I have a question regarding Sector Decomposition, which is briefly introduced in this paper arXiv: 0803.4177.

So far I played around with a toy example and applied the Sector Decomposition method to it. Let $I$ be the integral of interest for $\epsilon$ close to zero.

\begin{align} I= & \int_0^1 dz (z(1-z))^{-\epsilon}\int_0^1 dy (1-y)^{1-2\epsilon}y^{-1-\epsilon}\frac{2}{1-z(1-y)} \\ = & \int_0^1 dz (z(1-z))^{-\epsilon}\int_0^1 dy (1-y)^{1-2\epsilon}y^{-1-\epsilon}\frac{2}{1-(1-z)(1-y)} \end{align}

First question: The singular factor has a pole in $z=1$, $y=0$. Why do we have to shift the pole to the origin or at least a point, that lies on the line $x=y$?

\begin{align} I= & \underbrace{\int_0^1 dz \int_0^z dy (z(1-z))^{-\epsilon} (1-y)^{1-2\epsilon}y^{-1-\epsilon}\frac{2}{1-(1-z)(1-y)}}_{\textbf{(1)}} \\ &+\underbrace{\int_0^1 dy \int_0^y dz (z(1-z))^{-\epsilon} (1-y)^{1-2\epsilon}y^{-1-\epsilon}\frac{2}{1-(1-z)(1-y)}}_{\textbf{(2)}} \end{align}

Afterwards we decompose the integration domain in two triangular regions, such that the pole is "splitted" in some sense.

\begin{align} \textbf{(1)} &\overset{y=zt}{=} \int_0^1 dz \int_0^1 dt (z(1-z))^{-\epsilon} (1-zt)^{1-2\epsilon}(zt)^{-1-\epsilon}\frac{2}{1+t-zt} \\ \textbf{(2)} &\overset{z=yt}{=} \int_0^1 dy \int_0^1 dt (yt(1-yt))^{-\epsilon} (1-y)^{1-2\epsilon}y^{-1-\epsilon}\frac{2}{1+t-yt} \end{align}

The final step is technically the remapping of the triangular regions back to the unit square. The fraction has now poles outside of the integration domain and the singularities factorised.

To me it is still some sort of magic that the original integral, which has a singular fraction, can be rewritten in a way, that all singularities are just monomial factors. Here in particular $z^{-1-2\epsilon}$ and $t^{-1-\epsilon}$ for the first and $y^{-1-2\epsilon}$ for the second region. The paper just provides the method. Can someone help me to get insight in the method and a better understanding what is going on in there?


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