The connected amplitude $T(s)$ satisfies the unitarity condition

$$2\mathrm{Im}T(s) = \frac{1}{\sqrt{s(s-4m^2)}} |T(s)|^2, \qquad (C.3)$$

as well as crossing

$$T(s) = T(4m^2 - s). \qquad (C.4)$$

At large energies $|s| \to \infty$, adding the correct $i\epsilon$ prescription, one sees easily that $T(s)$ goes to a constant given by

$$c_{\infty} = -2 \sum_{i=1}^{N_{\text{tot}}} \sqrt{m_i^2(4m^2 - m_i^2)}. \qquad (C.5)$$

In this way we can write a single subtracted dispersion relation

$$T(s) = c_{\infty} + \int_{4m^2}^{\infty} \frac{ds'}{\pi} \mathrm{Im}T(s') \left( \frac{1}{s' - s} + \frac{1}{s' - (4m^2 - s)} \right), \qquad (C.6)$$

where at large energies $\mathrm{Im}T(s') \sim \frac{1}{s'}$ so that the dispersive integral in (C.6) converges. The representation (C.6) is clearly very similar to (3.1) which we will be using in the higher dimensions.

It is also interesting to consider the behavior of the amplitude close to the two-particle threshold

$$T(s) = 4im\sqrt{s-4m^2} + O(s-4), \quad N_{\text{tot}} \text{ odd}, \qquad (C.7)$$

$$T(s) = ic_0(s - 4m^2)^{3/2} + O(s - 4), \quad N_{\text{tot}} \text{ even}. \qquad (C.8)$$

One can ask if we can construct the S-matrices above using iterations of unitarity and dispersion relations. In $d=2$, we analyzed this question in our previous work [26]. One simple scenario is to fix $c_\infty$ and try to iterate unitarity, starting from $T^{(0)}(s) = c_\infty$. Curiously, for the problem above, this will lead to divergent results. Therefore, while the 2d was reasonably easy to understand, it happens so that the class of problem which we are interested in this paper (with no bound states) is not easily amenable to dispersive iterations in $d=2$. It would be interesting to understand how to generalize our methods to this case and if this generalization can be useful in higher dimensions. Consequently, our approach in this previous work was to solve the problem with bound-states.

The situation is better in higher dimensions ($d=3,4$) where the simple iterations of unitarity and dispersion relations leads to convergent results shown in the text.

## D Analytic Regge limit

To derive the analytic Regge solution in $d=4$ given by (3.29), we need to evaluate the large $s$ expansion of the following integrals

$$I_1 \left( \alpha, \frac{s}{m^2} \right) = 2 \int_{4m^2}^{\infty} \frac{ds'}{\pi} \frac{1}{(\log \frac{s'}{m^2})^{\alpha}} \frac{\log \frac{s'+s-4m^2}{s'}}{s-4m^2}, \\ I_2 \left( \alpha, \frac{s}{m^2} \right) = \int_{4m^2}^{\infty} \frac{ds'}{\pi} \frac{1}{(\log \frac{s'}{m^2})^{\alpha}} \left[ \frac{1}{s'-s} \right]_{P.V.} \qquad (D.1)$$
This can be done for example by introducing the Mellin representation for the integration kernels, then performing the integral over $s'$, and finally deforming the Mellin contour to generate the large $s$ expansion. The result for the first few terms take the following form

$$ I_1 \left( \alpha, \frac{s}{m^2} \right) = \frac{2}{\pi} \frac{(\log s/m^2)^{1-\alpha}}{\alpha-1} \left( 1 + \frac{\alpha-1}{\log s/m^2} + \left( 1 + \frac{\pi^2}{6} \right) \frac{\alpha(\alpha-1)}{(\log s/m^2)^2} + \left( 1 + \frac{\pi^2}{6} \right) \frac{(\alpha+1)\alpha(\alpha-1)}{(\log s/m^2)^3} + \dots \right) + \dots, \quad (D.2) $$

$$ I_2 \left( \alpha, \frac{s}{m^2} \right) = \frac{1}{\pi} \frac{(\log s/m^2)^{1-\alpha}}{\alpha-1} \left( 1 - \frac{\pi^2}{3} \frac{\alpha(\alpha-1)}{(\log s/m^2)^2} - \frac{\pi^2}{45} \frac{(\alpha+2)(\alpha+1)\alpha(\alpha-1)}{(\log s/m^2)^4} + \dots \right) + \dots. \quad (D.3) $$

Similarly, by taking the derivatives with respect to $\alpha$ we can generate the integrands of the type $\frac{(\log \log \frac{s'}{m^2})^k}{(\log \frac{s'}{m^2})^\alpha}$. Writing down the expansion (3.29), plugging it into the unitarity relation (3.22), and using the formulas from this appendix one can check our claim.

## E Proof of convergence à la Atkinson for toy-model in $3 < d < 4$

In this appendix we prove the contracting properties of the unitarity map for the toy-model amplitude in $3 < d < 4$. Many of the estimates of various integrals closely follow the $d = 4$ case analyzed in [24].

Consider the following amplitude

$$ T(s,t) = \lambda + \int_{4m^2}^{\infty} \frac{ds'}{\pi} \frac{\rho(s')}{s' - \frac{4m^2}{3}} \left( \frac{s - \frac{4}{3}m^2}{s' - s} + \frac{t - \frac{4m^2}{3}}{s' - t} + \frac{u - \frac{4m^2}{3}}{s' - u} \right), \quad s+t+u = 4m^2. \quad (E.1) $$

We introduce $s = 4m^2/x$ so that $x \in [0, 1]$ and consider class of $\rho(x)$ such that

$$ \rho(0) = \rho(1) = 0. \quad (E.2) $$

We also consider $\rho(x)$ to be Hölder-continuous

$$ |\rho(x_1) - \rho(x_2)| \le \xi|x_1 - x_2|^\mu, \quad 0 < \mu < \min\left(\frac{d-3}{2}, \frac{4-d}{2}\right), \quad (E.3) $$

where $\mu < \frac{d-3}{2}$ will arise from imposing unitarity close to the two-particle threshold, whereas $\mu < \frac{4-d}{2}$ comes from unitarity in the Regge limit. Hölder-continuity is natural because it is preserved under iterations of unitarity and dispersion relations.

Following Atkinson, introduce the following norm in this functional space

$$ \|\rho\| = \sup_{0 \le x_1, x_2 \le 1} \frac{|\rho(x_1) - \rho(x_2)|}{|x_1 - x_2|^\mu}. \quad (E.4) $$

We re-write unitarity $|S_0| = 1$ in terms of a map $\Phi$ as:

$$ \rho = \Phi[\rho]. \quad (E.5) $$
which we solve iteratively. If $\rho$ is given, let us define

$$ \rho' := \Phi[\rho]. \qquad (E.6) $$

Given a bound on the norm $||\rho|| \le B$, in this appendix, we would like to show that $||\rho'| \le B$ to make sure that the iterated spectral density stays in the same space defined by $||\rho|| \le B$. Via continuity of the map, this will guarantee that some solutions exist. We then would like to show that the map is actually contracting, which will prove uniqueness.

Let us write down the iteration equation and thus the map $\Phi$ explicitly

$$ \begin{aligned} \rho'(x) &= \frac{\pi^{\frac{3-d}{2}}}{2^{d+2}\Gamma(\frac{d-1}{2})} x^{\frac{4-d}{2}} (1-x)^{\frac{d-3}{2}} \left( (c_d \mathrm{Re}f_0[\rho](x))^2 + \rho(x)^2 \right), \\ c_d \mathrm{Re}f_0[\rho](x) &= \lambda + \int_0^1 \frac{dx'}{\pi} \rho(x') \left( \mathrm{PV} \frac{3-x}{3-x'} \frac{1}{x-x'} + K_d(x,x') \right), \\ K_d(x,x') &= \frac{2}{x'} \left( {}_2F_1 \left( 1, \frac{d-2}{2}, d-2, -\frac{(1-x)x'}{x} \right) - \frac{3}{3-x'} \right), \end{aligned} \qquad (E.7) $$

where $c_d = 2^{2d-3}\pi^{\frac{d-3}{2}}\Gamma(\frac{d-1}{2})$. Note that $K_d(x, x')$ is regular when $x' = 0$. The universal prefactor $x^{\frac{4-d}{2}}(1-x)^{\frac{d-3}{2}}$ in the unitarity relation is the source of the bounds on $\mu$ in (3.34).

## E.1 Bounding the norm $||\rho'||$

Our first task is to bound

$$ \begin{aligned} & \left|\rho'(x_1) - \rho'(x_2)\right| \left(\frac{\pi^{\frac{3-d}{2}}}{2^{d+2}\Gamma(\frac{d-1}{2})}\right)^{-1} \\ &= |x_1^{\frac{4-d}{2}}(1-x_1)^{\frac{d-3}{2}}\left((c_d \mathrm{Re}f_0[\rho](x_1))^2 + \rho(x_1)^2\right) - x_2^{\frac{4-d}{2}}(1-x_2)^{\frac{d-3}{2}}\left((c_d \mathrm{Re}f_0[\rho](x_2))^2 + \rho(x_2)^2\right)| \\ & \le |x_1^{\frac{4-d}{2}}(1-x_1)^{\frac{d-3}{2}} - x_2^{\frac{4-d}{2}}(1-x_2)^{\frac{d-3}{2}}| \left|\left(c_d \mathrm{Re}f_0[\rho](x_1)\right)^2 + \rho(x_1)^2\right| \\ & \qquad + x_2^{\frac{4-d}{2}}(1-x_2)^{\frac{d-3}{2}} \left(\left|\left(c_d \mathrm{Re}f_0[\rho](x_1)\right)^2 - \left(c_d \mathrm{Re}f_0[\rho](x_2)\right)^2\right| + \left|\rho(x_1)^2 - \rho(x_2)^2\right|\right). \end{aligned} \qquad (E.8) $$

Let us now bound each of the terms. First of all, we notice that

$$ |x_1^{\frac{4-d}{2}}(1-x_1)^{\frac{d-3}{2}} - x_2^{\frac{4-d}{2}}(1-x_2)^{\frac{d-3}{2}}| \le |x_1 - x_2|^\mu, \quad 0 < \mu \le \min\left(\frac{d-3}{2}, \frac{4-d}{2}\right). \qquad (E.9) $$

Second, using the fact that $\rho(0) = 0$ we get

$$ \frac{|\rho(x)|}{|x|^{\mu}} \le ||\rho|| \le B, \qquad (E.10) $$

which in particular implies

$$ |\rho(x)| \le B. \qquad (E.11) $$

For the difference we also have $|\rho(x_1) - \rho(x_2)| \le B|x_1 - x_2|^\mu$. Finally, we need to bound the terms that involve the real part $\mathrm{Re}f_0$. We do it in two parts by first bounding the principal value integral, and then the part that involves $K_d(x, x')$.
### E.1.1 Bounding the principal value integral

Here we use Atkinson tricks. We first extend the principal value integral

$$ \int_{-2}^{2} \frac{dx'}{\pi} \rho(x') \text{PV} \frac{3-x}{3-x'} \frac{1}{x-x'} \quad (E.12) $$

where we simply continuously set $\rho(x') = 0$ for $x \notin (0, 1)$. This extension preserves Holder continuity

$$ |\rho(x_1) - \rho(x_2)| \le B|x_1 - x_2|^\mu, \quad -2 \le x_1, x_2 \le 2. \quad (E.13) $$

The only non-trivial case to check is when one of the variables, for example, $x_1 \in [0, 1]$ and $x_2 \notin [0, 1]$.

This allows us to use a simple trick

$$ \begin{aligned} \left| \int_{-2}^{2} \frac{dx'}{\pi} \rho(x') \text{PV} \frac{3-x}{3-x'} \frac{1}{x-x'} \right| &= \left| \int_{-2}^{2} \frac{dx'}{\pi} \frac{3-x}{3-x'} \frac{\rho(x') - \rho(x)}{x-x'} + \frac{\rho(x)}{\pi} \log \frac{2+x}{10-5x} \right| \\ &\le \frac{B}{\pi} \left( \frac{2^{\mu} ( {}_2F_1(1, \mu, 1+\mu, -2/3) + {}_2F_1(1, \mu, 1+\mu, 2/3))}{\mu} + \log 5 \right). \end{aligned} \quad (E.14) $$

Note that the bound becomes weak as $\mu \to 0$.

Similarly, we can bound

$$ \left| \int_{0}^{1} \frac{dx'}{\pi} \rho(x') K_{d}(x, x') \right| \le \frac{B}{\pi} \left( \frac{2}{\mu} + \frac{2}{3} \frac{{}_2F_1(1, 1+\mu, 2+\mu, 1/3)}{\mu+1} \right). \quad (E.15) $$

In this way we get the following bound

$$ \begin{aligned} &|(c_d \text{Re} f_0[\rho](x_1)| \le |\lambda| + \frac{B}{\pi} C_{\text{Re} f_0} \\ &C_{\text{Re} f_0} = \frac{2^{\mu} ( {}_2F_1(1, \mu, 1+\mu, -2/3) + {}_2F_1(1, \mu, 1+\mu, 2/3))}{\mu} \\ &\qquad + \log 5 + \frac{2}{\mu} + \frac{2}{3} \frac{{}_2F_1(1, 1+\mu, 2+\mu, 1/3)}{\mu+1}. \end{aligned} \quad (E.16) $$

It is clear that this bound could be highly sub-optimal given the fact that it does not allow for cancellations.

Finally, we need to bound $|(c_d \text{Re} f_0[\rho](x_1) - (c_d \text{Re} f_0[\rho](x_2))|$. Consider first the principal value part

$$ \begin{aligned} &\left| \int_{-2}^{2} \frac{dx'}{\pi} \rho(x') \text{PV} \left( \frac{3-x_1}{3-x'} \frac{1}{x_1-x'} - \frac{3-x_2}{3-x'} \frac{1}{x_2-x'} \right) \right| \le B_1 + B_2, \\ &B_1 = \left| \int_{-2}^{2} \frac{dx'}{\pi} \frac{3-x_1}{3-x'} \frac{\rho(x') - \rho(x_1)}{x_1-x'} - \int_{-2}^{2} \frac{dx'}{\pi} \frac{3-x_2}{3-x'} \frac{\rho(x') - \rho(x_2)}{x_2-x'} \right| \\ &B_2 = \left| \frac{\rho(x_1)}{\pi} \log \frac{2+x_1}{10-5x_1} - \frac{\rho(x_2)}{\pi} \log \frac{2+x_2}{10-5x_2} \right|. \end{aligned} \quad (E.17) $$

To bound $B_1$ let us assume that $x_2 \ge x_1$ (the case $x_1 \ge x_2$ leads to the same bound) and split the integration into two regions: $\Omega: |x' - x_1| \le 2(x_2 - x_1)$, and $\bar{\Omega}$ which is the
rest. Both points $x' = x_1$ and $x' = x_2$ lie inside $\Omega$. We also have $|x_1 \pm 2(x_2 - x_1)| \le 2$ so that $\Omega$ is inside the integration interval for any $x_1, x_2 \in [0, 1]$.

In this way we get

$$
\begin{align}
B_1 &\le B_{11} + B_{12} + B_{13}, \\
B_{11} &= \left| \int_{\Omega} \frac{dx'}{\pi} \frac{3-x_1}{3-x'} \left| \frac{\rho(x') - \rho(x_1)}{x_1 - x'} \right| + \int_{\Omega} \frac{dx'}{\pi} \frac{3-x_2}{3-x'} \left| \frac{\rho(x') - \rho(x_2)}{x_2 - x'} \right| \right|, \\
B_{12} &= \left| \int_{\bar{\Omega}} \frac{dx'}{\pi} (\rho(x') - \rho(x_1)) \left( \frac{3-x_1}{3-x'} \frac{1}{x_1 - x'} - \frac{3-x_2}{3-x'} \frac{1}{x_2 - x'} \right) \right|, \\
B_{13} &= \left| \int_{\bar{\Omega}} \frac{dx'}{\pi} (\rho(x_1) - \rho(x_2)) \frac{3-x_2}{3-x'} \frac{1}{x_2 - x'} \right|. \tag{E.18}
\end{align}
$$

For $B_{11}$ we directly use the Holder condition to get

$$
B_{11} \le \frac{B}{\pi\mu} (3 \times 2^{1+\mu} + 2(1+3^{\mu})) |x_1 - x_2|^{\mu}. \quad (E.19)
$$

For $B_{12}$ we get

$$
\begin{align}
B_{12} &\le B \int_{\bar{\Omega}} \frac{dx'}{\pi} |x' - x_1|^\mu \frac{|x_1 - x_2|}{|x' - x_1||x' - x_2|} = \frac{B|x_1 - x_2|}{\pi} \int_{\bar{\Omega}} dx'|x' - x_1|^{\mu-1}|x' - x_2|^{-1} \nonumber \\
&\le \frac{2^\mu B|x_1 - x_2|^\mu}{(1 - \mu)\pi}. \tag{E.20}
\end{align}
$$

For $B_{13}$ the bound takes the form

$$
B_{13} \le \frac{B|x_1 - x_2|^\mu}{\pi} \left| \int_{\bar{\Omega}} dx' \frac{3-x_2}{3-x'} \frac{1}{x_2-x'} \right| \le \log 15 \frac{B|x_1 - x_2|^\mu}{\pi}. \quad (E.21)
$$

Next we bound $B_2$ as follows

$$
\begin{align}
B_2 &\le \frac{|\rho(x_1) - \rho(x_2)|}{\pi} \left| \log \frac{2+x_1}{10-5x_1} \right| + \left| \frac{\rho(x_2)}{\pi} \right| \left| \log \frac{2+x_1}{10-5x_1} - \log \frac{2+x_2}{10-5x_2} \right| \nonumber \\
&\le \log 5 \frac{B}{\pi} |x_1 - x_2|^\mu + \frac{B}{\pi} \log \left(1 + \frac{4(x_2 - x_1)}{(2+x_1)(2-x_2)}\right). \tag{E.22}
\end{align}
$$

Using

$$
\log(1 + A) \le A, \quad A \ge 0, \tag{E.23}
$$

we finally get

$$
B_2 \le \frac{B}{\pi} (2 + \log 5) |x_1 - x_2|^\mu. \quad (E.24)
$$

Our next step is to bound the difference term that involves $K_d(x, x')$. We get that

$$
\left| \int_0^1 \frac{dx'}{\pi} \rho(x') (K_d(x_1, x') - K_d(x_2, x')) \right| \le \frac{B|x_1 - x_2|^\mu}{\pi} \frac{\sqrt{\pi} 2^{d-2} \Gamma\left(\frac{d-1}{2}\right) \csc(\pi\mu) \Gamma\left(\frac{d}{2} - \mu - 1\right)}{\Gamma(d-\mu-2)}. \quad (E.25)
$$

To derive this formula we used (E.10), after which the integral can be taken exactly.
To summarize, in this part of the computation we have derived the following bound

$$ \|(c_d \mathrm{Re} f_0[\rho])\| \le \frac{B}{\pi} C_{\mathrm{norm}}, \qquad (\mathrm{E}.26) $$

$$ C_{\mathrm{norm}} = \frac{\sqrt{\pi} 2^{d-2} \Gamma\left(\frac{d-1}{2}\right) \csc(\pi \mu) \Gamma\left(\frac{d}{2} - \mu - 1\right)}{\Gamma(d - \mu - 2)} + \frac{2(3\ 2^\mu + 3^\mu + 1)}{\mu} + \frac{2^\mu}{1 - \mu} + 2 + \log 75. \qquad (\mathrm{E}.27) $$

Putting everything together we get the desired bound on the norm of the spectral density after the unitarity iteration

$$ \|\rho'\| \le B', \qquad (\mathrm{E}.28) $$

$$ B' = \frac{\pi^{\frac{3-d}{2}}}{2^{d+2}\Gamma(\frac{d-1}{2})} \left( \left( |\lambda| + \frac{B}{\pi} C_{\mathrm{Re} f_0} \right)^2 + 2B^2 \frac{C_{\mathrm{norm}} C_{\mathrm{Re} f_0}}{\pi^2} + 3B^2 \right), \qquad (\mathrm{E}.29) $$

where $C_{\mathrm{Re} f_0}$ and $C_{\mathrm{norm}}$ are only functions of $\mu$ and $d$. One can easily check using the explicit formula above that for small enough $\lambda$ and $3 < d < 4$ we have $B' \le B$.

## E.2 Proving the contracting property of the map

As a next step we would like to bound the norm of the images in terms of the norm of the pre-images. In other words, we would like to derive a bound of the type

$$ \|\rho'_2 - \rho'_1\| \le k \|\rho_2 - \rho_1\|. \qquad (\mathrm{E}.30) $$

To derive such a bound we start by noticing that the following bound on the real part holds

$$ |c_d \mathrm{Re} f_0[\rho_1](x) - c_d \mathrm{Re} f_0[\rho_2](x)| = \left| \int_0^1 \frac{dx'}{\pi} (\rho_1(x') - \rho_2(x')) \left( \mathrm{PV} \frac{3-x}{3-x'} \frac{1}{x-x'} + K_d(x,x') \right) \right| \\ \le C_{\mathrm{Re} f_0} \frac{\|\rho_2 - \rho_1\|}{\pi} \qquad (\mathrm{E}.31) $$

where we trivially used the results of the previous section.

Similarly, we have

$$ \|c_d \mathrm{Re} f_0[\rho_1] - c_d \mathrm{Re} f_0[\rho_2]\| \le C_{\mathrm{norm}} \frac{\|\rho_2 - \rho_1\|}{\pi} \qquad (\mathrm{E}.32) $$

We are now ready to derive the desired bound

$$ \begin{aligned} & (\rho'_1(x_1) - \rho'_2(x_1) - (\rho'_1(x_2) - \rho'_2(x_2))) \left( \frac{\pi^{\frac{3-d}{2}}}{2^{d+2}\Gamma(\frac{d-1}{2})} \right)^{-1} \\ &= x_1^{\frac{4-d}{2}} (1 - x_1)^{\frac{d-3}{2}} \left( (c_d \mathrm{Re} f_0[\rho_1](x_1))^2 + \rho_1(x_1)^2 - (c_d \mathrm{Re} f_0[\rho_2](x_1))^2 - \rho_2(x_1)^2 \right) \quad (\mathrm{E}.33) \\ & \quad - x_2^{\frac{4-d}{2}} (1 - x_2)^{\frac{d-3}{2}} \left( (c_d \mathrm{Re} f_0[\rho_1](x_2))^2 + \rho_1(x_2)^2 - (c_d \mathrm{Re} f_0[\rho_2](x_2))^2 - \rho_2(x_2)^2 \right) \end{aligned} $$
As before we split the r.h.s. in three parts

$$
\begin{align}
I_1 &= (x_1^{\frac{4-d}{2}} (1-x_1)^{\frac{d-3}{2}} - x_2^{\frac{4-d}{2}} (1-x_2)^{\frac{d-3}{2}}) \left( (c_d \mathrm{Re} f_0[\rho_1](x_1))^2 + \rho_1 (x_1)^2 - (c_d \mathrm{Re} f_0[\rho_2](x_1))^2 - \rho_2 (x_1)^2 \right) \nonumber \\
I_2 &= x_2^{\frac{4-d}{2}} (1-x_2)^{\frac{d-3}{2}} \left( (c_d \mathrm{Re} f_0[\rho_1](x_1))^2 - (c_d \mathrm{Re} f_0[\rho_2](x_1))^2 - (c_d \mathrm{Re} f_0[\rho_1](x_2))^2 + (c_d \mathrm{Re} f_0[\rho_2](x_2))^2 \right) \nonumber \\
I_3 &= x_2^{\frac{4-d}{2}} (1-x_2)^{\frac{d-3}{2}} \left( \rho_1 (x_1)^2 - \rho_2 (x_1)^2 - \rho_1 (x_2)^2 + \rho_2 (x_2)^2 \right). \tag{E.34}
\end{align}
$$

and bound each part separately.

Using the results that we have already derived we get

$$
|I_1| \le |x_1 - x_2|^\mu |\rho_2 - \rho_1| \left( 2 \frac{C_{\mathrm{Re}f_0}}{\pi} \left( |\lambda| + \frac{B}{\pi} C_{\mathrm{Re}f_0} \right) + 2B \right). \quad (\mathrm{E}.35)
$$

Next we proceed to $I_3$ where we add and subtract $(\rho_1(x_2) + \rho_2(x_2))(\rho_1(x_1) - \rho_2(x_1))$ to get

$$
|I_3| \le 4B|x_1 - x_2|^\mu |\rho_2 - \rho_1|. \quad (E.36)
$$

Let us present this step in a bit more detail, we write

$$
\begin{align}
I_3 ={}& (\rho_1(x_1) - \rho_1(x_2))(\rho_1(x_1) - \rho_2(x_1)) + (\rho_2(x_1) - \rho_2(x_2))(\rho_1(x_1) - \rho_2(x_1)) \nonumber \\
& + (\rho_1(x_2) + \rho_2(x_2))(\rho_1(x_1) - \rho_2(x_1) - \rho_1(x_2) + \rho_2(x_2)), \tag{E.37}
\end{align}
$$

and then bound each of the three terms separately.

And finally we get using the same trick for the real part of the partial wave

$$
|I_2| \le |x_1 - x_2|^\mu |\rho_2 - \rho_1| \left( \frac{2}{\pi} \left( \frac{B}{\pi} C_{\text{norm}} C_{\text{Ref}_0} + \left( |\lambda| + \frac{B}{\pi} C_{\text{Ref}_0} \right) C_{\text{norm}} \right) \right). \quad (\text{E.38})
$$

In this way we get for $k$ in (3.37)

$$
k = \frac{\pi^{\frac{3-d}{2}}}{2^{d+2}\Gamma(\frac{d-1}{2})} \left( |\lambda| \frac{2(C_{\text{Ref}_0} + C_{\text{norm}})}{\pi} + \frac{B}{\pi^2} (4C_{\text{Ref}_0}C_{\text{norm}} + 2C_{\text{Ref}_0}^2) + 6B \right). \quad (\text{E.39})
$$

We can now find $d, \mu, B, \lambda$ such that

$$
B' \le B, \quad k < 1. \tag{E.40}
$$

Given such parameters we can show that iterations converge to a unique fixed point. First, let us recall that the space of Hölder continuous function satisfying $||\rho|| \le B$ is complete. In other words, if we find a Cauchy sequence of functions $||\rho_n|| \le B$, such that for any $\epsilon > 0$ there exists $N$ for which

$$
\| \rho_{N+p} - \rho_N \| < \epsilon, \quad p \in \mathbb{N}, \tag{E.41}
$$

then $\rho_n \to \rho_*$ and $||\rho_*|| \le B$.

To demonstrate (E.41) we proceed as follows

$$
\begin{equation}
\begin{split}
\|\rho_{N+p} - \rho_N\| \le \sum_{m=N}^{N+p-1} \|\rho_{m+1} - \rho_m\| \le \|\rho_1 - \rho_0\| \sum_{m=N}^{N+p-1} k^m \le \|\rho_1 - \rho_0\| \frac{k^N}{1-k} \le 2B \frac{k^N}{1-k}.
\end{split}
\tag{E.42}
\end{equation}
$$

Therefore, given $\epsilon$ we choose $N$ so that $2B \frac{k^N}{1-k} < \epsilon$ is satisfied. This is always possible since $k < 1$. By setting $p = \infty$ in the formula above we converge to the limiting solution as follows

$$
\| \rho_* - \rho_N \| \le 2B \frac{k^N}{1-k}. \qquad (\text{E.43})
$$
### E.3 Generalization to $d = 3$ and $d = 4$

Let us discuss possible generalizations of the argument above to $d = 3$ and $d = 4$. An immediate problem is that from (3.34) it follows that $\mu = 0$, and many of the bounds of the previous section blow up for $\mu = 0$. Here we discuss a method to fix this problem inspired both by the success of our numerical analysis of these cases, as well as the trick used by Atkinson in [16].

Let us for concreteness discuss the case of $d = 4$.<fnref n="34" /> The basic idea is to modify the unitarity equation

$$ \rho'(x) = \frac{h(x)}{32\pi} (1-x)^{\frac{1}{2}} \left( (16\pi \text{Re} f_0[\rho](x))^2 + \rho(x)^2 \right), \qquad (\text{E.44}) $$

where we choose

$$ h(x) = \theta(x - x_0) + \frac{x}{x_0} \theta(x_0 - x). \qquad (\text{E.45}) $$

The role of $h(x)$ is to effectively soften the behavior of $\rho'(x)$ in the Regge limit $x \to 0$. With this modification we can consider a class of Hölder-continuous functions with $\mu = 1/2$ and run the argument of the previous section. All the bounds will now depend on $x_0$.

In a physical theory we would like to set $h(x) = 1$ or $x_0 = 0$. Naively, as explained above, the algorithm will stop converging (or at least we cannot argue for convergence). It is therefore interesting to see how we effectively solved this problem in the numerical implementation of the algorithm in $d = 4$.

Effectively, we considered *iteration-dependent* modification of the unitarity equation

$$ h_n(x) = \theta(x - x_n) + \frac{x}{x_n} \theta(x_n - x), \qquad (\text{E.46}) $$

where we chose $x_n$ such that $\lim_{n \to \infty} x_n = 0$. We observed convergence and smooth extrapolation of the solutions to the UV as we remove the cutoff $x_n$, where we could smoothly match to the analytic UV complete solution.

The fact that such an algorithm works is based on cancellation between various terms as we approach the solution better and better, see for example the sum rule (3.31), and therefore it clearly goes beyond the proof of the previous subsection based on a brute-force bounding of each term separately.

An alternative approach in $d = 4$ that does not require introduction of $h(x)$ and subtleties discussed here is to consider the Mandelstam equation with no subtractions (so that only the double spectral density is present), and construct amplitudes starting from a given $\rho_{\text{MP}}(s, t)$, as shown by Atkinson in [15], in this case one can again argue that the iterations of unitarity lead to a unique solution.

In $d = 3$ the modified unitarity equation takes the form

$$ \rho'(x) = \frac{\sqrt{x}h(x)}{32} \left( (8\text{Re}f_0)^2 + \rho(x)^2 \right), \qquad (\text{E.47}) $$

and in this case the regulator is needed close to the two-particle threshold

$$ h(x) = \theta(x_0 - x) + \frac{1-x}{1-x_0} \theta(x - x_0). \qquad (\text{E.48}) $$

<footnote n="34">³⁴The case of $d = 3$ is essentially identical, where the Regge limit is replaced by the near two-particle threshold region.</footnote>
# F Perturbative $\phi^4$

In this appendix we explicitly perform the first two unitarity iterations for the 2PR amplitude. It is completely equivalent to the two-loop computation in $\frac{\lambda}{4!}\phi^4$ theory.

## F.1 $d=4$

We expand the amplitude in the powers of $\lambda$

$$T(s,t) = \lambda T^{(0)}(s,t) + \lambda^2 T^{(1)}(s,t) + \lambda^3 T^{(2)}(s,t) + \dots \quad (F.1)$$

We would like to generate the usual perturbative expansion using dispersive iterations of unitarity with one subtraction described in the main text of the paper.

We initialize the iteration by setting

$$T^{(0)} = 1. \qquad (F.2)$$

From this we get using (2.13)

$$\rho^{(1)}(s) = \frac{1}{32\pi} \sqrt{1 - \frac{4m^2}{s}}, \qquad (F.3)$$

where we used the fact that $f_0^{(0)}(s) = \frac{1}{16\pi}$.

Plugging $\rho^{(1)}(s)$ into the Mandelstam representation (1.1) and using the fact that

$$\int_{4m^2}^{\infty} \frac{ds'}{\pi} \frac{(s-s_0)}{(s'-s)(s'-s_0)} \frac{1}{32\pi} \sqrt{1-\frac{4m^2}{s'}} = -\frac{1}{16\pi^2} (I(s)-I(s_0)), \quad (F.4)$$

$$I\left(\frac{4m^2}{1+y^2}\right) \equiv \int_0^1 dx \frac{1}{1+\frac{x^2}{y^2}} = \frac{iy}{2} \log \frac{y-i}{y+i}, \qquad (F.5)$$

we get that

$$T^{(1)}(s,t) = \frac{3}{16\pi^2} I(s_0) - \frac{1}{16\pi^2} (I(s) + I(u) + I(t)) \quad (F.6)$$

Next we compute $\rho^{(2)}(s)$ which takes the form

$$\rho^{(2)}(s) = 8\pi \frac{(s - 4m^2)^{1/2}}{\sqrt{s}} (f_0^{(0)}(s)(f_0^{(1)}(s))^* + (f_0^{(0)}(s))^* f_0^{(1)}(s)) = \frac{(s - 4m^2)^{1/2}}{\sqrt{s}} \mathrm{Re} f_0^{(1)}(s). \quad (F.7)$$

We can use (F.6) to get

$$\begin{aligned} \mathrm{Re} f_0^{(1)}(s) &= \frac{1}{32\pi} \int_{-1}^{1} dz \mathrm{Re} T^{(1)}(s, t(z)) \\ &= \frac{1}{16\pi} \left( \frac{3}{16\pi^2} I(s_0) - \frac{1}{16\pi^2} \mathrm{Re} I(s) - \frac{1}{16\pi^2} \int_{-1}^{1} dz \, I\left(-\frac{(s-4m^2)(1-z)}{2}\right) \right) \\ &= -\frac{1}{256\pi^3} \left( 3I(s_0) - \mathrm{Re} I(s) + \frac{m^2}{s-4m^2} \log^2 \left( \frac{1-\sqrt{1-\frac{4m^2}{s}}}{1+\sqrt{1-\frac{4m^2}{s}}} \right) \right. \\ & \qquad \left. - \frac{1}{\sqrt{1-\frac{4m^2}{s}}} \log \left( \frac{1-\sqrt{1-\frac{4m^2}{s}}}{1+\sqrt{1-\frac{4m^2}{s}}} \right) - 1 \right). \end{aligned} \quad (F.8)$$
Finally, we do the dispersive integrals (1.1) to compute $T^{(2)}(s, t)$. The result takes the following form

$$ T^{(2)}(s,t) = \frac{3}{8\pi^2} I(s_0)T^{(1)}(s,t) + \frac{1}{256\pi^4} (I(s)^2 + I(u)^2 + I(t)^2) \\ - \frac{1}{128\pi^4} (I_3(s) + I_3(u) + I_3(t)) - c_0^{(2)}, \qquad (F.9) $$

where

$$ I_3(\frac{4m^2}{1+y^2}) \equiv \frac{I(\frac{4m^2}{1+y^2})^2}{y^2} + \left( \frac{\pi^2 (y^2+1)}{12y^2} + 1 \right) I(\frac{4m^2}{1+y^2}) - \frac{(y^2+1) I(\frac{4m^2}{1+y^2})^3}{3y^4}, \qquad (F.10) $$

and the constant $c_0^{(2)}$ is fixed by the condition $T^{(2)}(\frac{4m^2}{3}, \frac{4m^2}{3}) = 0$. We see therefore that to the first two non-trivial orders dispersive iterations of unitarity and the usual Feynman perturbation theory coincide.

It is also interesting to examine the leading Regge behavior of the amplitude at each order computed so far. We get

$$ \lim_{s \to \infty} T(s,t) = \lambda \left( 1 - \frac{\lambda}{16\pi^2} \log s + \frac{3}{2} \left( \frac{\lambda}{16\pi^2} \log s \right)^2 + \dots \right). \qquad (F.11) $$

The behavior of the amplitude is correlated with the sign of the $\beta$-function. In our convention $\lambda < 0$ corresponds to the Landau pole in the UV and stable vacuum, whereas $\lambda > 0$ corresponds to asymptotic freedom, with the potential not bounded from below. We see from (F.11) that for negative $\lambda$ all the terms add up and the amplitude becomes big at high energies, whereas for $\lambda > 0$ we have cancelations.

At order $\lambda^4$ two new features appear. First of all, using $\rho^{(2)}(s)$ inside the Mandelstam equation (1.6), we get a nontrivial $\rho_{el}^{(4)}(s,t)$

$$ \rho_{\text{el}}^{(4)}(s,t) = \frac{(s - 4m^2)^{\frac{1}{2}}}{16\pi^2 \sqrt{s}} \frac{1}{(32\pi)^2} \int_{z_1}^{\infty} d\eta' \int_{z_1}^{\infty} d\eta'' \sqrt{1 - \frac{4m^2}{t(\eta')}} \sqrt{1 - \frac{4m^2}{t(\eta'')}} \frac{\theta(z - \eta_+)}{\sqrt{(z - \eta_ -)(z - \eta_ +)}}, \qquad (F.12) $$

which can be straightforwardly evaluated numerically. This corresponds to the “Aks” diagram figure 61.

Another novelty is that both the S-wave production amplitude $\eta_{\text{MP}}^{(4)}(s)$, as well as the multi-particle double spectral density are non-zero at this order $\rho_{\text{MP}}^{(4)}(s,t)$. These are due to the frog and open envelope graphs, see figure 1. Therefore, for our iteration method to continue agreeing with the Feynman graph computation at order $\lambda^4$, $\rho_{\text{MP}}^{(4)}(s,t)$ and $\eta_{\text{MP}}^{(4)}(s)$ should be independently computed and provided as an input.

## F.2 $d=3$

We now repeat the same exercise in $d=3$. We initialize the iteration by setting

$$ T^{(0)} = 1. \qquad (F.13) $$
From this we get using (2.13)

$$ \rho^{(1)}(s) = \frac{1}{16\sqrt{s}}, \qquad (F.14) $$

where we used the fact that $f_0^{(0)}(s) = \frac{1}{8}$.

Plugging $\rho^{(1)}(s)$ into the Mandelstam representation (1.1) and using the fact that

$$ \int_{4m^2}^{\infty} \frac{ds'}{\pi} \frac{(s-s_0)}{(s'-s)(s'-s_0)} \frac{1}{16\sqrt{s}} = \frac{1}{16\pi} (\tilde{I}(s) - \tilde{I}(s_0)), \qquad (F.15) $$

$$ \tilde{I}(s) = \int_{4m^2}^{\infty} \frac{ds'}{s'-s} \frac{1}{\sqrt{s'}} = \frac{\log \frac{1+\frac{\sqrt{s}}{2m}}{1-\frac{\sqrt{s}}{2m}}}{\sqrt{s}}, \qquad (F.16) $$

we get that

$$ T^{(1)}(s,t) = -\frac{3}{16\pi}\tilde{I}(s_0) + \frac{1}{16\pi}(\tilde{I}(s) + \tilde{I}(u) + \tilde{I}(t)). \qquad (F.17) $$

Next we compute $\rho^{(2)}(s)$ which takes the form

$$ \rho^{(2)}(s) = \frac{4}{\sqrt{s}} (f_0^{(0)}(s)(f_0^{(1)}(s))^* + (f_0^{(0)}(s))^* f_0^{(1)}(s)) = \frac{1}{\sqrt{s}} \mathrm{Re} f_0^{(1)}(s). \qquad (F.18) $$

We can use (F.17) to get

$$ \begin{aligned} \mathrm{Re} f_0^{(1)}(s) &= \frac{1}{16\pi} \int_0^{2\pi} d\theta \mathrm{Re} T^{(1)}(s, t(\theta)) \\ &= \frac{1}{128\pi} (\mathrm{Re}\tilde{I}(s) - 3\tilde{I}(s_0)) + \frac{1}{32\pi} \frac{\log \frac{\sqrt{s}}{2m} + \sqrt{\frac{s}{4m^2} - 1}}{\sqrt{s} - 4m^2}, \end{aligned} \qquad (F.19) $$

where $\mathrm{Re}\tilde{I}(s) = \frac{\log \frac{1+\frac{\sqrt{s}}{2m}}{\sqrt{s}}}{\sqrt{s}-1}$ for $s > 4m^2$.

Finally, we do the dispersive integrals (1.1) to compute $T^{(2)}(s,t)$. The result takes the following form

$$ T^{(2)}(s,t) = -\frac{3}{8\pi}\tilde{I}_1(s_0)T^{(1)}(s,t) + \frac{\tilde{I}_2(s) + \tilde{I}_2(t) + \tilde{I}_2(u)}{128\pi^2} + \frac{\tilde{I}_3(s) + \tilde{I}_3(t) + \tilde{I}_3(u)}{64\pi^2} - \tilde{c}_0^{(2)}, \qquad (F.20) $$

where

$$ s\tilde{I}_2(s) = \mathrm{Li}_2\left(\frac{1}{2} - \frac{\sqrt{s}}{4m}\right) + \mathrm{Li}_2\left(\frac{1}{2} + \frac{\sqrt{s}}{4m}\right) + \frac{1}{4}\log^2\left(\frac{1}{4} - \frac{s}{16m^2}\right) \\ + \frac{1}{4}\log^2\left(\frac{2m + \sqrt{s}}{2m - \sqrt{s}}\right) - \frac{\pi^2}{6}, \qquad (F.21) $$

and

$$ \tilde{I}_3(s) = \frac{\mathrm{Li}_2\left(\frac{-2m^2+s+\sqrt{s}\sqrt{s-4m^2}}{2m^2}\right) - \mathrm{Li}_2\left(\frac{2m^2}{-2m^2+s+\sqrt{s}\sqrt{s-4m^2}}\right)}{\sqrt{s}\sqrt{s-4m^2}}, \qquad (F.22) $$

and the constant $\tilde{c}_0^{(2)}$ is fixed by the condition $T^{(2)}(\frac{4m^2}{3}, \frac{4m^2}{3}) = 0$.
Expanding the amplitude at large $s$ we get

$$ 
\begin{aligned}
T(s, 0) & \approx \lambda - 0.04817\lambda^2 + 0.0026\lambda^3, \\
\mathcal{T}(s, \theta) & \approx \lambda - 0.06807\lambda^2 + 0.0038\lambda^3,
\end{aligned} 
\qquad (\text{F.23}) 
$$

where in the first line we consider the forward amplitude $t=0$, and in the second line we consider scattering at fixed angles $\theta \neq 0, \pi$. Setting $\lambda = -\pi$, we get good agreement with figure 16.

## G Aks graph inelasticity

In this appendix we first compute the simplest possible $\eta_{\text{MP}}(s)$ (multi-particle production) in $\phi^4$ in $d=3$ and $d=4$. Let us start by recalling the definition of $\eta_{\text{MP}}(s)$ given in (1.10). Using the fact that $S_J(s) = 1 + i \frac{(s-4m^2)^{\frac{d-3}{2}}}{\sqrt{s}} f_J(s)$, we can rewrite it as follows

$$ 2\text{Im}f_J(s) = \frac{(s-4m^2)^{\frac{d-3}{2}}}{\sqrt{s}}|f_J(s)|^2 + \frac{\sqrt{s}}{(s-4m^2)^{\frac{d-3}{2}}} \eta_{\text{MP}}(s). \qquad (\text{G.1}) $$

The leading order contribution to $\eta_{\text{MP}}(s)$ enters at order $\lambda^4$ and is given by two graphs, see figure (61), which we can call Aks and open envelope graphs. The complete $2 \to 4$ amplitude that is relevant for our computation is

$$ T(p_1, p_2; q_5, q_6, q_7, q_8) = -\lambda^2 \left( \frac{1}{(p_1 - q_{56})^2 - m^2} + \frac{1}{(p_1 - q_{57})^2 - m^2} + \frac{1}{(p_1 - q_{58})^2 - m^2} + (p_1 \leftrightarrow p_2) \right), \qquad (\text{G.2}) $$

$$ T^*(q_5, q_6, q_7, q_8; p_3, p_4) = -\lambda^2 \left( \frac{1}{(p_3 - q_{56})^2 - m^2} + \frac{1}{(p_3 - q_{57})^2 - m^2} + \frac{1}{(p_3 - q_{58})^2 - m^2} + (p_3 \leftrightarrow p_4) \right), \qquad (\text{G.3}) $$

where $q_{56}^{\mu} = q_5^{\mu} + q_6^{\mu}$, etc. Doing unitarity gluing with these amplitudes leads to

$$ \eta_{\text{MP}}(s) = 2 \times 6\eta_{\text{Aks}}(s) + 2 \times 12\eta_{\text{oe}}(s), \qquad (\text{G.4}) $$

where Aks stands for the Aks graph, and oe for the open envelope graph, see figure 61.35 To keep formulas shorter we do not write below the $\lambda$ factor.

### G.1 $d=3$

The four-body discontinuity takes the form

$$ 2T_s = \frac{1}{4!} \int \frac{d^2 \vec{q}_5}{(2\pi)^2 (2E_{\vec{q}_5})} \frac{d^2 \vec{q}_6}{(2\pi)^2 (2E_{\vec{q}_6})} \frac{d^2 \vec{q}_7}{(2\pi)^2 (2E_{\vec{q}_7})} \frac{d^2 \vec{q}_8}{(2\pi)^2 (2E_{\vec{q}_8})} \times (2\pi)^3 \delta(p_1 + p_2 - q_5 - q_6 - q_7 - q_8) T(p_1, p_2; q_5, q_6, q_7, q_8) T^*(q_5, q_6, q_7, q_8; p_3, p_4), \qquad (\text{G.5}) $$

<footnote n="35">35 We call it the Aks graph because it represents inelasticity that enters into the proof of the Aks theorem [27].</footnote>
<figure><img src="image_62.png" /><figcaption>**Figure 61.** Three-loop diagrams in $\phi^4$ that contribute to the multi-particle production. Double discontinuity of the Aks diagram is captured by our iteration process. The open envelope diagram is the leading contribution to $\rho_{\text{MP}}(s, t)$, which we do not compute in the present paper. Both diagrams contribute to $\eta_{\text{MP}}(s)$.</figcaption></figure>

where $s = (p_1 + p_2)^2$ and $t = (p_1 - p_3)^2$. We will be interested in the amplitudes

$$ T(p_1, p_2; q_5, q_6, q_7, q_8) = - \frac{1}{(q_5 + q_6 - p_1)^2 - m^2} \\ T^*(q_5, q_6, q_7, q_8; p_3, p_4) = - \frac{1}{(q_7 + q_8 - p_4)^2 - m^2}. \qquad (G.6) $$

Let us first transform the measure using the fact that the amplitudes (G.6) only depend on the combinations of momenta $q_5^\mu + q_6^\mu$ and $q_7^\mu + q_8^\mu$. We define the effective invariant masses of these pairs of particles $(q_5+q_6)^2 = m_{56}^2$, $(q_7+q_8)^2 = m_{78}^2$ and introduce integration over the corresponding 3-momenta $q_{56}^\mu \equiv q_5^\mu + q_6^\mu$, $q_{78}^\mu \equiv q_7^\mu + q_8^\mu$

$$ 2T_s = \frac{1}{4!} \int_{4m^2}^{(\sqrt{s}-2m)^2} dm_{56}^2 dm_{78}^2 \int d^3 q_{56} d^3 q_{78} \delta^+(q_{56}^2 - m_{56}^2) \delta^+(q_{78}^2 - m_{78}^2) T(p_1, q_{56}) T^*(p_4, q_{78}) \\ \times (2\pi)^3 \delta(p_1 + p_2 - q_{56} - q_{78}) \mathcal{I}, \qquad (G.7) $$

where the maximum value of $m_{56}^2$ is achieved by setting $\vec{q}_7 = \vec{q}_8 = 0$, and

$$ \mathcal{I} = \frac{1}{(2\pi)^8} \left( \int d^3 q_5 \delta^+(q_5^2 - m^2) \int d^3 q_6 \delta^+(q_6^2 - m^2) \delta^{(3)}(q_{56} - q_5 - q_6) \right) \\ \times \left( \int d^3 q_7 \delta^+(q_7^2 - m^2) \int d^3 q_8 \delta^+(q_8^2 - m^2) \delta^{(3)}(q_{78} - q_7 - q_8) \right). \qquad (G.8) $$

It is convenient to evaluate integrals in the brackets in the center-of-mass frame, for example $q_{78} = (m_{78}, \vec{0})$ to get

$$ \int d^3 q_7 \delta^+(q_7^2 - m^2) \int d^3 q_8 \delta^+(q_8^2 - m^2) \delta^{(3)}(q_{78} - q_7 - q_8) = \int \frac{d^2 \vec{q}_7}{4E_{\vec{q}_7}^2} \delta(m_{78} - 2E_{\vec{q}_7}) \\ = \frac{\pi}{2m_{78}}, \quad m_{78} \ge 2m. \qquad (G.9) $$
In this way we get $\mathcal{I} = \frac{1}{16} \frac{1}{(2\pi)^6} \frac{1}{m_{56}m_{78}}$ and we are left with the following unitarity integral

$$2T_s = \frac{(2\pi)^{-3}}{4!} \frac{1}{16} \int_{4m^2}^{(\sqrt{s}-2m)^2} \frac{dm_{56}^2}{m_{56}} \frac{dm_{78}^2}{m_{78}} \times \int d^3 q_{56} d^3 q_{78} \delta^+(q_{56}^2 - m_{56}^2) \delta^+(q_{78}^2 - m_{78}^2) T(p_1, q_{56}) T^*(p_4, q_{78}) \delta(p_1 + p_2 - q_{56} - q_{78}). \quad (G.10)$$

We proceed by evaluating the integral above in the center-of-mass frame $\vec{p}_1 + \vec{p}_2 = \vec{q}_{56} + \vec{q}_{78} = 0$

$$\int d^3 q_{56} d^3 q_{78} \delta^+(q_{56}^2 - m_{56}^2) \delta^+(q_{78}^2 - m_{78}^2) T(p_1, q_{56}) T^*(p_4, q_{78}) \delta(p_1 + p_2 - q_{56} - q_{78}) = \int \frac{d^2 \vec{q}_{56}}{4E_{\vec{q}_{56},m_{56}} E_{\vec{q}_{56},m_{78}}} \delta(\sqrt{s} - E_{\vec{q}_{56},m_{56}} - E_{-\vec{q}_{56},m_{78}}) T(p_1, q_{56}) T^*(p_4, q_{78}) \quad (G.11)$$

For the amplitudes we have

$$\begin{aligned} T(p_1, q_{56}) &= -\frac{1}{(q_{56} - p_1)^2 - m^2} = -\frac{1}{m_{56}^2 - 2q_{56} \cdot p_1} \\ T^*(p_4, q_{78}) &= -\frac{1}{(q_{78} - p_4)^2 - m^2} = -\frac{1}{m_{78}^2 - 2q_{78} \cdot p_4}. \end{aligned} \quad (G.12)$$

Note that due to the fact that $\vec{q}_{78} = -\vec{q}_{56}$ we have $z_{78,4} = -z_{56,4} = z_{56,3}$, where $z_{i,j} = \frac{\vec{q}_i \cdot \vec{q}_j}{|\vec{q}_i||\vec{q}_j|}$. As a result the unitarity integral takes the form

$$2T_s = \frac{(2\pi)^{-3}}{4!} \frac{1}{16} \int_{4m^2}^{(\sqrt{s}-2m)^2} \frac{dm_{56}^2}{m_{56}} \frac{dm_{78}^2}{m_{78}} \frac{\theta(\tilde{K})}{\tilde{K}} \frac{\sqrt{s}}{s-4m^2} \int_0^{2\pi} d\phi_{56} \frac{1}{z^0 - z_{56,1}} \frac{1}{z^0 - z_{56,3}}. \quad (G.13)$$

where we have introduced

$$\begin{aligned} \tilde{K} &= (m_{56}^2 - m_{78}^2)^2 - 2(m_{56}^2 + m_{78}^2)s + s^2, \\ z^0 &= \frac{\sqrt{s}}{\sqrt{s-4m^2}} \frac{s - m_{56}^2 - m_{78}^2}{\sqrt{\tilde{K}}} > 1. \end{aligned} \quad (G.14)$$

The constraint $\theta(K)$ arises from the fact that $|\vec{q}_{56}| \ge 0$. The integral over angles gives

$$\int_0^{2\pi} d\phi_{56} \frac{1}{z^0 - z_{56,1}} \frac{1}{z^0 - z_{56,3}} = \frac{4\pi z^0}{\sqrt{(z^0)^2 - 1} 2(z^0)^2 - 1 - z}, \quad (G.15)$$

where we have used that $z \equiv z_{1,3}$.

Restoring the overall normalization factor $\lambda^4$ we thus get

$$T_s = \frac{\lambda^4}{1536\pi^2} \int_{4m^2}^{(\sqrt{s}-2m)^2} \frac{dm_{56}^2}{m_{56}} \frac{dm_{78}^2}{m_{78}} \frac{\theta(\tilde{K})}{\tilde{K}} \frac{\sqrt{s}}{s-4m^2} \frac{z^0}{\sqrt{(z^0)^2-1} 2(z^0)^2-1-z} \quad (G.16)$$

We can now write the formula for $\eta_{\text{Aks}}(s)$, which is simply the projection of the discontinuity to the spin zero partial wave

$$\frac{\sqrt{s}}{2} \eta_{\text{Aks}}(s) = \frac{1}{8\pi} \int_{-1}^{1} dz \frac{1}{\sqrt{1-z^2}} T_s(s, z). \quad (G.17)$$
<figure><img src="image_63.png" /><figcaption>**Figure 62.** We plot S-wave inelasticity $\eta_{\text{Aks}}(s)$ in $d=3$ as a function of $s/m^2$ for $\lambda = -\pi$ using (G.19) (red). At large $s$, we have $\eta_{\text{Aks}}(s) \sim \frac{1}{s^{3/2}}$. With black dots we denote the result of initializing the iteration algorithm starting from $(\lambda, 0, 0)$ and applying the Froissart-Gribov formula to the first nontrivial result for $\rho_{\text{el}}(s, t)$. This serves us a nontrivial consistency check of many formulas used in the draft.</figcaption></figure>

The relevant integral takes the form

$$ \frac{1}{8\pi} \int_{-1}^{1} dz \frac{1}{\sqrt{1-z^2} 2(z^0)^2 - 1-z} = \frac{1}{16} \frac{1}{z^0 \sqrt{(z^0)^2 - 1}}. \qquad (G.18) $$

Putting everything together we get

$$ \eta_{\text{Aks}}(s) = \left(\frac{\lambda}{16}\right)^4 \frac{4}{3\pi^2} \int_{4m^2}^{(\sqrt{s}-2m)^2} \frac{\theta((m_{56}^2 - m_{78}^2)^2 - 2(m_{56}^2 + m_{78}^2)s + s^2)}{m_{56}^2 m_{78}^2 (m_{56}^2 m_{78}^2 - 2m^2(m_{56}^2 + m_{78}^2))s + m^2(m_{56}^2 - m_{78}^2)^2}. \qquad (G.19) $$

We plot $\eta_{\text{Aks}}(s)$ in figure 62. At large $s$ it decays as $\frac{1}{s^2}$.

## G.2 $d=4$

The four-body discontinuity takes the form

$$ 2T_s = \frac{1}{4!} \int \frac{d^3 \vec{q}_5}{(2\pi)^3 (2E_{\vec{q}_5})} \frac{d^3 \vec{q}_6}{(2\pi)^3 (2E_{\vec{q}_6})} \frac{d^3 \vec{q}_7}{(2\pi)^3 (2E_{\vec{q}_7})} \frac{d^3 \vec{q}_8}{(2\pi)^3 (2E_{\vec{q}_8})} (2\pi)^4 \delta(p_1 + p_2 - q_5 - q_6 - q_7 - q_8) T(p_1, p_2; q_5, q_6, q_7, q_8) T^*(q_5, q_6, q_7, q_8; p_3, p_4), \qquad (G.20) $$

where $s = (p_1 + p_2)^2$ and $t = (p_1 - p_3)^2$. We will be interested in the amplitudes

$$ T(p_1, p_2; q_5, q_6, q_7, q_8) = -\frac{1}{(q_5 + q_6 - p_1)^2 - m^2} \\ T^*(q_5, q_6, q_7, q_8; p_3, p_4) = -\frac{1}{(q_7 + q_8 - p_4)^2 - m^2}. \qquad (G.21) $$

Let us first transform the measure. We define $(q_5 + q_6)^2 = m_{56}^2$, $(q_7 + q_8)^2 = m_{78}^2$

$$ 2T_s = \frac{1}{4!} \int_{2m^2}^{(\sqrt{s}-2m)^2} dm_{56}^2 dm_{78}^2 \int d^4 q_{56} d^4 q_{78} \delta^+(q_{56}^2 - m_{56}^2) \delta^+(q_{78}^2 - m_{78}^2) T(p_1, q_{56}) T^*(p_4, q_{78}) \\ \times (2\pi)^4 \delta(p_1 + p_2 - q_{56} - q_{78}) \mathcal{I}, \qquad (G.22) $$
where

$$ \mathcal{I} \equiv \frac{1}{(2\pi)^{12}} \int d^4 q_5 \delta^+(q_5^2 - m^2) \int d^4 q_5 \delta^+(q_6^2 - m^2) \delta^{(4)}(q_56 - q_5 - q_6) \\ \times \int d^4 q_7 \delta^+(q_7^2 - m^2) \int d^4 q_8 \delta^+(q_8^2 - m^2) \delta^{(4)}(q_78 - q_7 - q_8). \qquad (G.23) $$

Let's go to the COM frame $q_{78} = (m_{78}, \vec{0})$ to get

$$ \int d^4 q_7 \delta^+(q_7^2 - m^2) \int d^4 q_8 \delta^+(q_8^2 - m^2) \delta^{(4)}(q_{78} - q_7 - q_8) \\ = \int \frac{d^3 \vec{q}_7}{4E_{\vec{q}_7}^2} \delta(m_{78} - 2E_{\vec{q}_7}) = \frac{\pi}{2} \sqrt{1 - \frac{4m^2}{m_{78}^2}}, \quad m_{78}^2 \ge 4m^2. \qquad (G.24) $$

So we are left with the following equation

$$ 2T_s = \frac{(2\pi)^{-8}}{4!} \frac{\pi^2}{4} \int_{4m^2}^{(\sqrt{s}-2m)^2} dm_{56}^2 dm_{78}^2 \sqrt{1 - \frac{4m^2}{m_{56}^2}} \sqrt{1 - \frac{4m^2}{m_{78}^2}} \qquad (G.25) \\ \times \int d^4 q_{56} d^4 q_{78} \delta^+(q_{56}^2 - m_{56}^2) \delta^+(q_{78}^2 - m_{78}^2) T(p_1, q_{56}) T^*(p_4, q_{78}) \delta(p_1 + p_2 - q_{56} - q_{78}). $$

Going to the COM frame we get

$$ \int d^4 q_{56} d^4 q_{78} \delta^+(q_{56}^2 - m_{56}^2) \delta^+(q_{78}^2 - m_{78}^2) T(p_1, q_{56}) T^*(p_4, q_{78}) \delta(p_1 + p_2 - q_{56} - q_{78}) \\ = \int \frac{d^3 \vec{q}_{56}}{4E_{\vec{q}_{56},m_{56}} E_{\vec{q}_{56},m_{78}}} \delta(\sqrt{s} - E_{\vec{q}_{56},m_{56}} - E_{-\vec{q}_{56},m_{78}}) T(p_1, q_{56}) T^*(p_4, q_{78}) \qquad (G.26) $$

We write

$$ T(p_1, q_{56}) = - \frac{1}{(q_{56} - p_1)^2 - m^2} = - \frac{1}{m_{56}^2 - 2q_{56} \cdot p_1} \\ T^*(p_4, q_{78}) = - \frac{1}{(q_{78} - p_4)^2 - m^2} = - \frac{1}{m_{78}^2 - 2q_{78} \cdot p_4}. \qquad (G.27) $$

Note that due to the fact that $\vec{q}_{78} = -\vec{q}_{56}$ we have $z_{78,4} = -z_{56,4} = z_{56,3}$.

For the angular measure we introduce

$$ d\Omega_{56} = 2 \frac{dz_{56,1} dz_{56,3} \theta(-K(z_{56,1}, z_{56,3}, z))}{\sqrt{-K(z_{56,1}, z_{56,3}, z)}}. \qquad (G.28) $$

As a result the integral takes the form

$$ 2T_s = \frac{(2\pi)^{-8}}{4!} \frac{\pi^2}{4} \int_{4m^2}^{(\sqrt{s}-2m)^2} dm_{56}^2 dm_{78}^2 \sqrt{1 - \frac{4m^2}{m_{56}^2}} \sqrt{1 - \frac{4m^2}{m_{78}^2}} \frac{\theta(\tilde{K})}{\sqrt{\tilde{K}(s-4m^2)}} \\ \times \int_{-1}^{1} dz_{56,1} dz_{56,3} \frac{\theta(-K(-z, z_{56,1}, z_{56,3}))}{\sqrt{-K(-z, z_{56,1}, z_{56,3})}} \frac{1}{z^0 - z_{56,1}} \frac{1}{z^0 - z_{56,3}}. \qquad (G.29) $$
where

$$ 
\begin{aligned}
\tilde{K} &= (m_{56}^2 - m_{78}^2)^2 - 2(m_{56}^2 + m_{78}^2)s + s^2, \\
z^0 &= \frac{\sqrt{s}}{\sqrt{s-4m^2}} \frac{s - m_{56}^2 - m_{78}^2}{\sqrt{\tilde{K}}} > 1.
\end{aligned} 
\qquad (G.30) 
$$

As before we can take one of the integrals easily using

$$ \int_{-1}^{1} dz_{56,1} \frac{\theta(-K(z, z_{56,1}, z_{56,3}))}{\sqrt{-K(z, z_{56,1}, z_{56,3})}} \frac{1}{z^0 - z_{56,1}} = \frac{\pi}{\sqrt{K(z, z^0, z_{56,3})}}. \qquad (G.31) $$

Finally we define the last integral

$$ 
\begin{aligned}
\mathcal{I}(z, z^0) &= \int_{-1}^{1} dz_{56,3} \frac{1}{\sqrt{K(z, z^0, z_{56,3})}} \frac{1}{z^0 - z_{56,3}} \\
&= \frac{\tanh^{-1}\left(\frac{2\sqrt{(1-z)(2(z^0)^2-1-z)((z^0)^2-z)}}{2z^2-4z(z^0)^2+(z^0)^4+2(z^0)^2-1}\right)}{\sqrt{(1-z)(2(z^0)^2-1-z)}}.
\end{aligned} 
\qquad (G.32) 
$$

In this way we get (where we also have restored the $\lambda$ factor)

$$ T_s = \frac{4\pi^3}{3} \left(\frac{\lambda}{16\pi^2}\right)^4 \int_{4m^2}^{(\sqrt{s}-2m)^2} dm_{56}^2 dm_{78}^2 \sqrt{1 - \frac{4m^2}{m_{56}^2}} \sqrt{1 - \frac{4m^2}{m_{78}^2}} \frac{\theta(\tilde{K})}{\sqrt{\tilde{K}}(s-4m^2)} \mathcal{I}(z, z^0). \qquad (G.33) $$

We can now write the formula for $\eta_{Aks}(s)$

$$ \frac{1}{2} \frac{\sqrt{s}}{\sqrt{s-4m^2}} \eta_{\text{Aks}}(s) = \frac{1}{32\pi} \int_{-1}^{1} dz T_s(s, z). \qquad (G.34) $$

We get that

$$ \frac{1}{32\pi} \int_{-1}^{1} dz \mathcal{I}(z, z^0) = \frac{1}{32\pi} \left( \log \frac{z^0 - 1}{z^0 + 1} \right)^2. \qquad (G.35) $$

In this way we get

$$ 
\begin{aligned}
\eta_{\text{Aks}}(s) = \frac{\pi^2}{12} \left(\frac{\lambda}{16\pi^2}\right)^4 \frac{1}{\sqrt{s}\sqrt{s-4m^2}} & \int_{4m^2}^{(\sqrt{s}-2m)^2} dm_{56}^2 dm_{78}^2 \\
& \times \sqrt{1 - \frac{4m^2}{m_{56}^2}} \sqrt{1 - \frac{4m^2}{m_{78}^2}} \frac{\theta(\tilde{K})}{\sqrt{\tilde{K}}} \left( \log \frac{z^0 - 1}{z^0 + 1} \right)^2
\end{aligned} 
\qquad (G.36) 
$$

We plot $\eta_{Aks}(s)$ in figure 63, where we also compare it to our numerics as a consistency check.
<figure><img src="image_64.png" /><figcaption>**Figure 63.** We plot S-wave inelasticity $\eta_{Aks}(s)$ in $d = 4$ as a function of $s/m^2$ for $\lambda = \pi$ using (G.36) (red). At large $s$, we have $\eta_{Aks}(s) \sim 1$. The black dots were produced by our numerical algorithm at first nontrivial iteration for double spectral density.</figcaption></figure>

## H Full coupling dependence of $d = 4$ 2QE and 2PR partial waves

In this appendix we present the complete set of data for partial waves of the 2QE and 2PR amplitudes in $d = 4$ for different values of $\lambda$. We identified three regions (small couplings, transition zone, and larger couplings), where the amplitude behaved in a slightly different way, see figure 64, figure 65, figure 66.

The most immediate explanation of the jaggedness would be that it is related to sensitivity of the amplitude in this region to the high-energy cut-off and, possibly, to the finiteness of the grid. We leave a more detailed exploration of these amplitudes for future work.

<figure><img src="image_65.png" /><figcaption>**Figure 64.** Inelasticities in the $J = 0$ sector for $d = 4$ 2QE amplitudes for different coupling constants $\lambda$. We identify three regions: small couplings, transition zone, and larger couplings. The later region is already plotted in figure 42 but we reproduce it here for convenience. The small couplings clearly seem to be dominated by some very large scale, inaccessible to our calculation. An irregular transition occurs between small and medium lambda, which would be interesting to understand.</figcaption></figure>
<figure><img src="image_66.png" /><figcaption>**Figure 65.** Inelasticities in the $J = 0$ sector for $d = 4$ 2PR amplitudes for different coupling constants $\lambda$. We did not generate data to probe accurately the transition range for the 2PR amplitudes, but the data for $\lambda = \pi/2$ shows a consistent behavior with the transition range in the $J = 2$ case for the 2QE dataset in figure 64.</figcaption></figure>

<figure><img src="image_67.png" /><figcaption>**Figure 66.** Maximum value of rescaled inelasticity in the partial waves of the 2PR amplitude in $d = 4$ as a function of $\lambda$, same color code as figure 64. The fitting parameters are given by $\tilde{b}_0 = 11$ and $\tilde{b}_2 = 1.4$.</figcaption></figure>

## I Impact parameter space and the Froissart-Gribov formula

In the main text we have analyzed the behavior of the partial waves in the impact parameter representation $S(s, b)$, see (1.8). Here we derive a formula for the phase shift $\delta(s, b)$ defined via

$$S(s, b) = e^{2i\delta(s, b)} \qquad (I.1)$$

using the Froissart-Gribov formula.
Recall that we consider the limit $J \to \infty$, $s \to \infty$ with the ratio

$$b \equiv \frac{2J}{\sqrt{s - 4m^2}} \qquad (I.2)$$

held fixed.

We would like to take the corresponding limit in the Froissart-Gribov representation which we write as follows

$$f_J(s) = \frac{2N_d}{\pi} \frac{2}{s - 4m^2} \int_{4m^2}^{\infty} dt (z^2 - 1)^{\frac{d-4}{2}} Q_J^{(d)} \left( 1 + \frac{2t}{s - 4m^2} \right) T_t(s, t), \qquad (I.3)$$

where we assume that as we take $s \to \infty$ the leading Regge intercept does not grow faster than $\sqrt{s}$ so that we have $J > J_0(s)$. This is manifestly true in the amplitudes that obey the Mandelstam representation since then $J_0(s)$ is bounded by some fixed number for any $s$.

The relevant limit of the Q-function takes the form

$$\lim_{J \to \infty} Q_J^{(d)} \left( 1 + \frac{tb^2}{2J^2} \right) = 2^{\frac{d-4}{2}} \Gamma \left( \frac{d-2}{2} \right) (\sqrt{tb})^{2-\frac{d}{2}} K_{\frac{d-4}{2}}(\sqrt{tb}). \qquad (I.4)$$

The relationship between the phase shift and the partial wave takes the form

$$e^{2i\delta(s,b)} = 1 + i \frac{(s - 4m^2)^{\frac{d-3}{2}}}{\sqrt{s}} f_J(s). \qquad (I.5)$$

So that we get

$$i(1 - e^{2i\delta(s,b)}) = \frac{1}{(2\pi)^{d/2}} \frac{1}{b^{d-4}} \int_{4m^2}^{\infty} \frac{dt}{s} (\sqrt{tb})^{\frac{d-4}{2}} K_{\frac{d-4}{2}}(\sqrt{tb}) T_t(s,t) + \dots, \qquad (I.6)$$

where $\dots$ stand for corrections which are subleading in $\frac{1}{\sqrt{s b}} \sim \frac{1}{J}$.

In the context of our paper high-energy phase shift is small. Therefore we can expand the l.h.s. of the formula above to get

$$2\delta(s,b) = \frac{1}{(2\pi)^{d/2}} \frac{1}{b^{d-4}} \int_{4m^2}^{\infty} \frac{dt}{s} (\sqrt{tb})^{\frac{d-4}{2}} K_{\frac{d-4}{2}}(\sqrt{tb}) T_t(s,t). \qquad (I.7)$$

Notice that $K_{\frac{d-4}{2}}(\sqrt{tb}) \sim e^{-\sqrt{tb}}$ and therefore the integral localizes close to its minimal value $t = 4m^2$ which produces the expected Yukawa potential $e^{-2bm}$.

In the main text we were interested in the behavior of $1 - |S(s,b)|$ which is controlled by the leading behavior of the imaginary part of the phase shift

$$1 - |S(s,b)| \approx 2\mathrm{Im}\delta(s,b) = \frac{1}{(2\pi)^{d/2}} \frac{1}{b^{d-4}} \int_{4m^2}^{\infty} \frac{dt}{s} (\sqrt{tb})^{\frac{d-4}{2}} K_{\frac{d-4}{2}}(\sqrt{tb}) \rho(s,t). \qquad (I.8)$$

This time the integral is localized close to the leading Landau curve in the crossed channel $\frac{4m^2s}{s-16m^2}$ which for $s \to \infty$ starts at $t = 4m^2$.

To get some feeling how the phase shift depends on the near-threshold behavior of the integrand consider the following ansatz $T_t(s,t) \sim c_0(\frac{t}{4m^2} - 1)^\alpha$, we then get

$$\delta(s,b) \sim \frac{c_0}{s} \frac{e^{-2bm}}{(bm)^{\frac{d-1}{2}+\alpha}}. \qquad (I.9)$$
## J Acnode evaluation

To derive (6.4) starting from (6.1), we first eliminate $p_7$ completely using the energy-momentum conservation and restore the on-shell conditions to get for the measure

$$ \frac{1}{3!} (2\pi) \int \frac{d^4 q_5}{(2\pi)^3} \frac{d^4 q_6}{(2\pi)^3} \delta^+(q_5^2 - m^2) \delta^+(q_6^2 - m^2) \delta^+((p_1 + p_2 - q_5 - q_6)^2 - m^2). \quad (J.1) $$

We next eliminate $|\vec{q}_i|$ integrals via

$$ \begin{aligned} \int d^4 q_i \delta^+(q_i^2 - m^2) &= \int \theta(q_i^0) dq_i^0 d|\vec{q}_i| |\vec{q}_i|^2 d\Omega_i \delta((q_i^0)^2 - m^2 - |\vec{q}_i|^2) \\ &= \int_m^\infty dq_i^0 \frac{\sqrt{(q_i^0)^2 - m^2}}{2} d\Omega_i. \end{aligned} \quad (J.2) $$

In this way we get for the measure

$$ \frac{1}{4} \frac{1}{3!} \frac{1}{(2\pi)^5} \int_m^\infty dq_5^0 \sqrt{(q_5^0)^2 - m^2} d\Omega_5 \int_m^\infty dq_6^0 \sqrt{(q_6^0)^2 - m^2} d\Omega_6 \delta^+((p_1 + p_2 - q_5 - q_6)^2 - m^2). \quad (J.3) $$

As a next step we write the angular measure in a more convenient way. First, we introduce the relative angles $z_{ij} \equiv \cos \vec{n}_i \cdot \vec{n}_j$, where $\vec{n}_i = \frac{\vec{q}_i}{|\vec{q}_i|}$. Second, we switch for the measure

$$ \frac{1}{4} d\Omega_5 d\Omega_6 = \frac{dz_{15} dz_{56} \theta(-K(z_{15}, z_{56}, z_{16}))}{\sqrt{-K(z_{15}, z_{56}, z_{16})}} \frac{dz_{26} dz_{46} \theta(-K(z_{24}, z_{26}, z_{46}))}{\sqrt{-K(z_{24}, z_{26}, z_{46})}}, \quad (J.4) $$

$$ -K(z, z', z'') = 1 - z^2 - z'^2 - z''^2 + 2zz'z''. \quad (J.5) $$

This formula holds if we use it for the integrand which only depends on the relative angles that appear in the r.h.s. This will be the case for us.

It is very convenient then to switch to the COM frame where $\vec{p}_2 = -\vec{p}_1$. In this frame we get

$$ \begin{aligned} (p_1 + p_2 - q_5 - q_6)^2 - m^2 &= (\sqrt{s} - q_5^0 - q_6^0)^2 - (\vec{q}_5 - \vec{q}_6)^2 - m^2 \\ &= 2\sqrt{(q_5^0)^2 - m^2} \sqrt{(q_6^0)^2 - m^2} (z_{56} - z_{56}^0), \\ z_{56}^0 &= \frac{s + m^2 + 2q_5^0 q_6^0 - 2\sqrt{s}(q_5^0 + q_6^0)}{2\sqrt{(q_5^0)^2 - m^2} \sqrt{(q_6^0)^2 - m^2}}. \end{aligned} \quad (J.6) $$

In this way we get for the phase space integral measure

$$ \frac{1}{2} \frac{1}{3!} \frac{1}{(2\pi)^5} \int_m^\infty dq_5^0 \int_m^\infty dq_6^0 \theta(\sqrt{s} - m - q_5^0 - q_6^0) \\ \times \int_{-1}^1 \frac{dz_{15} dz_{56} \theta(-K(z_{15}, z_{56}, -z_{26}))}{\sqrt{-K(z_{15}, z_{56}, -z_{26})}} \frac{dz_{26} dz_{46} \theta(-K(z_{24}, z_{26}, z_{46}))}{\sqrt{-K(z_{24}, z_{26}, z_{46})}} \delta(z_{56} - z_{56}^0), \quad (J.7) $$

where we used that in the COM frame $z_{16} = -z_{26}$. Next we introduce the invariant momenta of the pairs of particles $s_{57} = (q_5 + q_7)^2$ and $s_{67} = (q_6 + q_7)^2$ so that

$$ q_5^0 = \frac{s + m^2 - s_{57}}{2\sqrt{s}}, \quad q_6^0 = \frac{s + m^2 - s_{57}}{2\sqrt{s}}. \quad (J.8) $$
We get

$$ \int_m^\infty dq_5^0 \int_m^\infty dq_6^0 \theta(\sqrt{s}-q_5^0-q_6^0-m) = \frac{1}{4s} \int_{4m^2}^{(\sqrt{s}-m)^2} ds_{57} ds_{67} \theta(s_{57}+s_{67}-2m^2-2m\sqrt{s}). \quad (J.9) $$

If we set $\sqrt{s} = 3m$ the integration region shrinks to zero. Therefore we get the expected $\theta(s - 9m^2)$.

The limits of integrations could be further simplified and we finally get the following formula

$$ 2T_s = \frac{1}{2} \frac{1}{3!} \frac{1}{(2\pi)^5} \frac{1}{4s} \int_{4m^2}^{(\sqrt{s}-m)^2} ds_{57} ds_{67} \theta(1 - (z_{56}^0)^2) \times \int_{-1}^{1} \frac{dz_{15} \theta(-K(z_{15}, z_{56}^0, -z_{26}))}{\sqrt{-K(z_{15}, z_{56}^0, -z_{26})}} \frac{dz_{26} dz_{46} \theta(-K(z_{24}, z_{26}, z_{46}))}{\sqrt{-K(z_{24}, z_{26}, z_{46})}} T_{1,2 \to 5,6,7} T_{5,6,7 \to 3,4}^* \quad (J.10) $$

We next write the explicit expressions for the amplitudes that appear in the acnode graph

$$ T(p_1, p_2; q_5, q_6, q_7) = -\frac{1}{t_{15}-m^2} = -\frac{1}{2|\vec{p}_1||\vec{q}_5|} \frac{1}{z_{15}-z_{15}^0}, \quad (J.11) $$

$$ T^*(q_5, q_6, q_7; p_3, p_4) = -\frac{1}{t_{46}-m^2} = -\frac{1}{2|\vec{p}_4||\vec{q}_6|} \frac{1}{z_{46}-z_{46}^0}, \quad (J.12) $$

where

$$ z_{15}^0 = \frac{2p_1^0 q_5^0 - m^2}{2|\vec{p}_1||\vec{q}_5|}, \quad z_{46}^0 = \frac{2p_4^0 q_6^0 - m^2}{2|\vec{p}_4||\vec{q}_6|}. \quad (J.13) $$

It is easy to check using the on-shell condition that $z_{15}^0, z_{46}^0 > 1$.

Recall that

$$ z_{56}^0 = \frac{s(s_{57} + s_{67}) - s^2 + (s_{57} - m^2)(s_{67} - m^2)}{\sqrt{m^4 - 2m^2(s + s_{57}) + (s - s_{57})^2} \sqrt{m^4 - 2m^2(s + s_{67}) + (s - s_{67})^2}}. \quad (J.14) $$

We then get for the acnode

$$ 2T_s = \frac{1}{8s} \frac{1}{3!} \frac{1}{(2\pi)^5} \int_{4m^2}^{(\sqrt{s}-m)^2} \frac{ds_{57} ds_{67}}{4|\vec{p}_1||\vec{p}_5||\vec{p}_4||\vec{p}_6|} dz_{15} dz_{26} dz_{46} \theta(1 - (z_{56}^0)^2) \times \frac{\theta(-K(z_{15}, z_{56}^0, -z_{26})) \theta(-K(z, z_{26}, z_{46}))}{\sqrt{K(z_{15}, z_{56}^0, -z_{26}) K(z, z_{26}, z_{46})}} \frac{1}{z_{15} - z_{15}^0 z_{46} - z_{46}^0}. \quad (J.15) $$

Next step is to do integrals over $z_{15}$ and $z_{46}$. This is done using the following useful formula

$$ \int_{-1}^{1} dz_{15} \frac{\theta(-K(z_{15}, z_{56}^0, -z_{26}))}{\sqrt{-K(z_{15}, z_{56}^0, -z_{26})}} \frac{1}{z_{15}^0 - z_{15}} = \frac{\pi}{\sqrt{K(z_{15}^0, z_{56}^0, -z_{26})}}. \quad (J.16) $$

In this way we arrive at the final formula for the unitarity cut of the acnode (6.4).

**Open Access.** This article is distributed under the terms of the Creative Commons Attribution License (CC-BY 4.0), which permits any use, distribution and reproduction in any medium, provided the original author(s) and source are credited.
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