 \to 2$ scattering where we have
$\bf{p}_1 = - \bf{p}_2 = \bf{p}$, $\bf{p}_3 = - \bf{p}_4 = \bf{p}'$, $(E_1 + E_2) = (E_3 + E_4) =\sqrt{s} $ and
$$ |{\bf{p}}| = \tfrac{1}{2\sqrt{s}}\sqrt{\lambda(s,m_1^2,m_2^2)}, \quad \quad |{\bf{p}'}| = \tfrac{1}{2\sqrt{s}}\sqrt{\lambda(s^2,m_3^2,m_4^2)} $$ with $\lambda(x,y,z) = (x-y-z)^2 - 4y z$ being the Källen function. Furthermore, we have $$E_{1/2} = \frac{s \pm (m_1^2 - m_2^2)}{2\sqrt{s}} \quad \quad E_{3/4} = \frac{s \pm (m_3^2 - m_4^2)}{2\sqrt{s}} $$
As you have correctly observed, for $m_1=m_3$ and $m_2=m_4$ we indeed obtain $\frac{|\bf{p}'|}{|\bf{p}|} =1 $ and the formula reduces to
$$d \sigma_{\textrm{CM}} = \frac{1}{64 \pi^2 s } |\mathcal{M}|^2 d \Omega_{\textrm{CM}} $$
On page 164 Peskin is referring to Eq. 4.84 which reads $$ \frac{d \sigma_{\textrm{CM}} }{d \Omega_{\textrm{CM}}} = \frac{1}{2 E_1 2 E_2 v_{12}} \frac{|\bf{p}'|}{(2\pi)^2 4 \sqrt{s}}, $$ with $v_{12}$ being the absolute value of the relative velocity of incoming particles. This expression is actually the same as the Sredinicki formula, just written in a slightly different way. To see this, observe that in any frame $v_{12}$ can be rewritten as $$ v_{12} =|{\bf{v}}_1 - {\bf{v}}_2| = \left | \frac{{\bf{p}_1}}{E_1} - \frac{{\bf{p}_2}}{E_2} \right | = \frac{\sqrt{(E_1 {\bf{p}_2} - E_2 {\bf{p}_1})^2}}{E_1 E_2} $$ and when we go to the CM frame ${\bf{p}} \equiv \bf{p_1} = - \bf{p}_2$ we end up with $$ v_{12} = \frac{|{\bf{p}}|(E_1 + E_2)}{E_1 E_2} = \frac{|{\bf{p}}|\sqrt{s}}{E_1 E_2} $$ Plugging this into Peskin's formula you arrive to $$ \frac{d \sigma_{\textrm{CM}} }{d \Omega_{\textrm{CM}}} = \frac{1}{64 \pi^2 s} \frac{|\bf{p}'|}{|\bf{p}|} |\mathcal{M}|^2 $$ in perfect agreement with Srednicki.
So yes, it is actually equal.
By the way, notice that in any frame where the particle velocities are parallel or antiparallel, $v_{12}$ can be written in a covariant form since $$ (p_1 \cdot p_2)^2 - m_1^2 m_2^2 = E_1^2 E_2^2 {(1 - \bf{v}_1 \cdot \bf{v}_2)^2} - E_1^2 E_2^2 {(1- {\bf{v}}_1^2) (1- {\bf{v}}_2^2)} = E_1^2 E_2^2 ({\bf{v}_1} - {\bf{v}}_2)^2 $$ where we used $({\bf{v}_1} \cdot {\bf{v}_2})^2 = {\bf{v}_1^2} {\bf{v}_2^2}$, so that $$ v_{12} = \frac{\sqrt{ (p_1 \cdot p_2)^2 - m_1^2 m_2^2}}{E_1 E_2} $$
The main problem lies in the "large logarithms". Indeed, suppose you want to calculate some quantity in Quantum Field Theory, for instance a Green Function. In perturbation theory this is something like:
$$\tilde{G}(p_1,...,p_n)=\sum_k g^k F_k(p_1,...,p_n)$$
for some generic functions $F$ and $g$ is the coupling constant. It's not enough to require a small $g$. You need small $g$ AND small $F$, for every value of the momenta $p$ (so for every value of the energy scale of your system).
A nice little calculation to understand this point. It's obvious that:
$$\int_0^\infty \frac{dx}{x+a}=[log(x+a)]_0^\infty=\infty$$
Let's use a cutoff: $$\int_0^\Lambda \frac{dx}{x+a}=log\frac{(\Lambda+a)}{a}$$
This is still infinite if the (unphysical) cutoff is removed. The whole point of renormalization is to show that a finite limit exist (this is "Fourier-dual" to send the discretization interval of the theory to zero). This quantity is finite:
$$\int_0^\Lambda \frac{dx}{x+a}-\int_0^\Lambda \frac{dx}{x+b} \rightarrow log\frac{b}{a}$$
But if $a \rightarrow \infty $ the infinite strikes back! So for a generic quantity F(p) regularized to F(p)-F(0) we want at least two things: that the coupling is small at that momentum $p$ and that $p$ is not far away from zero. But zero is arbitrary, we can choose an arbitrary (subtraction) scale. So we can vary this arbitrary scale $\mu$ in such a way that it is always near the energy scale we are probing.
Is convenient to take this scale $\mu$ at the same value of the renormalization scale. This is the energy at which you take some finiteness conditions (usually two conditions on the two point Green function and one condition on the 4 point one). The finiteness conditions are real physical measures at an arbitrary energy scale, so they fix the universe in which you live. If you change $\mu$ and you don't change mass, charge, ecc. you are changing universe. The meaning of renormalization group equations is to span the different subtraction points of the theory, remaining in your universe. And of course every physical quantity is independent of these arbitrary scale.
EDIT: Some extra motivations for the running couplings and renormalization group equations, directly for Schwartz:
The continuum RG is an extremely practical tool for getting partial results for high- order loops from low-order loops. [...]
Recall [...] that the difference between the momentum-space Coulomb potential V (t) at two scales, t1 and t2 , was proportional to [...] ln t1 for t1 ≪ t2. The RG is able to reproduce this logarithm, and similar logarithms of physical quantities. Moreover, the solution to the RG equation is equivalent to summing series of logarithms to all orders in perturbation theory. With these all-orders results, qualitatively important aspects of field theory can be understood quantitatively. Two of the most important examples are the asymptotic behavior of gauge theories, and critical exponents near second-order phase transitions.
[...] $$e^2_{eff}(p^2)=\frac{e^2_R}{1-\frac{e^2_R}{12 \pi^2}ln\frac{p^2}{\mu^2}}$$ $$e_R=e_{eff}(\mu)$$  This is the effective coupling including the 1-loop 1PI graphs, This is called leading- logarithmic resummation. Once all of these 1PI 1-loop contributions are included, the next terms we are missing should be subleading in some expansion. [...] However, it is not obvious at this point that there cannot be a contribution of the form $ln^2\frac{p^2}{\mu^2}$ from a 2-loop 1PI graph. To check, we would need to perform the full zero order calculation, including graphs with loops and counterterms. As you might imagine, trying to resum large logarithms beyond the leading- logarithmic level diagrammatically is extremely impractical. The RG provides a shortcut to systematic resummation beyond the leading-logarithmic level.
Another example: In supersymmetry you usually have nice (theoretically predicted) renormalization conditions at very high energy for your couplings (this is because you expect some ordering principle from the underlying fundamental theory, string theory for instance). To get predictions for the couplings you must RG evolve all the couplings down to electroweak scale or scales where human perform experiments. Using RG equations ensures that the loop expansions for calculations of observables will not suffer from very large logarithms.
A suggested reference: Schwartz, Quantum Field Theory and the Standard model. See for instance pag. 422 and pag.313.
Best Answer
For a renormalizable theory all momentum-dependent divergences in diagrams one to three in the diagram are cancelled by diagrams four and five. Had this not been the case the theory would not be renormalizable and we would have to introduce counter-terms with additional derivatives.