FCFeynmanProjectiveQ[int, x] checks if the given Feynman parameter integral (without prefactors) depending on x[1], x[2], … is a projective form.
It is similar to FCFeynmanProjectivize but unlike the former it simply returns True or False depending on whether the integral is projective or not.
Overview, FCFeynmanParametrize, FCFeynmanPrepare, FCFeynmanProjectivize.
int = SFAD[{p3, mg^2}] SFAD[{p3 - p1, mg^2}] SFAD[{{0, -2 p1 . q}}]\frac{1}{(\text{p3}^2-\text{mg}^2+i \eta ) ((\text{p3}-\text{p1})^2-\text{mg}^2+i \eta ) (-2 (\text{p1}\cdot q)+i \eta )}
fp = FCFeynmanParametrize[int, {p1, p3}, Names -> x, Indexed -> True, FCReplaceD -> {D -> 4 - 2 ep},
Simplify -> True, Assumptions -> {mg > 0, ep > 0}, FinalSubstitutions -> {SPD[q] -> qq, mg^2 -> mg2}]\left\{(x(2) x(3))^{3 \;\text{ep}-3} \left((x(2)+x(3)) \left(\text{mg2} x(2) x(3)+\text{qq} x(1)^2\right)\right)^{1-2 \;\text{ep}},-\Gamma (2 \;\text{ep}-1),\{x(1),x(2),x(3)\}\right\}
FCFeynmanProjectiveQ[fp[[1]], x]\text{True}
FCFeynmanProjectiveQ[(x[1] + x[2])^(-2 + 2*ep)/(mb2*(x[1]^2 + x[1]*x[2] + x[2]^2))^ep, x]\text{True}
Feynman parametrization derived from propagator representation should be projective in most cases. However, arbitrary Feynman parameter integral do not necessarily have this property.
FCFeynmanProjectiveQ[x[1]^(x - 1) (x[2])^(y - 1), x]\text{False}