the same poles in a different order.

From the above exercise, we see that $I(S_r)$ obtains the form of a lengthy sum of terms of a similar kind: a product of residues $1/(2E_\alpha)$ from some set ${\mathcal S}_s$ of $N$ propagators, multiplied by a complicated sum of products of $\theta$ functions along with the product of the remaining propagators, with the ${\mathcal S}_s$ momenta placed on shell. Defining a function $A_{S_r}({\mathcal S}_s)$ to stand for the latter part of the result, we may write it in the form $$\begin{aligned}
I(S_r)&=&\sum_{{\mathcal S}_s} \prod_{k\in {\mathcal S}_s}\frac{1}{2E_k}\,A_{S_r}({\mathcal S}_s)\Big{|}_{\{q_0^k=iE_k\}}.\end{aligned}$$ A crucial realization is now the following: the sets of $N$ propagators obtained above, i.e. the ${\mathcal S}_s$, cannot contain any sets that are not part of the “superset” $S$, defined as all the possible choices of loop momenta in the original graph. This is a simple consequence of momentum conservation: we cannot take residues of a set of propagators whose momenta are linearly dependent. This implies that we may directly write the result in a form reminiscent of eq. (\[Iresmu0\]), $$\begin{aligned}
I(S_r)&=&\sum_{S_{r'}\in S}\;\prod_{i\in S_{r'}} \frac{1}{2E_i}\, \tilde{A}_{S_r}(S_{r'})\Big{|}_{\{q_0^i=iE_i\}}, \label{Iresmu1}\end{aligned}$$ where the tilde on $A$ highlights the fact that the summation now goes over the sets $S_{r'}$.

