Renormalization constants I would like to understand how to extract renormalization constants of vacuum polarization diagram in pseudoscalar Yukawa theory with interaction $ig\bar{\psi}\gamma^5\psi$. This diagram is represented by the following integral:
$$i\Pi(k^2)=-4g^2\int_p\frac{p^2+p\cdot k-m^2}{(p^2-m^2)((p+k)^2-m^2)},$$
where $\int_p=\int d^4p/(2\pi)^4$ and should give two renormalization constants:
$$\delta Z_{\phi}=\frac{g^2}{4\pi^2}\frac{1}{\epsilon};\quad \delta Z_m=-\frac{g^2}{2\pi^2}\frac{1}{\epsilon},$$
which comes from singular part of polarization:
$$\Pi_{\text{sing}}=\frac{g^2}{4\pi^2}\left[\frac{1}{\epsilon}(k^2-2m^2)\right]$$
I know the following statement:

It is enough to capture out divergence to find renormalization constants

I would like to avoid full calculation of the integral and only calculate these constants. Thus, I consider big momenta $p$ and expand
$$\frac{1}{(p+k)^2-m^2}=\frac{1}{p^2}-\frac{2(p\cdot k)}{p^4}+\mathcal{O}(p^{-4}).$$
Substituting this expansion, I try to define singularity:
$$i\Pi_{\text{sing}}=-4g^2\int_p\frac{p^2}{(p^2-m^2)p^2},$$
which can be calculated with dimension regularization and I find
$$\Pi_{\text{sing}}=-\frac{g^2m^2}{2\pi^2}\rightarrow \delta Z_m=-\frac{g^2}{2\pi^2}\frac{1}{\epsilon}.$$
Then I have cast doubts about my derivation and do not understand how to extract the singular term which is $\propto k^2$.
 A: The scalar boson self-energy integral you are looking at has a naive quadratic divergence.  This means that if you shift the integration variable to $\ell=p-xk$, the surface term you get is still divergent.  Since the integrand of the surface term has to be proportional to $(xk)^{2}$, the surface term has a logarithmic divergence (two powers of $p$ in the numerator replaced with two powers of the fixed external momentum $k$).  This term, with its proportionality to $k^{2}$, gives the field strength renormalization constant that you missed.
In an initially logarithmically divergent expression, you could just drop all other masses and momenta in comparison with the integration momentum $p$ and get the infinite residue that way.  However, with a quadratic divergence, there are two divergences.  In dimensional regularization, the residue of the divergence $\Gamma(1-d/2)$ (as $d\rightarrow 4$) is, for dimensional reasons, independent of $k$ and gives the mass renormalization.  After you introduce a counterterm to cancel that divergence, there is another divergence (proportional to $k^{2}$) coming from $\Gamma(2-d/2)$, which is canceled by a field strength renormalization counterterm.
