The one-loop propagator correction with cutoff in <PERSON> theory
On page 409 of <PERSON> and <PERSON>.
The one-loop propagator correction with dimensional regularization in <PERSON> theory is: $$\hspace{-190pt}i\cal{M}:=$$ $$\hspace{-150pt} = \frac{-2ig^2}{(4\pi)^2}(-p^2)\left( \frac{1}{2-d/2} +\log\frac{1}{-p^2}+C \right). \tag{12.33} $$ And the renormalization condition in this section is: \begin{cases} i\mathcal{M} + i(p^2\delta_Z-\delta_m) =0 \quad\text{at}\quad p^2=-M^2 \ \frac{d}{dp^2}\left[i\mathcal{M}+i(p^2\delta_Z-\delta_m)\right] =0 \quad\text{at}\quad p^2=-M^2 \tag{12.30} \end{cases} where $M$ is the renormalization scale, $\delta_Z$ the field strength renormalization counterterm, and $\delta_m$ the mass renormalization counterterm.
In the following discussion, we will assume the use of dimensional regularization. However, to emphasize the physical role of the cutoff, we will write expressions of the form (12.33) as $$i\mathcal{M}= \frac{-2ig^2}{(4\pi)^2}(-p^2)\left( \log\frac{\Lambda^2}{-p^2}+C \right) \tag{12.34}$$ The logarithmically divergent terms proportional to $p^2$ will agree with the divergences obtained with a momentum cutoff; the constant terms will not agree, but these will drop out of our nal results.
By dimensional analysis one can guess the pole $\frac{1}{2-d/2}$ corresponds to $\log \Lambda^2$ in (12.34).
And I calculated with cutoff.
First from $$ \hspace{-80pt} i\mathcal{M}= -4g^2 \int_0^1 dx \int\frac{d^4l}{(2\pi)^4}\frac{l^2+\Delta}{(l^2-\Delta)^2} \tag{10.33} $$ (where $\Delta=m_f^2-x(1-x)p^2$, and here I've already set $d=4$), performing a <PERSON> rotation $l^0=il_E^0$, \begin{alignat}{2} &=&& -4g^2 \int_0^1 dx \frac{-l_E^2+\Delta}{(l_E^2+\Delta)^2} \ &=&& -4ig^2 \left(\int\frac{d\Omega_4}{(2\pi)^4} \right) \int_0^1 dx \int_0^{\infty}\,dl_E l_E^3 \frac{-l_E^2+\Delta}{(l_E^2+\Delta)^2} \end{alignat} setting $s=l_E^2+\Delta$, $$ l_Edl_E=\frac{1}{2}ds $$ $$ \therefore \int_0^{\infty}\,dl_E l_E^3 \frac{-l_E^2+\Delta}{(l_E^2+\Delta)^2} = \frac{1}{2}\int_{\Delta}^{\Lambda^2+\Delta}ds \frac{(s-\Delta)(-s+2\Delta)}{s^2} \ = \frac{1}{2}\int_{\Delta}^{\Lambda^2+\Delta}ds \left( \frac{(-s^2+3\Delta s-2\Delta^2)}{s^2} \right) = \frac{1}{2} \left[ -s+3\Delta\log s+\frac{2\Delta^2}{s} \right]_{\Delta}^{\Lambda^2+\Delta} \sim -\frac{1}{2}\Lambda^2 $$ This doesn't agree with (12.34). 818