$varphi^4$ via renormalization group with hard cut-off

I am studying the application of the renormalization group to the $varphi^4$ theory:

$$mathcal{L} = -frac{1}{2} varphi (Box + m^2)varphi -frac{lambda}{4!}varphi^4.$$

In particular I wanted to follow two different regularization methods, and verify that the resulting critical exponents $nu$ did not differ, as I expect.

If you want to calculate the contributions up to one loop, you encounter two diagrams which diverge: the tadpole for the 2-points correlation function, and the loop for the 4-points correlation function. The first goes like $int frac{d^4k}{k^2-m^2}$, the second like $int frac{d^4k}{(k^2-m^2)^2}$.

In every textbook I know (e.g. Schwartz, Quantum Field Theory and the Standard Model, 23.5.1), you see the treamtment of the $varphi^4$ and of the renormalization group with the dimensional regularization, which leads to the following equations:

$$beta = -varepsilonlambda + frac{3lambda^2}{16pi^2}$$

$$mufrac{d}{dmu}m^2 = frac{lambda}{16pi^2}m^2$$

where $mu$ is the renormalization scale, $lambda$ and $m^2$ are the renormalized constant of interaction and mass, and $d=4-varepsilon$ is the dimension.

Solving for the Wilson-Fisher fixed point, we find $lambda^* = frac{16pi^2varepsilon}{3}$ and $m^{2*} =0$.

At the fixed point, the anomalous dimension of the mass is then $gamma_m=frac{lambda^{*}}{16pi^2}=frac{varepsilon}{3}$ giving $nu=frac{1}{2-gamma_m}=frac{3}{6-varepsilon}$.

Fine. Now, if I try to introduce an explicit cut-off $Lambda$ in the divergent integrals, in four dimensions I get something like:
$int frac{d^4k}{k^2-m^2} propto frac{Lambda^2}{m^2} + log(1+frac{Lambda^2}{m^2})$

$$int frac{d^4k}{(k^2-m^2)^2} propto log(1+frac{Lambda^2}{m^2}).$$

This is something worrisome, since we have two very different behaviours between the two integrals, and in the critical exponent they enter as a ratio.

But let’s get into the calculation. I switch to the Euclidean integrals, and I define:

$f(Lambda,m^2,d) = int_Lambda frac{d^dk}{k^2+m^2}$

and

$g(Lambda,m^2,d) = int_Lambda frac{d^dk}{(k^2+m^2)^2}$

Taking into account that

$Lambda frac{d}{dLambda} f(Lambda,m^2,d) = S_d frac{Lambda^d}{Lambda^2+m^2} + frac{partial f(Lambda,m^2,d)}{partial m^2}Lambda frac{d}{dLambda}m^2$

and

$Lambda frac{d}{dLambda} g(Lambda,m^2,d) = S_d frac{Lambda^d}{(Lambda^2+m^2)^2} + frac{partial g(Lambda,m^2,d)}{partial m^2}Lambda frac{d}{dLambda}m^2$,

where $S_d$ is the area of the d-dimensional unit sphere,

I obtain the following equations for d-dimensions:

$beta = -(4-d)lambda – frac{3}{2^{d+1}pi^d}lambda^2(-(4-d)f + frac{Lambda^d}{Lambda^2+m^2}S_d)$

$Lambdafrac{d}{dLambda}m^2 = -frac{lambda}{2^{d+1}pi^d}m^2 (-(4-d)g + frac{Lambda^d}{(Lambda^2+m^2)^2}S_d)$

If I now try to evaluate $gamma_m$ at the fixed point, I find something like:

$gamma_m = frac{4-d}{3}frac{-(4-d)g + frac{Lambda^d}{(Lambda^2+m^2)^2}S_d}{-(4-d)f + frac{Lambda^d}{Lambda^2+m^2}S_d}$

If I did not make any error, which is an assumption, that $gamma_m$ is not equivalent to the one obtained via dimensional regularization. I am probably missing something.

Any suggestions?

EDIT: I realized (thanks @TehMeh) that I defined the functions $f$ and $g$ differently from my pen and paper calculation, and came up with a mixed notation and a lot of mess, which ended up in a lot of errors. Sorry to everyone. Let me now correct.

$f(Lambda,m^2,d) = int_Lambda frac{d^dk}{(k^2+m^2)^2}$

and

$g(Lambda,m^2,d) = frac{1}{m^2}int_Lambda frac{d^dk}{k^2+m^2}$

Taking into account that

$Lambda frac{d}{dLambda} f(Lambda,m^2,d) = S_d frac{Lambda^d}{(Lambda^2+m^2)^2} + frac{partial f(Lambda,m^2,d)}{partial m^2}Lambda frac{d}{dLambda}m^2$,

and

$Lambda frac{d}{dLambda} g(Lambda,m^2,d) = frac{S_d}{m^2} frac{Lambda^d}{Lambda^2+m^2} + frac{partial g(Lambda,m^2,d)}{partial m^2}Lambda frac{d}{dLambda}m^2$

where $S_d$ is the area of the d-dimensional unit sphere,

I obtain the following equations for d-dimensions:

$beta = -(4-d)lambda – frac{3}{2^{d+1}pi^d}lambda^2(-(4-d)f + frac{Lambda^d}{(Lambda^2+m^2)^2}S_d)$

$Lambdafrac{d}{dLambda}m^2 = -frac{lambda}{2^{d+1}pi^d}m^2 (-(4-d)g + frac{Lambda^d}{Lambda^2+m^2}frac{S_d}{m^2})$

If I now try to evaluate $gamma_m$ at the fixed point, I find something like:

$gamma_m = frac{4-d}{3}frac{-(4-d)g + frac{Lambda^d}{Lambda^2+m^2}frac{S_d}{m^2}}{-(4-d)f + frac{Lambda^d}{(Lambda^2+m^2)^2}S_d}$

$f$ and $g$ are representable with the hypergeometric function, but if we take the limit for small $4-d$ it should not anyway matter their expression.