# Manipulation of momentum factors in Feynman integrals (solved)

My current understanding: Consider a massless Feynman graph with propagators of the form $$1/p^2$$ and some momentum factors in the numerator. Now consider a vertex, which for simplicity of notation is assumed to be trivalent. By momentum conservation we can write the momenta of the three propagators that are attached to this vertex as $$p_1,p_2,-p_1-p_2$$. For a factor $$p^\mu_1$$ in the numerator I am free to write it as $$p^\mu_1=-(-p_1-p_2)^\mu-p^\mu_2,$$ i.e. I can swap the momentum factor from one of the legs to the other two. This is also one of the essential features of IBP that many programs use to rewrite Feynman diagrams in a basis.

My Problem: I have used this relation extensively and in many cases it works. But I found some instances, where this doesn't seem to give the correct result and I don't understand, why it fails. Is there some subtlety in the general principle that I'm missing?

Example: The easiest case I have encountered is the following. Consider the topology (called FA in the Mincer manual) and define a Feynman integral with a numerator depending on the momenta $$I(N(P_i,Q))=\int\frac{\mathrm{d}^d k_1}{(2\pi)^{d}}\int\frac{\mathrm{d}^d k_2}{(2\pi)^{d}}\int\frac{\mathrm{d}^d k_3}{(2\pi)^{d}} \frac{N(P_i,Q)}{\prod_{i=1}^7 P_i^2}$$ with some choice of loop momenta $$k_i$$. In particular since $$P_2=P_1+P_6$$ it should be possible to write $$I(P_1 \cdot P_2)=I(P_1 \cdot P_6)+I(P_1^2)$$ however this is not the case as a test with Mincer shows. I tried to generalize this to $$I(P_1 \cdot P_2)=a\, I(P_1 \cdot P_6)+b\, I(P_1^2)$$ for some rational coefficients $$a,b$$, but this has no solution either.

For completeness here are the $$\epsilon$$-expansions for the respective integrals $$I(P_1 \cdot P_2)=\frac{1}{6 \epsilon ^2}+\frac{3}{2 \epsilon }+\frac{49}{6}+\mathcal{O}(\epsilon)$$ $$I(P_1 \cdot P_6)=-\frac{1}{4 \epsilon ^3}-\frac{25}{12 \epsilon ^2}-\frac{47}{4 \epsilon }-\frac{673}{12}+\frac{41 \zeta _3}{4}+\mathcal{O}(\epsilon)$$ $$I(P_1^2)=\frac{1}{6 \epsilon ^3}+\frac{3}{2 \epsilon ^2}+\frac{55}{6 \epsilon }+\frac{95}{2}-\frac{53 \zeta _3}{6}+\mathcal{O}(\epsilon)$$

• Do you see the same problem in a simpler Feynman diagram that you can do yourself without relying on a program? Jan 4, 2019 at 21:33
• No, unfortunately, I have not found an easier example. All the one- and two-loop diagrams I have looked at, work as expected. At three loops I have to rely on a program due to the complexity of the calculation.
– JPC
Jan 5, 2019 at 0:03
• Tip: Consider to submit the solution as an answer post rather than as part of the question post. Jan 22, 2019 at 18:05
• Ok, done. I did not know, whether putting this as an answer would be a little cheap.
– JPC
Jan 22, 2019 at 18:42