Pauli Villars Regularization Consider the t-channel diagram of phi-4 one loop diagrams. Evaluated it is, with loop momenta p, 
$\frac{\lambda^2}{2}\displaystyle\int\frac{d^4p}{(2\pi)^4}\frac{1}{(p+q)^2-m^2}\frac{1}{p^2-m^2}$
If I want to regularize this using Pauli-Villars regularization, which is the correct method? The procedure is to make the replacement $\frac{1}{p^2-m^2}\rightarrow \frac{1}{p^2-m^2}-\frac{1}{p^2-\Lambda^2}$. 
My question is do I apply the reguarization to one term in the integral or both terms? 
I've seen variations where the propagators become 
$\frac{1}{p^2-m^2}\frac{1}{(p+q)^2-m^2}\rightarrow \frac{1}{p^2-m^2}\frac{1}{(p+q)^2-m^2}-\frac{1}{p^2-\Lambda^2}\frac{1}{(p+q)^2-\Lambda^2}$
and also where we have
$\frac{1}{p^2-m^2}\frac{1}{(p+q)^2-m^2}\rightarrow (\frac{1}{p^2-m^2}-\frac{1}{p^2-\Lambda^2})(\frac{1}{(p+q)^2-m^2}-\frac{1}{(p+q)^2-\Lambda^2})$
In the latter case one ends up with four terms and each term is then evaluated using a Feynman parameter and integrating over wick rotated momenta, obtaining a logarithmic expression.
I'm pretty sure I've also seen where it was only applied to one of the terms.
Which is correct? (or are they equivalent?)
 A: The first thing that you need to do is put the integral variable $p$ in a more clear way. You can doo this with Feynman Parameters.
$$
\frac{1}{AB}=\int_{0}^{1} dx \frac{1}{[A+(B-A)x]^2}
$$
I will give you an exercise: find the $A$ and $B$, as well as $\Delta$ and $C$, such that:
$$
\int\frac{d^4p}{(2\pi)^4}\frac{1}{(p+q)^2-m^2}\frac{1}{p^2-m^2}=\int_{0}^{1}dx\,C(x,q,m)\int\frac{d^4k}{(2\pi)^4}\frac{1}{[k^2-\Delta(x,q,m)]^2}
$$
You see that this is a loop of mass $\sqrt{\Delta}$. Applying the Pauli-Villars regularization to this loop integral is simple now:
$$
\frac{1}{[k^2-\Delta]^2}\rightarrow\frac{1}{[k^2-\Delta]^2}-\frac{1}{[k^2-\Lambda^2]^2}
$$
Using the Wick rotation we have:
$$
\int\frac{d^4k}{(2\pi)^4}\left[\frac{1}{(k^2-\Delta)^2}-\frac{1}{(k^2-\Lambda^2)^2}\right]=\frac{i}{8\pi^2}\int_{0}^{\infty}dk_E\left[\frac{k_E^3}{(k_E^2-\Delta)^2}-\frac{k_E^3}{(k_E^2-\Lambda^2)^2}\right]
$$
$$
=\frac{-i}{16\pi^2}\ln\left(\frac{\Delta}{\Lambda^2}\right)
$$
Then, the result is:
$$
\int\frac{d^4p}{(2\pi)^4}\frac{1}{(p+q)^2-m^2}\frac{1}{p^2-m^2}=\int_{0}^{1}dx\,\frac{-iC(x,q,m)}{16\pi^2}\ln\left(\frac{\Delta(x,q,m)}{\Lambda^2}\right)
$$
A: Actually, in Pauli - Villars regularization, a divergence arising from a loop integral is modulated by a spectrum of auxiliary particles added to the propagator. In a loop, we have two propagators, so we need to modulate two of them.
