Calculating the beta function through wavefunction renormalization

Question

The ways I'm used to calculating $\beta$ functions is by either the Callan-Symanzik equation or by taking the total derivative with respect to the renormalization scale of a coupling and simplifying.

In these notes by Manohar (pg 20), he calculates the QED $\beta$ function through a different approach which doesn't use the three point function at all. He considers the wave function renormalization of the photon through and gets a renormalized two point function, 

$$\begin{equation}  - i \frac{ e ^2 }{ 2 \pi ^2 } ( p _\mu p _\nu - p ^2 g _{ \mu \nu } ) \left[ \int _0 ^1 d x x ( 1 - x ) \log \frac{ m ^2 - p ^2 x ( 1 - x ) }{ m ^2 + M ^2 x ( 1 - x ) } \right]  \end{equation}$$  
He then says that the $\beta$ function is given by acting on the coefficient of $i ( p _\mu p _\nu - p ^2 g _{ \mu \nu } )$. He of course gets the right anwer. But why does this work?

Answer 1

Preceding answer suppressed because wrong. New version coming soon.

EDIT. New answer.

Due to quantum corrections, the propagator of the photon is some multiple $(1+F(M))$ of the freestandard propagator. At the one loop order, we have $F(M) = e(M)^{2}G(M)$ for some function $G$. The free standard propagator of the photon  in $1/p^{2}$ gives rise to a static potential between two electrons of charges $e$ separated by a large distance $r$ in $e^{2}/r$, this is the standard Coulomb law. Given the quantum corrections to the photon propagator, this static potential at long distances is modified in $e(M)^{2}(1+F(M))/r$. This shows that the function $e(M)^{2}(1+F(M))$ is a physical quantity which can be experimentally measured. In particular, it can not depend of the renormalization scale $M$ i.e. $\frac{d}{dM}(e(M)^{2}(1+F(M)))=0$.

At the one loop order, this implies

$\beta(e(M)) = M \frac{d}{dM} e(M) = - \frac{e(M)}{2} M \frac{d}{dM}F(M)$.

Comments

You're answer seemed correct to me. What was wrong with it?