quantum field theory - Cutkosky rules for a two-loop diagram

Let's say we want to calculate the imaginary part of the following scalar diagram in $\varphi^3$ theory: This amplitude is given by the expression

$$i \mathcal{M} = i^5 \int \frac{d^4 \ell_1}{(2\pi)^4} \int \frac{d^4 \ell_2}{(2\pi)^4} \frac{1}{D_1 D_2 D_3 D_4 D_1},$$ where $D_k$ is the $k$-th denominator as in the figure. To take the imaginary part, we have to perform cuts according to Cutkosky. Here we have three possible cuts - we can cut the following lines : $12$, $234$, and again $12$. Employing the cutting rules, for the cut $k$-th line we put $$\frac{1}{D_k} \to -2\pi i \delta (D_k)$$ and get $$\Im \mathcal{M} = 2 \int \frac{d^4 \ell_1}{(2\pi)^4} \int \frac{d^4 \ell_2}{(2\pi)^4} \frac{(-2\pi i)^2}{D_3 D_4} \frac{\delta(D_1)}{D_1} \delta(D_2) + \int \frac{d^4 \ell_1}{(2\pi)^4} \int \frac{d^4 \ell_2}{(2\pi)^4} \frac{(-2\pi i)^3}{D_1^2} \delta(D_2) \delta(D_3) \delta(D_4).$$ The problem is, of course, the first term. How to interpret the explicitly divergent term $\delta(D_1)/D_1$?

Answer

Going back to Cutkosky's original paper (http://aip.scitation.org/doi/10.1063/1.1703676), it is clear he derives his result via residue theorem, as QuantumDot pointed out in his comment. Therefore, it seems natural that the generalization of the Cutkosky's cutting rule would have to analogous to the formula for the residue of a pole of order higher than one. Explicitly, if the cut propagator is raised to the $n$-th power, we should substitute

$$\frac{1}{D^n} \to (-2\pi i) \frac{(-1)^{n-1}}{(n-1)!} \delta^{(n-1)} (D).$$ In case $n=2$, we would then have $$\frac{1}{D^2} \to (2\pi i) \delta'(D).$$ While I have not yet checked this substitution rule in a real calculation, I suspect that it will hold.