Solving two specific integrals with so-called Feynman parameters

Question

I am trying to solve some integrals which appear in the context of renormalization in quantum field theory and integrals with so-called Feynman parameters, but I am unable to reproduce what is (according to the lecture notes) the correct answer. In particular, there are two integrals where I am stuck.

1. The first relation I am unable to prove is \begin{align} \int^1_0 \int^1_0 \int^1_0 \frac{1}{(x_1x_2 + x_2x_3 + x_3x_1)^{2-\epsilon}} \delta(1-x_1-x_2-x_3) \mathrm dx_1\mathrm dx_2\mathrm dx_3 \simeq\\ 3 \int^1_0 \int^1_0 \frac{1}{(x_1+x_2)^{2-\epsilon}}\mathrm dx_1\mathrm dx_2 \text. \end{align} Here, $\epsilon \ll 1$ and $\delta$ is the Dirac delta-function. I have tried to use that \begin{align}\tag{1}\label{1} \frac 1{a^{2-\epsilon}} = \frac{a^\epsilon}{a^2} \simeq \frac{1 + \epsilon \ln(a)}{a^2} \end{align} for a dimensionless quantity $a$, but it doesn't seem to help much when applying this for the integrand. Can anyone see how the above approximate equality holds?

2. The second relation I am unable to prove is \begin{align}\tag{2}\label{2} \int^1_0 \int^1_0 \int^1_0 \frac{x_1x_2x_3}{(x_1x_2 + x_2x_3 + x_3x_1)^{3-\epsilon}} \delta(1-x_1-x_2-x_3) \mathrm dx_1\mathrm dx_2\mathrm dx_3 =\\ \frac{1}{2}(1 + \epsilon C) \end{align} where $C$ is the (supposedly finite) integral: \begin{align} C = \int^1_0 \int^1_0 \int^1_0 \frac{x_1x_2x_3 \ln(x_1x_2 + x_2x_3 + x_3x_1)}{(x_1x_2 + x_2x_3 + x_3x_1)^{3}} \delta(1-x_1-x_2-x_3)\mathrm dx_1\mathrm dx_2\mathrm dx_3 \end{align} By using the expansion in Eq. \eqref{1}, I can get the $C$-term, but without the factor $\frac{1}{2}$, so I am wondering where this prefactor comes from. Secondly, when I try to integrate \begin{align} \int^1_0 \int^1_0 \int^1_0 \frac{x_1x_2x_3}{(x_1x_2 + x_2x_3 + x_3x_1)^{3}} \delta(1-x_1-x_2-x_3)\mathrm dx_1\mathrm dx_2\mathrm dx_3 \end{align} using an online integrator, it tells me that the integral does not converge. This means I am unable to obtain the first term on the right hand side of Eq. \eqref{2}.

In case someone is interested to see the origin of these two problematic integrals, please see equations 3.3.32 to 3.3.37 here: https://www.physics.uu.se/digitalAssets/405/c_405910-l_1-k_qft.pdf

Thank you for considering my question.

Answer 1

TL;DR: It seems OP's troubles are caused by not constraining the integration region of the Feynman parameters $(x_1,x_2,x_3)$ properly.

The constraint is $$ x_1+x_2+x_3~=~1.\tag{A}$$ The denominator is $$\Delta~:=~x_1x_2+x_2x_3+x_3x_1~\geq~ 0, \qquad x_1,x_2,x_3~\geq~0. \tag{B}$$ Clearly $$\Delta~=~0\qquad \Leftrightarrow\qquad (x_1,x_2,x_3)\text{ belongs to a corner of the integration plane (A)}.\tag{C}$$ Note that $$\begin{align}\Delta~\stackrel{(A)}{=}~&x_1x_2+(x_1+x_2)(1-x_1-x_2) ~=~ (x_1\leftrightarrow x_2)\cr ~=~&x_1+x_2-(x_1^2+x_2^2+x_1x_2)\cr ~=~&x_1+x_2+\text{ higher orders}.\end{align}\tag{D}$$

1. If we remove 3 neighborhoods around the 3 corners, OP's first integral $$\begin{align} I_1(\epsilon)~:=~&\iiint_{\mathbb{R}_+^3}\! \mathrm{d}x_1~\mathrm{d}x_2~\mathrm{d}x_3 ~\Delta^{\epsilon-2}~\delta(1-\sum_{j=1}^3x_j)\cr ~\stackrel{(A)}{=}~ &\int_0^1\! \mathrm{d}x_1\int_0^{1-x_1}\!\mathrm{d}x_2~\Delta^{\epsilon-2}\cr ~\stackrel{(x_1\leftrightarrow x_2)}{=}& 2\int_0^1\! \mathrm{d}x_1\int_{x_1}^{1-x_1}\!\mathrm{d}x_2~\Delta^{\epsilon-2} \end{align}\tag{E}$$ becomes finite.

   The integral of 1 corner neighborhood $[0,\delta_1]\times[0,\delta_2]$ in the $(x_1,x_2)$ plane yields (up to higher orders) $$\begin{align}\int_0^{\delta_1}\! \mathrm{d}x_1\int_0^{\delta_2}\!\mathrm{d}x_2~(x_1+x_2)^{\epsilon-2} ~=~&\frac{1}{\epsilon-1}\int_0^{\delta_1}\! \mathrm{d}x_1\left[(x_1+x_2)^{\epsilon-1}\right]^{x_2=\delta_2}_{x_2=0}\cr ~=~&\frac{1}{\epsilon(\epsilon-1)}\left[(x_1+\delta_2)^{\epsilon}-x_1^{\epsilon}\right]^{x_1=\delta_1}_{x_1=0}\cr ~=~&\epsilon^{-1}+O(\epsilon^0). \end{align}\tag{F}$$ Here $1\gg \delta_1,\delta_2 \gg \epsilon\to 0.$ Eq. (F) explains OP's first integral, eq. (3.3.34) in Ref. 1.

2. OP's second integral is $$\begin{align} I_2(\epsilon)~:=~&\iiint_{\mathbb{R}_+^3}\! \mathrm{d}x_1~\mathrm{d}x_2~\mathrm{d}x_3\frac{x_1x_2x_3}{\Delta^{3-\epsilon}}\delta(1-\sum_{j=1}^3x_j)\cr ~=~&\iiint_{\mathbb{R}_+^3}\! \mathrm{d}x_1~\mathrm{d}x_2~\mathrm{d}x_3\frac{x_1x_2x_3}{\Delta^{3}}e^{\epsilon\ln\Delta}\delta(1-\sum_{j=1}^3x_j)\cr ~=~&I_2(\epsilon\!=\!0)+\epsilon C + O(\epsilon^2), \end{align}\tag{G}$$ where $C$ is defined in eq. (3.3.38) of Ref. 1. The coefficient in front of the next-to-leading $\epsilon$-term in eq. (G) differs from OP's eq. (2) by $1/2$.

   Note that potential singularities must come from the 3 corners. An investigation of the corner neighborhoods reveals that the integral is in fact finite.

   The leading term is a half: $$\begin{align} I_2(\epsilon\!=\!0)~:=~&\iiint_{\mathbb{R}_+^3}\! \mathrm{d}x_1~\mathrm{d}x_2~\mathrm{d}x_3\frac{x_1x_2x_3}{\Delta^3}\delta(1-\sum_{j=1}^3x_j)\cr ~\stackrel{(A)}{=}~& \int_0^1\! \mathrm{d}x_1\int_0^{1-x_1}\!\mathrm{d}x_2~\frac{x_1x_2(1-x_1-x_2)}{\Delta^3}\cr ~=~& \int_0^1\! \mathrm{d}x_1\frac{1-4x_1+3x_1^2+8x_1\sqrt{\frac{1-x_1}{1+3x_1}} {\rm artanh}\sqrt{\frac{1-x_1}{1+3x_1}}}{(1-x_1)^2(1+3x_1)^2}\cr ~=~& \left[\frac{-1+x_1+12x_1^2\sqrt{\frac{1-x_1}{1+3x_1}} {\rm artanh}\sqrt{\frac{1-x_1}{1+3x_1}}}{3(1+2x_1-3x_1^2)}\right]_0^1\cr ~=~&\frac{1}{2}. \end{align}\tag{H}$$ The integrals (H) were found using Mathematica.

References:

1. J.A. Minahan, 2011 MIT 8.323 QFT notes; subsection 3.3.3.