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I am trying to show for the one-loop integral with three propagators with different internal masses $m_1$, $m_2$, $m_3$, and all off-shell external momenta $p_1$, $p_2$, $p_3$ the following formula appearing in 't Hooft(1979), Bardin (1999), Denner (2007): (unfortunate metric $-,+,+,+$)

$$\int d^d q\frac{1}{(q^2+m_1^2)((q+p1)^2+m_2^2)((q+p_1+p_2)^2+m_3^2)} $$ $$=i\pi^2\int_0^1dx\int_0^xdy\frac{1}{ax^2+by^2+cxy+dx+ey+f}$$

where $a$, $b$, $c$, ... are coefficients depending on the momenta in the following way:







I don't really care about factors in fromt like $i\pi^2$. My simple problem is: I am totally unable to reproduce coefficients $d$, $e$ and $f$. The problem is, when I integrate over the third Feynman parameter, $m_3$ appears in all three coefficients $d$, $e$ and $f$. How do I squeeze the denominators to reproduce this formula?

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I am a little confused - you are indicating you are having trouble with a 3rd Feynman parameter, when the identity you wrote only has 2? – DJBunk Feb 24 '13 at 23:00
Ah yes. When I attempted to derive the identity, I had a third parameter $z$ accompanied by the delta function $\delta(1-x-y-z)$. After integrating over $z$ (which I thought would be trivial), I got something like the identity in my post, but the coefficients came out wrong. I need help with the coefficients. – QuantumDot Feb 24 '13 at 23:42
I think the limits on the remaining 2 Feynman parameter integrals should instead be: $\int_0^1 dx \int_0^{1-x} dy $. – DJBunk Feb 25 '13 at 0:03
Isn't $q$ the variable you should be integrating over in your first formula? – Learning is a mess Feb 25 '13 at 0:19
Maybe you should post your answer for future users? It's received 5 upvotes so there seems to be interest in it. – DJBunk Feb 25 '13 at 17:29
up vote 5 down vote accepted

Define the LHS of the equation above:

$$I=\int d^d q\frac{1}{(q^2+m_1^2)((q+p_1)^2+m_2^2)((q+p_1+p_2)^2+m_3^2)}$$

The first step is to squeeze the denominators using Feynman's trick:

$$I=\int_0^1 dx\,dy\,dz\,\delta(1-x-y-z)\int d^d q\frac{2}{[y(q^2+m_1^2)+z((q+p_1)^2+m_2^2)+x((q+p_1+p_2)^2+m_3^2)]^3}$$

The square in $q^2$ may be completed in the denominator by expanding:

$$[\text{denom}]=q^2+2q.(z p_1+x(p_1+p_2))+y m_1^2+z (p_1^2+m_2^2)+x(m_3^2+(p_1+p_2)^2)$$ $$=q^2+2q.Q+A^2\,$$

where $Q^\mu=z p_1^\mu+x(p_1+p_2)^\mu$ and $A^2=y m_1^2+z (p_1^2+m_2^2)+x(m_3^2+(p_1+p_2)^2)$, and by shifting the momentum, $q^\mu=(k-Q)^\mu$ as a change of integration variables. Upon performing the $k$ integral, we are left with integrals over Feynman parameters (because this integral has three propagators, it is UV finite):

$$I=i\pi^2\int_0^1 dx\,dy\,dz\,\delta(1-x-y-z)\frac{1}{[-Q^2+A^2]}$$

Now integrate over $z$ with the help of the Dirac delta:

$$I=i\pi^2\int_0^1 dx\int_0^{1-x}dy \frac{1}{[-Q^2+A^2]_{z\rightarrow1-y-z}}$$

To arrive at the RHS of the OP's equation(which is the part I forgot to do), we make a final change of variables: $x=1-x'$:

So that the denominator reads $ax^2+by^2+cxy+dx+ey+f$, with the coefficients $a,b,c,\ldots$ exactly defined in the question of OP. Note the change in the range of integration in $y$.


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