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?


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