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 $A=\omega-1$?


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.

  • $\begingroup$ Voting to move to the math stack exchange, this is only about performing an integration, and you will probably at least find a neat complex contour integration solution to this. $\endgroup$ – JamalS Oct 3 '19 at 20:00
  • $\begingroup$ What values can $A$ have ? $\endgroup$ – StephenG Oct 3 '19 at 20:45
  • $\begingroup$ @StephenG In the case where I am interested, $A=\omega-1$ with $\omega \rightarrow 2$, therefore $A \approx 1$. $\endgroup$ – Jxx Oct 3 '19 at 20:47

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.

  • $\begingroup$ Thanks for your answer! But won't I end up with the same expression as $(2)$ in the OP? My problem is precisely to take the $p$-dimensional integral, the loop integral is already gone. $\endgroup$ – Jxx Oct 4 '19 at 11:10
  • $\begingroup$ @Jxx I think it isn't always possible to take the $p$-dimensional integral analytically. Quite frequently QFT amplitudes are written in the form of the integral. $\endgroup$ – Prof. Legolasov Oct 4 '19 at 20:48
  • $\begingroup$ Yes that is what I think too. Incidentally the integral I wrote is a part of a larger integral, involving a Wilson line, and that can also be used to make progress in the integral. Still working on it though! $\endgroup$ – Jxx Oct 4 '19 at 21:20

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.

  • $\begingroup$ Thank you for your answer and for the link. Do you know how they get their $\Delta$ in the link? When I do the substitution and write it in the desired form, I find $\Delta = -m^2(1-z) + q^2 y (1-y) + p^2 z(1-z) + 2 p\cdot q y z$, while they somehow do not have any $p$'s left. Is there some on-shell assumption? If yes, then this method is inapplicable to my problem, since the x's are coordinates, and doing the integral as they do should just give back the equation labeled $(2)$ in the OP. $\endgroup$ – Jxx Oct 4 '19 at 9:29
  • $\begingroup$ Sorry, I just saw the comment. It's a pretty busy day for me today, and I am finishing late from the uni. I will try to have a more thorough look later! $\endgroup$ – A_user_with_NoName Oct 4 '19 at 13:23

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