Pretending photon has a small mass in soft bremsstrahlung

Question

In Peskin and Schroeder chapter 6, on page 184 when discussing the infrared divergence problem in perturbative QED, the book says we can make the following equation

$$\tag{6.25} \text{Total probability}\approx\frac{\alpha}{\pi}\int_0^{|\boldsymbol{q}|}dk\frac{1}{k}I(\boldsymbol{v},\boldsymbol{v}').$$

well defined by pretending the photon has a small mass $\mu$, giving us

$$\tag{6.26} d\sigma(p\rightarrow p'+\gamma(k))=d\sigma(p\rightarrow p')\cdot\frac{\alpha}{2\pi}\log\bigg(\frac{-q^2}{\mu^2}\bigg)I(\boldsymbol{v},\boldsymbol{v}')$$

Here we are considering the bremsstrahlung of an electron with momentum $p$, resulting an electron with momentum $p'$ and a soft photon with momentum $k$, $q=p-p'$. $I(\boldsymbol{v},\boldsymbol{v}')$ is an expression independent of $k$.

How did we get equation 6.26 from 6.25? What do we mean when we say "pretending the photon has a small mass"?

Answer 1

We have: \begin{align} \int_0^{E} \frac{1}{k}dk \stackrel{(\ast)}{\leadsto} \int_0^E \frac{1}{\sqrt{k^2+\mu^2}}dk \stackrel{\mu \rightarrow 0}{\simeq} \int_0^E \frac{1}{\sqrt{k^2+\mu^2}}d\left( \sqrt{k^2+\mu^2} \right) \end{align} Which gives something like $\simeq \frac{1}{2}\ln \left( \frac{E^2}{\mu^2} \right)$. $(\ast)$ is precisely what is meant by 'considering the photon to have a small mass'. After some manipulations, you may eventually convert the energy squared $E^2$ into the Lorentz invariant $-q^2$.