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"?
Best Answer
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 $-q^2$.