2
$\begingroup$

I want to know the closed form of the following master integral in (any) $D$ dimension \begin{equation} \int\frac{d^D k}{(2\pi)^D}\frac{1}{k^2(k+r)^2(k+p)^2}. \end{equation} The references that I can find have formulas for the integral with very general propagators $\frac{1}{p^{2\alpha}}$, but they involve some hideous looking hypergeometric functions and are half-page long, which makes the formulas almost impossible to use. I wonder if there is a simpler fomula that is specific to the simplest propagator $\frac{1}{p^2}$. Thanks in advance!

BTW, the real integral that I am interested in is the following \begin{equation} \int\frac{d^D k}{(2\pi)^D}\frac{k^\mu k^\nu}{k^2(k+r)^2(k+p)^2}. \end{equation} I know that it is related to the master integral by the Passarino-Veltman reduction.

$\endgroup$
4
  • $\begingroup$ There is no constraint on r and p and you are right, it is just the three point scalar integral. $\endgroup$ Commented Feb 19, 2019 at 2:39
  • $\begingroup$ Sorry I wanted to reword that before posting, have you checked A. Denner's or R.K.Ellis' papers on radiative corrections? They typically contain the results of these integrals $\endgroup$
    – Triatticus
    Commented Feb 19, 2019 at 2:42
  • $\begingroup$ Actually I found a good resource here, on page 178 phy.pku.edu.cn/~qhcao/resources/cpy/2014/CAO_scalar_thesis.pdf $\endgroup$
    – Triatticus
    Commented Feb 19, 2019 at 2:50
  • $\begingroup$ Thank you very much! I guess this is indeed as close as I can get $\endgroup$ Commented Feb 19, 2019 at 2:52

1 Answer 1

1
$\begingroup$

The integral to evaluate is

\begin{equation} I = \int\frac{d^D k}{(2\pi)^D}\frac{1}{k^2(k+r)^2(k+p)^2}. \end{equation}


The usual way to do this to use Feynman parameters. The formula is given by

$$ {\displaystyle {\frac {1}{A_{1}\cdots A_{n}}}=\left(n-1\right)!\int _{0}^{1}d\alpha _{1}\cdots \int _{0}^{1}d\alpha _{n}{\frac {\delta \left(1-\alpha _{1}-\cdots -\alpha _{n}\right)}{[\alpha _{1}A_{1}+\cdots +\alpha _{n}A_{n}]^{n}}}.} $$

Using this for the case of $n=3$ (your case) we have that

$$ \frac{1}{k^2(k+r)^2(k+p)^2} = 2! \int dx_1 \int dx_2 \int dx_3 \frac{\delta(1 - x_1 - x_2 - x_3)}{[x_1 k^2 + x_2(k+r)^2 + x_3 (k+p)^2]^3} $$

Then swap the order of integration so that

$$ I = 2!\iiint dx_1 dx_2 dx_3\delta(1 - x_1 - x_2 - x_3) \int\frac{d^D k}{(2\pi)^D}\frac{1}{[x_1 k^2 + x_2(k+r)^2 + x_3 (k+p)^2]^3} $$

Then expand the denominator into an order 2 polynomial in $k$ so it is of the form

$$ I = 2!\iiint dx_1 dx_2 dx_3 \delta(1 - x_1 - x_2 - x_3) \int\frac{d^D k}{(2\pi)^D}\frac{1}{[f(x_i) k^2 + g(x_i)k + h(x_i)]^3} \tag{a} $$

where $f(x_i):= f(x_1, x_2, x_3)$. Now, the innermost integral in $(a)$ is given in closed form in the appendix of Peskin in terms of $\Gamma$ functions and other stuff.


Now is when I leave you, as integrating over the feynman parameters is usually the ugliest part of the problem. Moreover, if all you want is a perturbative answer then this is certainly the way to go.

I could work through it, but it would be of little value to you (and me). So I will merely wish you good luck and a welcome you to the QFT Integral Club.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.