Regularization: Evaluating the one-loop $\phi^4$ integral to order $\lambda^2$ I'm currently on the chapter of regularization on Zee's QFT book. For the $\phi^4$ theory, an amplitude for a single loop correction to order $\lambda^2$ is given by a diagram

Following the Feynman rules, we obtain the amplitude
$$
\mathcal{M}=\frac{i^2}{2}(-i\lambda)^2\int\frac{d^4k}{(2\pi)^4}\frac{1}{k^2-m^2+i\epsilon}\frac{1}{(K-k)^2-m^2+i\epsilon}
$$
where $K=k_1+k_2$. Assuming that $m<<k$ and having the cut off $\Lambda$, he said that this is equal to
$$
\mathcal{M}=iC\lambda^2log\bigg(\frac{\Lambda^2}{K^2}\bigg) 
$$
Where $C$ is just some constant after evaluation. I was trying to evaluate this integral to obtain the same answer as he got, but I am getting nowhere close. I don't even understand what the bounds are in the 4 integrals if we replace them by $\Lambda$. Are all the bounds $\Lambda$?
How did he get that answer?
 A: For the regulation of the amplitude $M = \frac{i^2}{2}(-i\lambda)^2 I$ with
$$I:=\int \frac{d^4k}{(2\pi)^4} \frac{1}{k^2-m^2}\frac{1}{(k-K)^2-m^2}$$
the following the strategy is applied (according to R.Feynman). With $A=(k-K)^2-m^2$ and $B=k^2-m^2$
$$I = \int \frac{d^4k}{(2\pi)^4} \frac{1}{A}\frac{1}{B}=\int_0^1 dx\int \frac{d^4k}{(2\pi)^4}\frac{1}{[xA + (1-x) B]^2}$$
Expanding the denominator we get:
$$I=\int_0^1 dx\int \frac{d^4k}{(2\pi)^4}\frac{1}{[k^2 - 2xK\cdot k + xK^2 -m^2]^2}$$
Next step is a substitution $l =k-xK$: ($l$ is a 4-vector with the components $(l^0,l^1,l^2,l^3)$):
$$I=\int_0^1 dx\int \frac{d^4l}{(2\pi)^4}\frac{1}{[l^2+x(1-x)K^2 -m^2]^2}$$
followed by a Wick rotation:  $l^0_E = -il^0$ and $\vec{l_E} = \vec{l}$
$$I = i\int_0^1 dx\int \frac{d^4l_E}{(2\pi)^4}\frac{1}{[l_E^2-x(1-x)K^2 +m^2]^2}= i\int_0^1 dx\int \frac{d^4l_E}{(2\pi)^4}\frac{1}{[l_E^2+\Delta]^2}$$
with $\Delta = m^2-x(1-x)K^2$  Apparently Zee applied Pauli-Villars regularisation which consists of replacing
$$ \frac{1}{(l_E^2 +\Delta)^2} \longrightarrow \frac{1}{(l_E^2 +\Delta)^2}-\frac{1}{(l_E^2 +\Lambda^2)^2}$$
Applying it on our integral $I$ it becomes $I_\Lambda$:
$$I_\Lambda=i\int_0^1 dx\int \frac{d^4l_E}{(2\pi)^4}(\frac{1}{[l_E^2+\Delta]^2}-\frac{1}{[l_E^2+\Lambda^2]^2}) = i\int_0^1 dx\int d\Omega_4 \int_0^{\infty} \frac{dl_E }{(2\pi)^4} \left( \frac{l_E^3}{[l_E^2+\Delta]^2} -  \frac{l_E^3}{[l_E^2+\Lambda^2]^2}\right)$$
A final substitution $z= l_E^2$ with  $dz = 2l_E dl_E $ yields ($\int d\Omega_4 = 2\pi^2$):
$$I_\Lambda=i\int_0^1 dx\int_0^{\infty}\frac{dz}{(4\pi)^2} \left( \frac{z}{[z+\Delta]^2} -  \frac{z}{[z+\Lambda^2]^2}\right)=\frac{i}{(4\pi)^2}\int_0^1 dx\,\, \log(\frac{\Lambda^2}{\Delta}) $$
Therefore we get for the regularized amplitude $M_\Lambda$:
$$ M_\Lambda = \frac{i^2}{2}(-i\lambda)^2 I_\Lambda =\frac{i\lambda^2}{32\pi^2} \int_0^1 dx\,\, \log\left(\frac{\Lambda^2}{m^2-x(1-x)K^2}\right)$$
which corresponds to Zee's formula (14) on page 152. In the computation the terms $+i\epsilon$ were omitted for simplicity.
A: If you do not need to know the constant $C$ then all you have to do is observe that the integral is Log divergent at large momentum, so it must contain a $\ln \Lambda^2$, but the argument of the log has to be dimensionless and the only dimensionful quantity that the integral dependens on is $K^2$ so it contains a $\ln (K^2/\Lambda^2)$. To get the factor of $i$ I assume that a Wick rotation is being made so that the $dk_0$ is being rotated to $i dk_0$.
If you need the actual number $C$ use the Feynman parameter method. If $m=0$ it is much easier to work  in Euclidan signature and in configuation space. 
