# Basic cut-off regularization

I've been reading these notes on regularization by Hitoshi Murayama here, and on page 3 there's a few lines of calculations on a quick method of regularizing an integral. But I can't follow the steps - where does the $z$ come from? Why is a second integration variable suddenly introduced, in the step before the original integration variable disappears?

I'd really appreciate any help, I've looked absolutely everywhere to find another example like this, but in general other methods like Paul-Villars renormalization seem to be used.

So it says the propagator is multiplied by a factor of $\frac{\Lambda^2}{\Lambda^2+p^2}$ so that it becomes (I think the initial integration limits should be infinity and zero for the first line, but it doesn't say):

$$\int \frac{d^2 p}{(2\pi)^2} \frac{\Lambda^2}{(p^2+\Lambda^2)(p^2+m^2)}$$ $$=\int^1_0 dz \frac{d^2 p}{(2\pi)^2}\frac{\Lambda^2}{(p^2+z\Lambda^2+(1-z)m^2)^2}$$ $$=\int^1_0dz \frac{1}{4\pi}\frac{\Lambda^2}{z\Lambda^2+(1-z)m^2}$$ $$=\frac{1}{4\pi}\frac{\Lambda^2}{\Lambda^2-m^2}\ln\frac{\Lambda^2}{m^2}$$

• Minor comment to the post (v4): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. – Qmechanic Mar 11 '18 at 11:01
• Besides Feynman parameters, you should learn to read off the coefficients of log-divergences by inspection. It's easy as pi, and it gets you useful things like the leading term of the beta function (running coupling roll-off). – Bert Barrois Mar 11 '18 at 12:04

$$\frac1{AB} = \int_o^1dz\ \frac1{(Az+(1-z)B)^2}$$
Substitute $A= (p^2 + \Lambda^2)$ and $B= p^2+m^2$ and see what follows.