About the Feynman parameterization--when the 'Delta' in the denominator is minus, how should I do? 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.

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.
 A: 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!
