Renormalization with hard cutoff of a loop diagram with single vertex

[theittsco]

TL;DR

Confusion on how to renormalize a loop diagram with only a single vertex. Feynman parameters don't work.

Trying to solve for the loop contribution when renormalizing a one loop $\frac{\lambda}{4!}\phi^4$ diagram with two external lines in $d=4$ dimensions. After writing down the Feynman rule I see that:

$$\frac{(-i\lambda)}{2}\int d^4q \frac{i}{q^2-m_{\phi}^2+i\epsilon} $$

But I see no way to convert this to Feynman parameters like in $1/AB$. When I Wick rotate I get:

$$\frac{(\lambda)}{2}\int dq_E \frac{q_E^3}{q_E^2-m_{\phi}^2} $$

Which when integrated from $0,\Lambda$ (a hard cut off) via the great Mathematica yields:

$$\lambda/2 (1/2)(\Lambda^2-m^2\log(-m^2)+m^2\log(\Lambda^2-m^2))$$

My renormalization conditions are $\Pi(m^2)=\Pi'(m^2)=0$, so when solving for counter terms:

$$i\Pi(p^2) = \lambda/2 (1/2)(\Lambda^2-m^2\log(-m^2)+m^2\log(\Lambda^2-m^2)) + i(\delta_Zp^2-\delta_m)$$

my counter terms still end up with $\Lambda$ dependence. So I messed up somewhere but I can't figure out where. Help please!

 

[HomogenousCow]

Your counter terms should absolutely depend on the cut-off, that's how they cancel the divergences. There's no momentum dependence so $\delta_Z = 0$ and $\delta_m$ cancels out everything. Also I'm pretty sure you made a mistake when you did your wick-rotation, the sign in the denominator should be positive.

The whole idea behind renormalized perturbation theory is that you deform your theory with some parameter $\Lambda$ that makes the observables finite, then you ascribe $\Lambda$-dependence to your constants which are selected to cancel the infinites and satisfy the renormalization conditions.