How do I do this loop integral? I'm having some trouble doing a loop integral. It is the loop integral that gives neutrinos their mass in the famous Zee model (http://dx.doi.org/10.1016/0370-2693(80)90349-4).
The integral is
$$
\int \frac{d^4 k}{(2\pi)^4} \frac{1}{k^2-m_h^2} \frac{1}{k^2-m_\phi^2} \frac{1}{k^2}.
$$
The integral is convergent so there is no need for any regularisation scheme. I have tried to do the integral using the Feynman parameter:
$$
\frac{1}{ABC} = \int_0^1 dx \ dy \ dz \ \delta(x+y+z-1)\frac{2}{[xA+yB+zC]^3}
$$
and then the master equation for loop diagrams:
$$
  \int \dfrac{d^4 k}{(2\pi)^4} \dfrac{1}{\left( k^2 - \Delta + i\epsilon  \right)^n} = \dfrac{(-1)^n i}{16\pi^2} \dfrac{\Gamma(n-2)}{\Gamma(n)}\left(\dfrac{1}{\Delta}\right)^{n-2}.
$$
Unfortunately none of the answers I'm getting look like those given in the paper. The integral should come out to something like:
$$
m_h^2 \log(m_h^2/m_\phi^2)
$$
for $m_h > m_\phi$.
Any help is very much appreciated.
 A: I think i got it: You have $$I=\int \frac{d^4k}{(2\pi)^4}\frac{1}{k^2-m^2_h}\frac{1}{k^2-m^2_\phi}\frac{1}{k^2}$$ in order to calculate this integral in an easy way we make a wick rotation to put ourselves in 4-D euclidean space i.e. $k_0\rightarrow ik_0$ this means that $d^4k=i d^4k$ and $k^2=k_0^2-\vec{k}^2=-k_0^2-\vec{k}^2$ so $k_0^2+\vec{k}^2=-k^2$ if you substitute in the integral you will have: $$-i\int \frac{d^4k}{(2\pi)^4}\frac{1}{k^2+m^2_h}\frac{1}{k^2+m^2_\phi}\frac{1}{k^2}$$ ok now we use 4-D spherical coordinates$d^4k=d\Omega k^3dk$:  $$-\frac{2\pi^2}{\Gamma(2)}i\int_0^\Lambda\frac{dk}{(2\pi)^4}\frac{k}{k^2+m^2_h}\frac{1}{k^2+m^2_\phi}$$ now let's split the integral: $$\frac{k}{k^2+m^2_h}\frac{1}{k^2+m^2_\phi}=\frac{A}{k^2+m^2_h}+\frac{B}{k^2+m^2_\phi}$$ you will find that $$A=\frac{1}{m^2_\phi-m^2_h}\\B=-A=\frac{1}{m^2_h-m^2_\phi}$$ so you have $$-\frac{2A\pi^2}{\Gamma(2)(2\pi)^4}i\int_0^\Lambda dk k [\frac{1}{k^2+m^2_h}-\frac{1}{k^2+m^2_\phi}]$$ which is now trivial: $$\int_0^\Lambda dk \frac{k}{k^2+m^2_i} =\frac{1}{2}log\left(\frac{\Lambda}{m_i^2}\right)$$ so $$I=-\frac{2A\pi^2}{\Gamma(2)(2\pi)^4}i\cdot \frac{1}{2}log\left(\frac{m^2_\phi}{m^2_h}\right)=\frac{i}{16\pi^2}\frac{1}{m^2_\phi-m^2_h}log\left(\frac{m^2_h}{m^2_\phi}\right)$$
i hope that helped!
EDIT:
Well the simplest justification that comes to my mind is that $k_0=ik_E$ is invertible and as regular as a linear transformation can be so every operation that was allowed before is allowed after the trnasformation. Ok so consider these two facts: the pole is not in the origin as far as $k_0$ is concerned but it's in $k_0^2=\vec{k}^2$ whichy gets mapped to the immaginary axis that means you have no ambiguity in the choice of an integration path (for $k_0$) since the poles don't overlap with the real axis after the rotation, second: if you choose to integrate in spherical coordinates then you have to take into account the presence of measure which gives a $k^3$ contribution that removes the singularity in $k_0=\vec{k}^2$. 
A more rigorous approach would require you to do the Wick rotation at the path integral level, on $Z[J/..]$ which is a prefectly legitimate operation on the action integral $x_0=i x_E$ so when you derive the green functions from the generating functional they already are Wick rotated and so are their poles without any ambiguity. 
I hope that clears your doubt a bit let me know if it doesn't unfortunatley these days i have very few time spare.
