# Two ways to calculate imaginary part of vacuum polarization in pseudoscalar Yukawa theory

I consider vacuum polarization in pseudoscalr Yukawaw theory with interaction $$ig\bar{\psi}\gamma^5\phi\psi$$. The expression for diagram is $$i\Pi(k^2)=-g^2\int\frac{d^4p}{(2\pi)^4}\frac{\mathrm{Tr}[i\gamma^5(\gamma^{\mu}p_{\mu}+M)i\gamma^5(\gamma^{\nu}(p+k)_{\nu}+M)]}{(p^2-M^2+i\epsilon)((p+k)^2-M^2+i\epsilon)},$$ where $$M$$ is the mass of fermionic field. I can calculate this integral with help of dimensional regularization, determine counter-term, etc. I am interested in the imaginary part of $$\Pi(k^2)$$ which can be obtained from the following integral $$\int_{0}^{1}dx\left[M^2-3k^2x(1-x)\right]\log\frac{M^2-k^2x(1-x)}{\lambda^2}$$ with assumption that $$k^2>4M^2$$ and $$\lambda$$ is the parameter which comes from dimensional regularization. After calculation of this integral, I find $$\boxed{\mathrm{Im}\,\Pi(k^2)=\frac{g^2}{8\pi}(k^2+4M^2)\sqrt{1-\frac{4M^2}{k^2}}.}$$ This seems true but for satisfaction I am trying to calculate $$\mathrm{Im}\,\Pi(k^2)$$ another way. I write $$\mathrm{Im}\,\Pi(k^2)=\mathrm{Im}\frac{4g^2i}{(2\pi)^4}\int d^4p\frac {F(p)}{(p^2-M^2+i\epsilon)((p+k)^2-M^2+i\epsilon)}(*),$$ where $$F(p)=p^2+(p\cdot k)-M^2$$ which comes from $$\mathrm{Tr}$$: $$\mathrm{Tr}[i\gamma^5(\gamma^{\mu}p_{\mu}+M)i\gamma^5(\gamma^{\nu}(p+k)_{\nu}+M)]=4[p^2+(p\cdot k)-M^2].$$ Then I break the integral $$(*)$$ as integral over $$p^0$$ and integral over $${\bf p}$$, introduce $$\omega^2={\bf p}^2+m^2$$, use reference frame where $$k=(k^0,\,0)$$ and write down $$I=\int_{-\infty}^{\infty}dp^0\frac{F(p_0,\,{\bf p})}{(p_0^2-\omega^2+i\epsilon)((p_0+k_0)^2-\omega^2+i\epsilon)}$$ It is straightforward to calculate jump $$\Delta I$$, then use $$\Delta I=2i\mathrm{Im}\,I$$ to evaluate $$\mathrm{Im}\,I$$ and finally obtain $$\mathrm{Im}\,\Pi(k^2)$$. But I have the problem. In $$\Delta I$$ calculation I should compute $$F(p_0,\,{\bf p})$$ for $$p_0=k_0/2$$ and $${\bf p}=k_0^2/4-M^2$$: $$F(p_0,\,{\bf p})=\frac{k_0^2}{4}-\frac{k_0^2}{4}+M^2+\frac{k_0^2}{2}-M^2=\frac{k_0^2}{2},$$ which gives the "wrong" answer: $$\mathrm{Im}\,\Pi(k^2)\propto k^2\sqrt{1-\frac{4M^2}{k^2}}$$ i.e. without $$4M^2$$ term...

I am totally disappointed because I do not understand where I made a mistake. Therefore, the question is: is it possible to calculate $$\mathrm{Im}\,\Pi(k^2)$$ by extraction of imaginary part of Feynman diagram or it is wrong?