How to solve integrals with $3$ Feynman parameters?

Question

I would like to evaluate integrals of the following type (in position space):

$$\int \frac{d^{2\omega}z}{\left[(x_1-z)^2 (x_2-z)^2 (x_3-z)^2 \right]^A} \tag{1}$$

I can introduce three Feynman parameters using the last equation of the section "Formulas" in this wikipedia article, integrate over $z$, then integrate over one Feynman parameter with the delta function in order to obtain:

$$\pi^\omega \frac{\Gamma(3A-\omega)}{\Gamma^3(A)} \int_0^1 d\gamma \int_0^{1-\gamma} d\beta \frac{\left[ \gamma\beta (1-\gamma-\beta) \right]^{A-1}}{\left[\gamma\beta x_{12}^2+\gamma(1-\gamma-\beta)x_{13}^2 +\beta(1-\gamma-\beta)x_{23}^2\right]^{3A-\omega}} \tag{2}$$

where $x_{ij}:=x_i-x_j$. Is it possible to go further with this integral? If yes, in what way? If not, is there at least a way to extract the divergent part for $\omega \rightarrow 2$ and $A=\omega-1$?

EDIT:

So far the answers seem to not be addressing my problem, so I would like to emphasize the issue: I can do the loop integral, but am stuck at solving the remaining integral with the Feynman parameters. I have also posted the same question at math.stackexchange but did not receive any attention so far. I am starting to believe that this integral is solvable only numerically, because of this quote of Sidney Coleman from the book "Lectures of Sidney Coleman on Quantum Field Theory":

In principle you can reduce any Feynman graph to an integral over Feynman parameters. At that point, typically, you are stuck, but you can always work it out numerically with a computer.

Answer 1

I think that this is a useful reference for this example Feynman Trick. Example 2 is relevant to yours. If you think that my reply is not useful or relevant, let me know and I will delete it.

Since you have the link, let me mention the very basic steps.

Starting from the standard

$\begin{align} \frac{1}{AB}=\int_0^1dx \frac{1}{(x A + (1-x)B)^2} = \int_0^1 dx \int_0^1 \frac{\delta(x+y-1)}{(xA + yB)^2} \end{align}$

and performing differentiation repeatedly we arrive at

$\begin{align} \frac{1}{A_1 A_2 \cdots A_n}=\int_0^1 dx_1 \int_0^1 dx_2 \cdots \int_0^1 dx_n \frac{(n-1)!\delta(x_1+x_2+\cdots + x_n-1)}{(x_1 A_1 + \cdots x_n A_n)^n} \end{align}$

Let me re-write your original integral in the following way

$$\int d^4 k \int_0^1 dx dy dz \frac{2 \delta(x+y+z-1)}{D^3}$$

where D is given by

$\begin{align} \begin{aligned} D &= x (x_1 - k)^2 + y (x_2 - k)^2 + z (x_3-k)^2 \\ &= k^2 - 2k (x x_1 + y x_2 + z x_3) + x x_1^2+ y x_2^2+ z x_3^2 \end{aligned} \end{align}$

having used $x+y+z=1$. And now the idea is to introduce to new variables such that you re-write the above D-factor in the form

$$D = \ell - \Delta $$

and then you continue the computation as demonstrated in the aforementioned example.

Answer 2

You can perform the Feynman trick like Konstantinos suggested.

Alternatively, you can perform the Schwinger proper-time trick, which I personally find simpler and easier in practice.

The Schwinger trick is basically substituting

$$ \frac{1}{a} = \intop_0^{\infty} d\tau e^{-a \tau}. $$

The denominator $a$ usually comes from a propagator of a virtual particle. The integration variable $\tau$ is called the proper time, and it actually corresponds the proper time of the virtual particle in the first-quantized formalism.

After expanding all propagators in this way, you will end up with an $p$-dimensional integral (each propagator has its own proper time) that you can take after you take the (regularized) loop integrals.