Calculating Self-Energy Correction to Electron Propagator

[evilcman]

If one wants, to calculate the self energy correction to the electron propagator(using the approach where one introduces a photon mass $$\mu$$ to deal with IR divergences), one gets after some work an integral like this (this is from the Itzykson Zuber book equ. 7-34):

$$\int_ 0 ^ 1 d\beta \beta \left[ ln \left( \frac{\beta \Lambda^2}{(1-\beta)^2m^2} \right) - \frac{2(2-\beta)(1-\beta)}{(1-\beta)^2+\beta \frac{\mu^2}{m^2}} \right] = \frac{1}{2} ln \left( \frac{\Lambda^2}{m^2}\right) + ln \left( \frac{\mu^2}{m^2}\right) + \frac{9}{4} + O\left(\frac{\mu}{m}\right)$$

What I don't know is how this equation was obtained, that is how can I extract the divergence from the second term in the integral, which gave the $$ln \left( \mu^2 / m^2\right)$$ term. I would appreciate it if someone told me how a calculation like this goes.

 

[blechman]

The IR divergence is for $\beta\rightarrow 1$, so to compute the divergence we can set $\beta=1$ everywhere except for the $1-\beta$ terms, which are the problem when $\mu=0$. If we look at that second term, it is straightforward to show (let $u=1-\beta$) that this integral gives you a logarithm.