Although writing the Pomeron expression as a sum of four terms is a purely mathematical operation, it is nevertheless useful for the discussion to associate the indices in the interval $I_{Nk}$ to “Pomerons of type N”. This allows one to consider $r_{Nk}$ as the number of Pomerons of type N associated to nucleon-nucleon pair $k$. With the above definitions and interpretations, one gets

$$ W_{AB} = \sum_{r_{11}=0}^{\infty} \cdots \sum_{r_{41}=0}^{\infty} \cdots \sum_{r_{1AB}=0}^{\infty} \cdots \sum_{r_{4AB}=0}^{\infty} \\ \times \prod_{k=1}^{AB} \frac{(-\alpha_1)^{r_{1k}}}{r_{1k}!} \frac{(-\alpha_2)^{r_{2k}}}{r_{2k}!} \frac{(-\alpha_3)^{r_{3k}}}{r_{3k}!} \frac{(-\alpha_4)^{r_{4k}}}{r_{4k}!} \\ \times U^+(\{r_{Nk}\}, \{x_i^+\}) U^-(\{r_{Nk}\}, \{x_i^-\}), \quad (B19) $$

seen as the sum over all possible numbers of Pomerons of all possible types (1,2,3,4), with

$$ U^+(\{r_{Nk}\}, \{x_i^+\}) = \int d\tilde{X}_{AB}^{+} \prod_{k=1}^{AB} \left[ \prod_{\lambda \in I_{1k}} (\tilde{x}_{k\lambda}^{+})^{\beta_1} \cdots \prod_{\lambda \in I_{4k}} (\tilde{x}_{k\lambda}^{+})^{\beta_4} \right] \\ \times \prod_{i=1}^{A} \left( x_i^+ - \sum_{\substack{k=1 \\ \pi(k)=i}}^{AB} \sum_{\lambda=1}^{l_k} \tilde{x}_{k\lambda}^{+} \right)^{\alpha_{\text{remn}}} . \quad (B20) $$

The expression for $U^-(\{r_{Nk}\}, \{x_i^-\})$ is identical, just with $x_j^-$ instead of $x_i^+$, with $\prod_{j=1}^B$ instead of $\prod_{i=1}^A$, and with $\tau(k) = j$ instead of $\pi(k) = i$. Using

$$ \prod_{k=1}^{AB} = \prod_{i=1}^{A} \prod_{\pi(k)=i}^{A}, \quad (B21) $$

and defining

$$ \epsilon_{\lambda k} = \begin{cases} \beta_1 & \text{for } \lambda \in I_{1k} \\ \beta_2 & \text{for } \lambda \in I_{2k} \\ \beta_3 & \text{for } \lambda \in I_{3k} \\ \beta_4 & \text{for } \lambda \in I_{4k}, \end{cases} \quad (B22) $$

one may separate the contributions for the different nucleons $i$ as

$$ U^+(\{r_{Nk}\}, \{x_i^+\}) = \prod_{i=1}^{A} U_i^+(\{r_{Nk}\}, x_i^+) $$

with an expression for given $i$,

$$ U_i^+(\{r_{Nk}\}, x_i^+) = \int \prod_{\substack{k=1 \\ \pi(k)=i}}^{AB} \left( \prod_{\lambda=1}^{l_k} d\tilde{x}_{k\lambda}^+ \right) \prod_{\substack{k=1 \\ \pi(k)=i}}^{AB} \left[ \prod_{\lambda=1}^{l_k} (\tilde{x}_{k\lambda}^+)^{\epsilon_{\lambda k}} \right] \\ \times \left( x_i^+ - \sum_{\substack{k=1 \\ \pi(k)=i}}^{AB} \sum_{\lambda=1}^{l_k} \tilde{x}_{k\lambda}^+ \right)^{\alpha_{\text{remn}}} . \quad (B23) $$

Let me rename, for given $i$, the $\tilde{x}_{k\lambda}^+$ linked to nucleon $i$ as $x_1, x_2, \dots, x_L$, the $\epsilon_{\lambda k}$ as $\epsilon_1, \epsilon_2, \dots, \epsilon_L$, where $L$ is per definition the number of Pomerons linked to $i$, and let us define

$x = x_i^+$. Then one gets

$$ U_i^+(\{r_{Nk}\}, x) = \int \prod_{\Lambda=1}^{L} dx_{\Lambda} \prod_{\Lambda=1}^{L} (x_{\Lambda})^{\epsilon_{\Lambda}} (x - \sum_{\Lambda=1}^{L} x_{\Lambda})^{\alpha_{\text{remn}}}. \quad (B24) $$

I define new variables,

$$ u_{\Lambda} = \frac{x_{\Lambda}}{x - x_1 - \cdots - x_{\Lambda-1}}, \quad (B25) $$

$$ du_{\Lambda} = \frac{dx_{\Lambda}}{x - x_1 - \cdots - x_{\Lambda-1}}, \quad (B26) $$

which have the property

$$ \prod_{a=1}^{\Lambda-1} (1 - u_a) = \prod_{a=1}^{\Lambda-1} \frac{x - \cdots - x_a}{x - \cdots - x_{a-1}} = \frac{x - \cdots - x_{\Lambda-1}}{x}, \quad (B27) $$

and therefore

$$ x_{\Lambda} = xu_{\Lambda} \prod_{a=1}^{\Lambda-1} (1-u_a), \quad dx_{\Lambda} = xdu_{\Lambda} \prod_{a=1}^{\Lambda-1} (1-u_a). \quad (B28) $$

This leads to

$$ U_i^+(\{r_{Nk}\}, x) = x^{\alpha_{\text{remn}} + \sum_{\Lambda} \tilde{\epsilon}_{\Lambda}} \int \prod_{\Lambda=1}^{L} du_{\Lambda} \\ \times \left\{ \prod_{\Lambda=1}^{L} \left[ u_{\Lambda}^{\epsilon_{\Lambda}} \prod_{a=1}^{\Lambda-1} (1-u_a)^{\tilde{\epsilon}_{\Lambda}} (1-u_{\Lambda})^{\alpha_{\text{remn}}} \right] \right\}, \quad (B29) $$

where I used $\tilde{\epsilon}_{\Lambda} = \epsilon_{\Lambda} + 1$, for convenience. One has

$$ \prod_{\Lambda=1}^{L} \prod_{a=1}^{\Lambda-1} (1-u_a)^{\tilde{\epsilon}_{\Lambda}} = \prod_{a=1}^{L} \prod_{\Lambda=a+1}^{L} (1-u_a)^{\tilde{\epsilon}_{\Lambda}} \quad (B30) $$

$$ = \prod_{\Lambda=1}^{L} \prod_{a=\Lambda+1}^{L} (1-u_{\Lambda})^{\tilde{\epsilon}_{a}} \quad (B31) $$

$$ = \prod_{\Lambda=1}^{L} (1 - u_{\Lambda})^{\sum_{a=\Lambda+1}^{L} \tilde{\epsilon}_{a}}, \quad (B32) $$

which provides

$$ U_i^+(\{r_{Nk}\}, x) = x^{\alpha_{\text{remn}} + \sum_{\Lambda} \tilde{\epsilon}_{\Lambda}} \int \prod_{\Lambda=1}^{L} du_{\Lambda} \\ \times \left\{ \prod_{\Lambda=1}^{L} \left[ u_{\Lambda}^{\epsilon_{\Lambda}} (1-u_{\Lambda})^{\sum_{a=\Lambda+1}^{L} \tilde{\epsilon}_{a}} (1-u_{\Lambda})^{\alpha_{\text{remn}}} \right] \right\}. \quad (B33) $$

Defining

$$ \alpha = \alpha_{\text{remn}} + \sum_{\Lambda=1}^{L} \tilde{\epsilon}_{\Lambda}, \quad (B34) $$

$$ \gamma_{\Lambda} = \alpha_{\text{remn}} + \sum_{a=\Lambda+1}^{L} \tilde{\epsilon}_{a}, \quad (B35) $$
one finds

$$U_i^+(\{r_{Nk}\}, x) = x^\alpha \prod_{\Lambda=1}^{L} \int_0^1 du_\Lambda u_\Lambda^{\epsilon_\Lambda} (1 - u_\Lambda)^{\gamma_\Lambda}. \quad (B36)$$

The integral can be done $[\int_0^1 t^{x-1}(1-t)^{y-1}dt$ is the Euler beta function], and one gets, with $x = x_i^+$,

$$U_i^+(\{r_{Nk}\}, x_i^+) = (x_i^+)^\alpha \prod_{\Lambda=1}^{L} \frac{\Gamma(1 + \epsilon_\Lambda)\Gamma(1 + \gamma_\Lambda)}{\Gamma(2 + \epsilon_\Lambda + \gamma_\Lambda)}. \quad (B37)$$

Using the relation $1 + \epsilon_\Lambda + \gamma_\Lambda = \gamma_{\Lambda-1}$, one gets

$$U_i^+(\{r_{Nk}\}, x_i^+) = (x_i^+)^\alpha \prod_{\Lambda=1}^{L} \frac{\Gamma(1 + \epsilon_\Lambda)\Gamma(1 + \gamma_\Lambda)}{\Gamma(1 + \gamma_{\Lambda-1})}. \quad (B38)$$

In the following, the different factors in this expression will be discussed. One has

$$\prod_{\Lambda=1}^{L} = \prod_{k=1}^{AB} \prod_{\lambda=1}^{AB} l_k. \quad (B39)$$

which actually indicates the relation between the indices $\Lambda$ and the pair of indices $k$ and $\lambda$. So one gets

$$\prod_{\Lambda=1}^{L} \Gamma(1 + \epsilon_{\Lambda}) = \prod_{k=1}^{AB} \prod_{\lambda=1}^{AB} l_k \Gamma(1 + \epsilon_{\lambda k}) \quad (B40)$$

$$= \prod_{k=1}^{AB} \prod_{\pi(k)=i}^{4} \prod_{\lambda \in I_{Nk}} \Gamma(1 + \beta_N), \quad (B41)$$

where I split $\prod_{\lambda=1}^k l_k$ into four products corresponding to the four Pomeron types 1–4, and where I used Eq. (B22). So one finds

$$\prod_{\Lambda=1}^{L} \Gamma(1 + \epsilon_{\Lambda}) = \prod_{k=1}^{AB} \prod_{\pi(k)=i}^{4} \Gamma(1 + \beta_N)^{r_{Nk}}. \quad (B42)$$

Concerning $(x_i^+)^\alpha$, I use Eq. (B34) and $\tilde{\epsilon}_\Lambda = \epsilon_\Lambda + 1$, and I define $\tilde{\beta}_N = \beta_N + 1$, to get

$$\alpha = \alpha_{\text{remn}} + \sum_{\Lambda=1}^{L} \tilde{\epsilon}_{\Lambda} \quad (B43)$$

$$= \alpha_{\text{remn}} + \sum_{k=1}^{AB} \sum_{\lambda=1}^{AB} \sum_{\pi(k)=i}^{k} \tilde{\epsilon}_{\lambda k} \quad (B44)$$

$$= \alpha_{\text{remn}} + \sum_{k=1}^{AB} \sum_{\pi(k)=i}^{4} \sum_{\lambda \in I_{Nk}} \tilde{\beta}_N \quad (B45)$$

$$= \alpha_{\text{remn}} + \sum_{k=1}^{AB} \sum_{\pi(k)=i}^{4} r_{Nk} \tilde{\beta}_N, \quad (B46)$$

which gives

$$(x_i^+)^\alpha = (x_i^+)^{\alpha_{\text{remn}}} \prod_{k=1}^{AB} \prod_{\pi(k)=i}^{4} (x_i^+)^{r_{Nk} \tilde{\beta}_N}. \quad (B47)$$

Furthermore, one has

$$\prod_{\Lambda=1}^{L} \frac{\Gamma(1 + \gamma_{\Lambda})}{\Gamma(1 + \gamma_{\Lambda-1})} = \frac{\Gamma(1 + \gamma_L)}{\Gamma(1 + \gamma_0)} \quad (B48)$$

$$= \frac{\Gamma(1 + \alpha_{\text{remn}})}{\Gamma(1 + \alpha_{\text{remn}} + \sum_{\Lambda=1}^{L} \tilde{\epsilon}_{\Lambda})} \quad (B49)$$

$$= \frac{\Gamma(1 + \alpha_{\text{remn}})}{\Gamma(1 + \alpha_{\text{remn}} + \sum_{\pi(k)=i}^{AB} \sum_{N=1}^{4} r_{Nk} \tilde{\beta}_N)}. \quad (B50)$$

Defining a function $g(z)$ as

$$g(z) = \frac{\Gamma(1 + \alpha_{\text{remn}})}{\Gamma(1 + \alpha_{\text{remn}} + z)}, \quad (B51)$$

one gets

$$\prod_{\Lambda=1}^{L} \frac{\Gamma(1 + \gamma_{\Lambda})}{\Gamma(1 + \gamma_{\Lambda-1})} = g\left(\sum_{k=1}^{AB} \sum_{\pi(k)=i}^{4} r_{Nk} \tilde{\beta}_N\right). \quad (B52)$$

Inserting Eqs. (B42), (B47), and (B52) into Eq. (B38), one gets

$$U_i^+(\{r_{Nk}\}, x_i^+) = (x_i^+)^{\alpha_{\text{remn}}} \prod_{k=1}^{AB} \prod_{\pi(k)=i}^{4} (x_i^+)^{r_{Nk} \tilde{\beta}_N} \\ \times \prod_{k=1}^{AB} \prod_{\pi(k)=i}^{4} \Gamma(1 + \beta_N)^{r_{Nk}} g\left(\sum_{k=1}^{AB} \sum_{\pi(k)=i}^{4} r_{Nk} \tilde{\beta}_N\right). \quad (B53)$$

A corresponding expression can be found for $U_j^-(\{r_{Nk}\}, x_j^-)$. The two expressions

$$U^+(\{r_{Nk}\}, \{x_i^+\}) = \prod_{i=1}^{A} U_i^+(\{r_{Nk}\}, x_i^+), \quad (B54)$$

$$U^-(\{r_{Nk}\}, \{x_j^\dagger\}) = \prod_{j=1}^{B} U_j^-(\{r_{Nk}\}, x_j^\dagger), \quad (B55)$$
may be inserted into Eq. (B19), using Eq. (B53) and the corresponding $U_j^-$ expression, and one gets

$$ W_{AB} = \prod_{i=1}^{A} (x_i^+)^{\alpha_{\text{remn}}} \prod_{j=1}^{B} (x_j^-)^{\alpha_{\text{remn}}} \sum_{\{r_{Nk}\}} \left\{ \begin{aligned} & \prod_{k=1}^{AB} \prod_{N=1}^{4} \frac{(-\alpha_N)^{r_{Nk}}}{r_{Nk}!} \\ & \times \prod_{i=1}^{A} \left[ \prod_{k=1}^{AB} \prod_{\pi(k)=i}^{4} (\Gamma(\tilde{\beta}_N)(x_i^+)^{\tilde{\beta}_N})^{r_{Nk}} g\left(\sum_{k=1}^{AB} \sum_{\pi(k)=i}^{4} r_{Nk} \tilde{\beta}_N\right) \right] \\ & \times \prod_{j=1}^{B} \left[ \prod_{k=1}^{AB} \prod_{\tau(k)=j}^{4} (\Gamma(\tilde{\beta}_N)(x_j^-)^{\tilde{\beta}_N})^{r_{Nk}} g\left(\sum_{k=1}^{AB} \sum_{\tau(k)=j}^{4} r_{Nk} \tilde{\beta}_N\right) \right] \end{aligned} \right\}, \quad (B56) $$

where $\sum_{\{r_{Nk}\}}$ means summing all the indices $r_{Nk}$, with $1 \le N \le 4$ and with $1 \le k \le AB$, from zero to infinity. I use $\tilde{\beta}_N = \beta_N + 1$. A similar formula (with different notations) has been found in Ref. [5].

## 2. Doing the infinite sums

Equation (B56) is the best one can do, without further assumptions. At least the integrations could be done, and (although time consuming) the infinite sums can be performed on a powerful computer (they converge).

However, a “small” modification would allow one to simplify the expression enormously, and this modification concerns the function $g(z) = \Gamma(1 + \alpha_{\text{remn}})/\Gamma(1 + \alpha_{\text{remn}} + z)$. The argument of $g$ is of the form $\sum_{\lambda} \tilde{\beta}_{\lambda}$ with given coefficients $\tilde{\beta}_{\lambda}$. If one would have

$$ g\left(\sum_{\lambda} \tilde{\beta}_{\lambda}\right) = c_1 \prod_{\lambda} c_2 g(c_3 \tilde{\beta}_{\lambda}), \quad (B57) $$

with three parameters $c_\mu$, then one obtains

$$ g\left(\sum_{k=1}^{AB} \sum_{\pi(k)=i}^{4} r_{Nk} \tilde{\beta}_N\right) = c_1 \prod_{\pi(k)=i}^{AB} \prod_{N=1}^{4} (c_2 g(c_3 \tilde{\beta}_N))^{r_{Nk}}, \quad (B58) $$

$$ g\left(\sum_{k=1}^{AB} \sum_{\tau(k)=j}^{4} r_{Nk} \tilde{\beta}_N\right) = c_1 \prod_{k=1}^{AB} \prod_{\tau(k)=j}^{4} (c_2 g(c_3 \tilde{\beta}_N))^{r_{Nk}}, \quad (B59) $$

and one finds for $W_{AB}$ the expression

$$ W_{AB} = \prod_{i=1}^{A} c_1(x_i^+)^{\alpha_{\text{remn}}} \prod_{j=1}^{B} c_1(x_j^-)^{\alpha_{\text{remn}}} \sum_{\{r_{Nk}\}} \prod_{k=1}^{AB} \prod_{N=1}^{4} \\ \times \left\{ \frac{(-\alpha_N)^{r_{Nk}}}{r_{Nk}!} [(\Gamma(\tilde{\beta}_N)(x_{\pi(k)}^+)^{\tilde{\beta}_N})^{r_{Nk}} (c_2 g(c_3 \tilde{\beta}_N))^{r_{Nk}}] \right. \\ \left. \times [(\Gamma(\tilde{\beta}_N)(x_{\tau(k)}^-)^{\tilde{\beta}_N})^{r_{Nk}} (c_2 g(c_3 \tilde{\beta}_N))^{r_{Nk}}] \right\}. \quad (B60) $$

Defining

$$ D_N = \Gamma(\tilde{\beta}_N) c_2 g(c_3 \tilde{\beta}_N) = \frac{\Gamma(\tilde{\beta}_N) c_2 \Gamma(1 + \alpha_{\text{remn}})}{\Gamma(1 + \alpha_{\text{remn}} + c_3 \tilde{\beta}_N)}, \quad (B61) $$

one gets

$$ W_{AB} = \prod_{i=1}^{A} c_1(x_i^+)^{\alpha_{\text{remn}}} \prod_{j=1}^{B} c_1(x_j^-)^{\alpha_{\text{remn}}} \sum_{\{r_{Nk}\}} \prod_{k=1}^{AB} \prod_{N=1}^{4} \frac{1}{r_{Nk}!} \\ \times [-\alpha_N(x_{\pi(k)}^+)^{\tilde{\beta}_N} (x_{\tau(k)}^-)^{\tilde{\beta}_N} D_N^2]^{r_{Nk}}. \quad (B62) $$

Exchanging sum and product, one gets

$$ W_{AB} = \prod_{i=1}^{A} c_1(x_i^+)^{\alpha_{\text{remn}}} \prod_{j=1}^{B} c_1(x_j^-)^{\alpha_{\text{remn}}} \prod_{k=1}^{AB} \prod_{N=1}^{4} \sum_{r_{Nk}!} \frac{1}{r_{Nk}!} \\ \times [-\alpha_N(x_{\pi(k)}^+)^{\tilde{\beta}_N} (x_{\tau(k)}^-)^{\tilde{\beta}_N} D_N^2]^{r_{Nk}}, \quad (B63) $$

where one recognizes the power series of the exponential function, and so one gets finally

$$ W_{AB} = \prod_{i=1}^{A} c_1(x_i^+)^{\alpha_{\text{remn}}} \prod_{j=1}^{B} c_1(x_j^-)^{\alpha_{\text{remn}}} \prod_{k=1}^{AB} \exp(-\tilde{G}(x_{\pi(k)}^+ x_{\tau(k)}^-)), \quad (B64) $$

with

$$ \tilde{G}(x) = \sum_{N=1}^{4} \tilde{\alpha}_{N} x^{\tilde{\beta}_{N}} \quad (B65) $$

and with

$$ \tilde{\alpha}_N = \alpha_N D_N^2 = \alpha_N \left( \frac{\Gamma(\tilde{\beta}_N) c_2 \Gamma(1 + \alpha_{\text{remn}})}{\Gamma(1 + \alpha_{\text{remn}} + c_3 \tilde{\beta}_N)} \right)^2, \quad (B66) $$
$$ \tilde{\beta}_N = \beta_N + 1. \quad (B67) $$

So the property in Eq. (B57) for the function $g$ provides infinite sums of the form of power series of the exponential function, and the exponentials make sure that one always has $W_{AB} \ge 0$.

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