 Symmetric two-propagator expansion

Set $A=p^2-m^2\ne 0$. The denominator product is exactly $(A+k^2{)}^2-4(p\cdot k{)}^2$. Its reciprocal therefore has the Taylor expansionThe coefficients follow by expanding $1/(1+x)=1-x+O(x^2)$ with $x=2k^2/A-4(p\cdot k{)}^2/A^2+O(k^4)$. The expansion is a local algebraic statement away from denominator zeros; integrating it through Feynman propagator poles requires a regulator and the correct Feynman i-epsilon prescription.