I have thought about this question for a very long time. The question will be described as the following.

I have an effective Feynman loop diagram to calculate, which describing B meson decay into 3 particles. There are 4 propagators. After using Feynman parameterization and completing the square and shifting the loop momentum, the denominator looks like $l^2-\Delta$, where $l$ is the loop momentum, $\Delta$ is a function of Feynman parameters and external momenta.

$$ \frac{1}{[k^2-m_4^2][(k+p_1)^2-m_1^2][(k+p_2+p_3)^2-m_2^2][(k+p_3)^2-m_3^2]}=... \\=\int_0^1dx_1dx_2dx_3dx_4\delta(\sum x_i-1)\frac{3!}{[l^2-\Delta]^4} $$

It is easy to handle it when $\Delta > 0$, but I have no idea about when it can be sometimes positive and sometimes negative as the Feynman parameters and external momenta change.

How should I do? Thank you.

Something new should be posted here. enter image description here

A little bit ugly, but you can see it clearly.

The exact integral that I am fighting is the following: $$ \int d^4 k \cdot EXP \frac {TR}{[k^2-m_4^2][(k+p_1)^2-m_1^2][(k+p_2+p_3)^2-m_2^2][(k+p_3)^2-m_3^2]}, $$ where $EXP$ stands for 3 exponential functions, which come from the 3 effective meson vertices, whose effect is to depress the UV part of the integral to make it finite. However they are fairly complicated, look like the following, $ EXP=\exp[\frac{k^2 p_1^2-(k\cdot p_1)^2}{c_1^2}]\cdot\exp[\frac{[k^2 p_2^2-(k\cdot p_2)^2]+[p_2^2p_3^2-(p_2\cdot p_3)^2]+[2k \cdot p_3 p_2^2 - 2k\cdot p_2 p_2\cdot p_3]}{c_2^2}]\cdot\exp[\frac{k^2 p_3^2-(k\cdot p_3)^2}{c_3^2}], $ $ TR=tr[\gamma^5(m_4+ k\!\!\!/)\gamma^5(m_3+ k\!\!\!/ + p\!\!\!/_3)\gamma^5(m_2+ k\!\!\!/ +p\!\!\!/_2+ p\!\!\!/_3)\gamma^{\mu}(1-\gamma^5)(m_1+k\!\!\!/+p\!\!\!/_1)] $

Numerical methods may be needed to get the value of the integral. Maybe I can try another easier meson vertices, which look like this, $EXP=\exp(c k^2)$, which seem that a analytic method can be applied. However I have tried several analytic methods, I failed all of them.

I will post the method that I want to carry out. Using Feynman parameterization + residue theorem.

Write the integral above as the following: $$ \int d^4 k \frac{f^{\mu}(k)}{[k^2-m_4^2][(k+p_1)^2-m_1^2][(k+p_2+p_3)^2-m_2^2][(k+p_3)^2-m_3^2]}\\ =3!\cdot \int d^4 k \int _0 ^1 dx_1 dx_2 dx_3 dx_4 \delta(\sum x_i -1) \frac{f^{\mu}(k)}{[(k+P)^2-\Delta]^4}\\ =3!\cdot \int d^3 k d k^0 \int _0 ^1 dx_1 dx_2 dx_3 dx_4 \delta(\sum x_i -1) \frac{f^{\mu}(k)}{[(k^0+P^0)^2-(\textbf{k}+\textbf{P})^2-\Delta]^4} $$

(where $f(k)=EXP \cdot TR$, which is very complicated, and which makes this integral so muddy.)

if $(\textbf{k}+\textbf{P})^2+\Delta < 0$, then close the contour above(for example) to enclose a pure imaginary root in it, then use residue theorem.

if $(\textbf{k}+\textbf{P})^2+\Delta \geq 0$, then close the contour above(for example) to enclose a root which looks like $a + i\epsilon$ in it, then use residue theorem.

There are several words to say.

1) Clearly I have to use ugly numerical method.

2) My tutors worry about the situation near 0, that is when $(\textbf{k}+\textbf{P})^2+\Delta$ is close to 0, what will happen?

3) Using residue theorem requires to differentiate a very very complicated function for the third order derivative, which is hard to realize it.

If this method is right, I will try it although it is hard to carry out.

I want to hear some confirmation or some criticism. Thank you.

  • $\begingroup$ Could you please include the relevant diagram? $\endgroup$
    – Danu
    Commented Sep 12, 2014 at 7:35
  • $\begingroup$ Ok, I will post the diagram. $\endgroup$
    – zzx2012
    Commented Sep 12, 2014 at 8:01

1 Answer 1


After getting to the above step you simply need to carry out the integration over the loop momenta $l$ which is highly convergent in your case (if you are working in four space-time dimensions). At this this step you really don't need to worry about the changes in the parameters inside $\Delta$ while carrying out the loop momentum integral (the Feynman parameters are frozen anyways when you are carrying out the loop integral). Just use the following formula: $$ \int \frac{d^dl}{(2\pi)^d}\frac{1}{(l^2-\Delta)^n}=\frac{(-1)^n i}{(4\pi)^{d/2}}\frac{\Gamma(n-\frac{d}{2})}{\Gamma(n)}\left(\frac{1}{\Delta}\right)^{n-\frac{d}{2}} $$ with $n=4$ and $d=4$. You can see that the integral is convergent for $n=4$ as $d\to 4$. After this step you really need to see how do the parameters behave the under the $x_i$ integral. This generally requires a bit of analysis of the integrand if it contains a Log term (which is not your case). But for sake of illustration let us suppose the case of $n=2,d=4$. In this situation you get a Log term containing $\Delta$ as its argument when you take the limit $d\to 4.$ You see the argument of Log cannot be negative. But $\Delta$ can take any value it likes as the Feynman parameters vary. In that case we isolate the domain of the integral of Feynman parameters under which the Log contains a negative argument (for the rest of the domain do the Feynman integrals as usual, or leave them as it is).

Now comes the important point. When $\Delta$ is negative inside the Log you need to restore the Feynman $i\epsilon$ prescription all over. The whole purpose of putting the $i\epsilon$ in the propagator of a particle is necessary for this step (in other cases it mostly doesn't matter much, but you should remember that it is there). The restoration is done by $$m^2\to m^2-i\epsilon.$$ You can trace it out! You see that the Log term now becomes: Log[$-\Delta-i\epsilon$] ($\Delta$ is now positive. I have explicitly pulled out the minus sign.). It is a small exercise to prove that: $$ \text{Lim}_{\epsilon\to 0}\left(\text{Log}[-\Delta\pm i\epsilon]\right)=\text{Log}|\Delta|\pm i\pi $$ The Feynman parameter-integral containing the first term will join the rest to form the usual integration over the full parameter range (i.e. from 0 to 1). The imaginary part $i\pi$ is a trivial integral. This contribution will ultimately show up in the formula for decay rates. All this analysis can be understood very rigorously for the case of the 1 loop propagator of a scalar field $\Phi$ with mass $M$ coupled to another scalar field $\phi$ of mass $m$ $(M>2m)$ with a coupling of the form $g\Phi\phi\phi$ (for $M>2m$, $\Delta$ indeed becomes negative in this case).

Hope this helps!

  • $\begingroup$ Thank you very much and I have learned a lot from your reply. But the problem that I am struggling with looks like much more difficult, which I will present it right now above--that is, I will say it more precisely. $\endgroup$
    – zzx2012
    Commented Sep 12, 2014 at 13:13
  • $\begingroup$ Can you also mention what is TR in the numerator? $\endgroup$
    – Orbifold
    Commented Sep 12, 2014 at 22:33
  • $\begingroup$ TR has been added. $\endgroup$
    – zzx2012
    Commented Sep 13, 2014 at 4:17
  • $\begingroup$ Is it not better to convert the integral into Euclidean form by performing a Wick rotation? These issues will not arise it such case. Moreover since the denominator would be rotationally invariant (in the Euclidean space) the numerator could be simplified considerably by either throwing away the odd terms in the loop momenta or simplifying tensors (into rotationally invariant forms) that contain even factors of the loop momenta. $\endgroup$
    – Orbifold
    Commented Sep 14, 2014 at 7:38
  • $\begingroup$ I do have considered the Wick rotation seriously. The reason that I give up this method for several reasons. 1) After Wick rotation, the denominator looks like (-l^2-Delta)^4 which seems to still have singular points since Delta can be positive or negative. 2) We cannot simplify the numerator even for a little, since f(k)=EXP*TR, EXP does not have good symmetry essentially because EXP contains terms like (k.p1)^2 so that we cannot complete the square about k. $\endgroup$
    – zzx2012
    Commented Sep 14, 2014 at 12:51

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