for $\mathbf{u} = (\mathbf{u}_q)_{q \in [1, Q] \setminus \{j\}}$, and for $j \in [1, Q]$,

$$
Q_{\mathbf{N},\mathbf{N}'}^0(j, \mathcal{P}, \mathbf{k}) = \sum_{\substack{\mathbf{w}=(\mathbf{w}_p)_{p \in \mathcal{P}} \in (\mathbb{N}_0^Q)^{\mathcal{P}^c} \\ \forall p \in \mathcal{P}^c | \mathbf{w}_p| = k_p \\ \sum_{\forall q \in [1,Q] \setminus \{j\}, N'_q = \mathbf{v}(q) + \mathbf{w}(q)} \mathbf{v}=(\mathbf{v}_p)_{p \in \mathcal{P}} \in (\mathbb{N}_0^{[1,Q] \setminus \{j\})^{\mathcal{P}}}} \left( \prod_{p \in \mathcal{P}^c} \binom{k_p}{\mathbf{w}_p} \prod_{q=1}^{Q} c_{q,p}^{w_{p,q}} \right) \cdot \left( \prod_{p \in \mathcal{P}} \binom{-N_p - 1}{|\mathbf{v}_p|} \binom{|\mathbf{v}_p|}{\mathbf{v}_p} \prod_{\substack{q=1 \\ q \neq j}}^{Q} c_{q,p}^{v_{p,q}} \right) \prod_{\substack{q=1 \\ q \neq j}}^{Q} (-1)^{N'_q} N'_q!,
$$

and

$$
Q_{\mu, \mu'}^{1, \mathbf{N}, \mathbf{N}'}(j, \mathcal{P}, \mathbf{k}) = (-1)^{|\mathbf{N}'| \cap [1, Q] \setminus \{j\}} \prod_{\substack{q=1 \\ q \neq j}}^{Q} N'_q! \sum_{\substack{\mathbf{w}=(\mathbf{w}_p)_{p \in \mathcal{P}} \in (\mathbb{N}_0^Q)^{\mathcal{P}^c} \\ \forall p \in \mathcal{P}^c | \mathbf{w}_p| = k_p \\ \sum_{\forall q \neq j, \mathbf{v}(q) + \mathbf{w}(q) = N'_q} \mathbf{v}=(\mathbf{v}_p)_{p \in \mathcal{P}} \in (\mathbb{N}_0^{[1,Q] \setminus \{j,f\})^{\mathcal{P}}}} \left( \prod_{p \in \mathcal{P}^c} \binom{k_p}{\mathbf{w}_p} \prod_{q=1}^{Q} c_{q,p}^{w_{p,q}} \right) \\
\left( \prod_{p \in \mathcal{P}} \binom{-N_p - 1}{|\mathbf{v}_p|} \binom{|\mathbf{v}_p|}{\mathbf{v}_p} \prod_{\substack{q=1 \\ q \neq j}}^{Q} c_{q,p}^{v_{p,q}} \right) \left( \sum_{p \in \mathcal{P}} \mu_p (h_{N_p} - h_{N_p + |\mathbf{v}_p|}) + \sum_{\substack{q=1 \\ q \neq j}}^{Q} \mu'_q (\gamma - h_{N'_q}) \right) \\
+ (-1)^{|\mathbf{N}'| \cap [1, Q] \setminus \{j\}} \left( \prod_{\substack{q=1 \\ q \neq j}}^{Q} N'_q! \right) \sum_{\substack{f=1 \\ f \neq j}}^{Q} \mu'_f \sum_{\substack{\mathbf{w}=(\mathbf{w}_p)_{p \in \mathcal{P}} \in (\mathbb{N}_0^Q)^{\mathcal{P}^c} \\ \forall p \in \mathcal{P}^c | \mathbf{w}_p| = k_p \\ \sum_{\forall q \neq j, \mathbf{v}(q) + \mathbf{w}(q) = N'_q} \mathbf{v}=(\mathbf{v}_p)_{p \in \mathcal{P}} \in (\mathbb{N}_0^{[1,Q] \setminus \{j,f\})^{\mathcal{P}}}} \mathbf{w}=(\mathbf{w}_p)_{p \in \mathcal{P}} \in (\mathbb{N}_0^Q)^{\mathcal{P}^c} \\
\left( \prod_{p \in \mathcal{P}^c} \binom{k_p}{\mathbf{w}_p} \prod_{q=1}^{Q} c_{q,p}^{w_{p,q}} \right) \left( \prod_{p \in \mathcal{P}} \binom{-N_p - 1}{|\mathbf{v}_p|} \binom{|\mathbf{v}_p|}{\mathbf{v}_p} \prod_{\substack{q=1 \\ q \neq j, f}}^{Q} c_{q,p}^{v_{p,q}} \right) F_{\mathcal{P}, j, f, \mathbf{v}, \mathbf{w}}(\mathbf{N}, \mathbf{N}'),
$$

with $F_{\mathcal{P},j,f,\mathbf{v},\mathbf{w}}(\mathbf{N}, \mathbf{N}')$ the constant defined in Proposition 1.3.2.

Let's observe that $Q_{\mathbf{N},\mathbf{N}'}^0(j, \mathcal{P}, \mathbf{k}) \in \mathbb{K}$. Moreover, the coefficient $F_{\mathcal{P},j,f,\mathbf{v},\mathbf{w}}(\mathbf{N}, \mathbf{N}')$ belongs in the vector space on $\mathbb{K}$ generated by coefficients $\ln\left(1 + \frac{c_{q,p}}{c_{j,p}}\right)$ with $1 \le q, j \le Q$ and $1 \le p \le P$. Therefore
$$F_{\mathcal{P},j,f,\mathbf{v},\mathbf{w}}(\mathbf{N}, \mathbf{N}') \in \mathrm{Vect}_{\mathbb{K}}\left[\left(\ln\left(1 + \frac{c_{q,p}}{c_{j,p}}\right)\right)_{\substack{a \le j, q \le Q \\ 1 \le p \le P}}\right]. \text{ We then find that}$$

$$
Q_{\mu, \mu'}^{1, \mathbf{N}, \mathbf{N}'}(j, \mathcal{P}, \mathbf{k}) \in \mathrm{Vect}_{\mathbb{K}} \left[ \gamma, \left( \ln \left( 1 + \frac{c_{q,p}}{c_{j,p}} \right) \right)_{\substack{a \le j, q \le Q \\ 1 \le p \le P}} \right].
$$

As the special values $\zeta(-N, d_p)$ belong in the field above $\mathbb{Q}$ generated by the coefficients $d_1, \dots, d_P$, we find that these special values belong in $\mathbb{K}$. Therefore we get that $Z'(-(\mathbf{N}, \mathbf{N}'))$ belong in the vector space
$$
\mu, \mu'
$$
on $\mathbb{K}$ generated by $\gamma$, by coefficient of the form $\ln\left(1 + \frac{c_{q,p}}{c_{j,p}}\right)$, by special values of the derivative of the auxiliary functions $\partial_s \varphi_{\mathbf{R}(\mathbf{N},j,\mathbf{u})}^j(s)|_{s=-N'_j}$, and by special values at nonpositive integers of the derivative with respect to $s$ of the Hurwitz zeta functions.

5.4.2 Proof of Corollary C1

We use the fact that, when the linear forms are rational, Corollary B1 provides an expression for
$\partial_s \varphi_{\mathbf{R}(\mathbf{N},j,\mathbf{u})}^j(s)|_{s=-N'_j}$ in terms of the derivative of the Hurwitz zeta function. We then replace the terms
of the form $\partial_s \varphi_{\mathbf{R}(\mathbf{N},j,\mathbf{u})}^j(s)|_{s=-N'_j}$ in Theorem C with terms of the form $\zeta'(-n, y)$.
### 5.4.3 Proof of Corollary D1

Corollary D1 is obtained by applying the formula found for $Z'(-(\mathbf{N}, \mathbf{N}'))$ in Theorem A, and replacing the terms of the form $\varphi_{\mathbf{R}}^{j'}(-N)$ with the formula provided by Corollary B1.

Since the linear forms $l_j$ have rational coefficients, we can set $c_{j,p} = \frac{a_{j,p}}{b_{j,p}}$ for all $1 \le p \le P$, assuming that this fraction is irreducible. We then define $x_j(\mathbf{c}) = \frac{\text{lcm}(a_{j,p})}{\text{gcd}(b_{j,p})} \in \mathbb{Q}_+^*$, and $\beta_{j,p}(\mathbf{c}) := \frac{x_j(\mathbf{c})}{c_{j,p}} \in \mathbb{N}$ for all $1 \le p \le P$.

By Corollary B1, for $\mathbf{R} \in \mathbb{N}_0^P$ we get that

$$ \varphi_{\mathbf{R}}^{j'}(-N_j') = $$

$$ x_j(\mathbf{c})^N \prod_{p=1}^{P} \beta_{j,p}(\mathbf{c})^{R_p} \sum_{\substack{0 \le v_1 \le \beta_{j,1}(\mathbf{c})-1 \\ 0 \le v_P \le \beta_{j,P}(\mathbf{c})-1}} \sum_{\mathcal{P} \subseteq [1,P]} \left( \prod_{p \in \mathcal{P}^c} R_p! \right) \sum_{\substack{\mathbf{k} \in \mathbb{N}_0^P, k' \in \mathbb{N}_0 \\ |\mathbf{k}|+k'=|\mathbf{R}_{||\mathcal{P}^c||}+|\mathcal{P}^c|-1}} (-1)^{|\mathbf{k}|} $$

$$ \prod_{p \in \mathcal{P}} \frac{\zeta\left(-\frac{R_p - k_p, \frac{d_p+v_p}{\beta_{j,p}(\mathbf{c})}}{k_p!}\right)}{k_p!} \left( \frac{\zeta'(-N - k', l_j(x_j(\mathbf{c})^{-1}(\mathbf{d}+\mathbf{v})))}{k'!} - \ln(x_j(\mathbf{c}) \frac{\zeta(-N - k', l_j(x_j(\mathbf{c})^{-1}(\mathbf{d}+\mathbf{v})))}{k'!}) \right). $$

$$ = x_j(\mathbf{c})^N \prod_{p=1}^{P} \beta_{j,p}(\mathbf{c})^{R_p} \sum_{\substack{0 \le v_1 \le \beta_{j,1}(\mathbf{c})-1 \\ 0 \le v_P \le \beta_{j,P}(\mathbf{c})-1}} \sum_{\mathcal{P} \subseteq [1,P]} \left( \prod_{p \in \mathcal{P}^c} \mathbf{R}(\mathbf{N}, j, \mathbf{u})_p! \right) \sum_{\substack{\mathbf{k} \in \mathbb{N}_0^P, k' \in \mathbb{N}_0 \\ |\mathbf{k}|+k'=|\mathbf{R}_{||\mathcal{P}^c||}+|\mathcal{P}^c|-1}} (-1)^{|\mathbf{k}|} $$

$$ \cdot \left( \prod_{p \in \mathcal{P}} \frac{\zeta\left(-\frac{\mathbf{R}(\mathbf{N}, j, \mathbf{u})_p - k_p, \frac{d_p+v_p}{\beta_{j,p}(\mathbf{c})}}{k_p!}\right)}{k_p!} \right) $$

$$ \cdot \left( \frac{\zeta'(-N_j' - k', l_j(x_j(\mathbf{c})^{-1}(\mathbf{d}+\mathbf{v})))}{k'!} - \ln(x_j(\mathbf{c}) \frac{\zeta(-N_j' - k', l_j(x_j(\mathbf{c})^{-1}(\mathbf{d}+\mathbf{v})))}{k'!}) \right). $$

By injecting this formula in the one obtained for $Z'(-(\mathbf{N}, \mathbf{N}'))$ in theorem D, we found the required result.

We obtain again Corollary C1, which corresponds to a qualitative reformulation of Corollary D1.

### 5.4.4 Proof of Corollary D2

By setting $(\mathbf{N}, \mathbf{N}') = (\mathbf{0}, \mathbf{0})$ in Theorem D, we can simplify several sums by noting the following facts:

1) Let $\mathbf{v} = (\mathbf{v}_p)_{p \in \mathcal{P}} \in (\mathbb{N}_0^{[1,Q]\setminus\{j\}})^P$ and $\mathbf{w} = (\mathbf{w}_p)_{p \in \mathcal{P}^c} \in (\mathbb{N}_0^Q)^{P^c}$ such that

$$ \begin{aligned} |\mathbf{w}_p| &= k_p & (p \in \mathcal{P}^c) \\ \mathbf{v}(q) + \mathbf{w}(q) &= 0 & (q \in [1, Q] \setminus \{j\}). \end{aligned} $$

Then we found that the index have many zero components:

$$ \forall q \in [1, Q] \setminus \{j\}, \forall p \in \mathcal{P}, \quad v_{p,q} = 0, $$

and

$$ \forall q \in [1, Q] \setminus \{j\}, \forall p \in \mathcal{P}^c, \quad w_{p,q} = 0. $$

Therefore, as $w_{p,j} = |w_p| = k_p$ for all $p \in \mathcal{P}^c$, we get that $\mathbf{w} = (k_p \mathbf{e}_j)_{p \in \mathcal{P}^c}$ with $\mathbf{e}_j \in \mathbb{N}_0^Q$ being the vector with a 1 in the $j$-th coordinate, and 0 elsewhere.
2) Let $\mathbf{v} = (\mathbf{v}_p)_{p \in \mathcal{P}} \in (\mathbb{N}_0^{[1,Q]} \setminus \{j,f\})^{\mathcal{P}}$ and $\mathbf{w} = (\mathbf{w}_p)_{p \in \mathcal{P}^c} \in (\mathbb{N}_0^Q)^{\mathcal{P}^c}$ such that

$$ 
\begin{aligned}
|\mathbf{w}_p| &= k_p & (p \in \mathcal{P}^c) \\
\mathbf{v}(q) + \mathbf{w}(q) &= 0 & (q \in [1, Q] \setminus \{j, f\}).
\end{aligned}
$$

We found that

$$ \forall q \in [1, Q] \setminus \{j, f\}, \forall p \in \mathcal{P}, v_{p,q} = 0, $$

and

$$ \forall q \in [1, Q] \setminus \{j, f\}, \forall p \in \mathcal{P}^c, w_{p,q} = 0. $$

Thus, we find that the index $\mathbf{w}$ takes the form $(w_{p,j}\mathbf{e}_j + w_{p,f}\mathbf{e}_f)_{p \in \mathcal{P}^c}$, with $w_{p,j} + w_{p,f} = k_p$ for all $p \in \mathcal{P}^c$. We thus find that the multinomial coefficient present in the expression for $Q^1$ is:

$$ \begin{pmatrix} k_p \\ \mathbf{w}_p \end{pmatrix} = \frac{k_p!}{w_{p,j}! w_{p,f}!}. $$

To simplify this expression, we will take the sum over $w'_p \in [0, k_p]$, and we will set $w_{p,f} = w'_p$ and $w_{p,j} = k_p - w'_p$.

3) Let $j \in [1, Q]$, and $\mathbf{u} = (\mathbf{u}_q)_{q \in [1, Q] \setminus \{j\}}$ such that

$$ |\mathbf{u}_q| = N'_q = 0 \quad (q \in [1, Q] \setminus \{j\}). $$

It trivially follows that $\mathbf{u} = \mathbf{0}$, and therefore $\mathbf{R}(\mathbb{N}, j, \mathbf{u}) := (N_p + \mathbf{u}(p))_{p \in [1, P]} = \mathbf{0}$. In particular, the auxiliary functions $\varphi^j_{(N_p+\mathbf{u}(p))_{p \in [1, P]}}(s)$ are actually Barnes zeta functions of the form

$$ \varphi^j_0(s) = \zeta^B(s, d'_j | c_{j,1}, \dots, c_{j,P}). $$

We recall that the derivative of this Barnes zeta function at $s = 0$ corresponds to $\ln(\Gamma_P(d'_j|c_{j,1}, \dots, c_{j,P}))$, where $\Gamma_P$ corresponds to the Barnes multigamma functions introduced in Definition 1.2.14.

We have already seen in the proof of Corollary A1 that

$$ Q_{0,0}^0(j, \mathcal{P}, \mathbf{k}) = \prod_{p \in \mathcal{P}^c} c_{j,p}^{k_p}. $$

Using these three preceding points, and the fact that the harmonic number $h_0$ is 0, we obtain:

$$ 
\begin{aligned}
Z'(-(\mathbf{0}, \mathbf{0})) = & \\
& \sum_{\mathcal{P} \subsetneq [1, P]} \sum_{j=1}^{Q} (-1)^{|\mathcal{P}|} \frac{\mu'_j}{|\boldsymbol{\mu}'| + |\boldsymbol{\mu}|_{\mathcal{P}}} \sum_{\substack{k=(k_p)_{p \in \mathcal{P}^c} \in \mathbb{N}_0^{\mathcal{P}^c} \\ |\mathbf{k}|=|\mathcal{P}|}} \prod_{p \in \mathcal{P}^c} \frac{\zeta(-k_p, d_p)}{k_p!} \left( \left( \prod_{p \in \mathcal{P}^c} c_{j,p}^{k_p} \right) \gamma(\mu'_j + |\boldsymbol{\mu}|_{\mathcal{P}}) + Q_{\mu, \mu'}^1(j, \mathcal{P}, \mathbf{k}) \right) \\
& + \sum_{\mathcal{P} \subsetneq [1, P]} \sum_{j=1}^{Q} (-1)^{|\mathcal{P}|} \frac{\mu'_j}{|\boldsymbol{\mu}'| + |\boldsymbol{\mu}|_{\mathcal{P}}} \sum_{\substack{k=(k_p)_{p \in \mathcal{P}^c} \in \mathbb{N}_0^{\mathcal{P}^c} \\ |\mathbf{k}|=|\mathcal{P}|}} \prod_{p \in \mathcal{P}^c} c_{j,p}^{k_p} \left( \sum_{i \in \mathcal{P}^c} \mu_i \frac{\zeta'(-k_i, d_i)}{k_i!} \prod_{p \in \mathcal{P}^c \setminus \{i\}} \frac{\zeta(-k_p, d_p)}{k_p!} \right) \\
& + \sum_{j=1}^{Q} \mu'_j \partial_s \zeta^B(s, d'_j | c_j)_{|s=0} - \sum_{j=1}^{Q} \gamma \mu'_j \sum_{\mathcal{P} \subsetneq [1, P]} (-1)^{|\mathcal{P}|} \prod_{p \in \mathcal{P}} c_{j,p}^{-1} \sum_{\substack{k=(k_p)_{p \in \mathcal{P}^c} \in \mathbb{N}_0^{\mathcal{P}^c} \\ |\mathbf{k}|=|\mathcal{P}|}} \prod_{p \in \mathcal{P}^c} c_{j,p}^{k_p} \frac{\zeta(-k_p, d_p)}{k_p!}.
\end{aligned}
$$

By the point 1) and the point 2) previously stated, we get:

$$ 
\begin{aligned}
Q_{\mu, \mu'}^1(j, \mathcal{P}, \mathbf{k}) = & \sum_{\substack{f=1 \\ f \neq j}}^{Q} \mu'_f \sum_{\mathbf{w}'=(w'_p)_{p \in \mathcal{P}^c} \in \prod_{p \in \mathcal{P}^c} [0, k_p]} F_{\mathcal{P}, j, f, 0, \mathbf{0}, (w'_p \mathbf{e}_f + (k_p - w'_p) \mathbf{e}_j)_{p \in \mathcal{P}^c}} (\mathbf{0}, \mathbf{0}) \\
& + \gamma |\boldsymbol{\mu}'|_{[1, Q] \setminus \{j\}} \prod_{p \in \mathcal{P}^c} c_{j,p}^{k_p}.
\end{aligned}
$$
5.4.5 Proof of Corollary D3

Applying Corollary D2, and by taking $\mu = (0, ..., 0)$, and $\mu' = (1, ..., 1)$, we get

$$
\begin{aligned}
Z'(-(\mathbf{0}, \mathbf{0})) = & \sum_{\mathcal{P} \subseteq [1, P]} \sum_{j=1}^{Q} \frac{(-1)^{|\mathcal{P}|}}{Q} \sum_{\mathbf{k}=(k_p)_{p \in \mathcal{P}^c} \in \mathbb{N}_0^{\mathcal{P}^c} \atop |\mathbf{k}|=|\mathcal{P}|} \prod_{p \in \mathcal{P}^c} \frac{\zeta(-k_p, d_p)}{k_p!} \left( \gamma \left( \prod_{p \in \mathcal{P}^c} c_{j,p}^{k_p} \right) + Q^1(j, \mathcal{P}, \mathbf{k}) \right) \\
& + \sum_{j=1}^{Q} \ln(\Gamma_P(d'_j | \mathbf{c}_j)) - \gamma \sum_{j=1}^{Q} \sum_{\mathcal{P} \subseteq [1, P]} (-1)^{|\mathcal{P}|} \prod_{p \in \mathcal{P}} c_{j,p}^{-1} \sum_{\mathbf{k}=(k_p)_{p \in \mathcal{P}^c} \in \mathbb{N}_0^{\mathcal{P}^c} \atop |\mathbf{k}|=|\mathcal{P}|} \prod_{p \in \mathcal{P}^c} c_{j,p}^{k_p} \frac{\zeta(-k_p, d_p)}{k_p!},
\end{aligned}
$$

with

$$
Q^1(j, \mathcal{P}, \mathbf{k}) = \sum_{\substack{f=1 \\ f \neq j}}^{Q} \sum_{\mathbf{w}'=(w'_p)_{p \in \mathcal{P}^c} \in \bigcap_{p \in \mathcal{P}^c} [0, k_p]} F_{\mathcal{P}, j, f, \mathbf{0}, (w'_p \mathbf{e}_f + (k_p - w'_p) \mathbf{e}_j)_{p \in \mathcal{P}^c}} (\mathbf{0}, \mathbf{0}) \left( \prod_{p \in \mathcal{P}^c} \binom{k_p}{w'_p} c_{f,p}^{w'_p} c_{j,p}^{k_p - w'_p} \right) \\
+ \gamma(Q-1) \prod_{p \in \mathcal{P}^c} c_{j,p}^{k_p}.
$$

By Definition 1.2.16, we know that $Z'(-(\mathbf{0}, \mathbf{0})) = \ln (\partial_s(\zeta^{Sh}(s, (\mathbf{c}_1, ..., \mathbf{c}_Q), \mathbf{d}')|_{s=0}))$, and by Definition

1.2.14, we know that $(\partial_s \zeta^B(s, d|c_1, ..., c_P))|_{s=0} = \ln(\Gamma_P(d|c_1, ..., c_P))$, and $\partial_s(\zeta^{Sh}(s, \mathbf{d}'|\mathbf{c}))|_{s=0} = \ln(\Gamma_P(\mathbf{d}'|\mathbf{c}))$.
By all the above, we get Corollary D3.
<figure><img src="image_20.png" /></figure>
# Chapter 6

## Witten zeta functions

We recall here the theory of Witten's zeta functions from the perspectives of Komori, Matsumoto, and Tsumura. Details of the proofs are available in the book [KMT24] which will be published in November 2023. Alternatively, one can also find details in the papers [KMT10b], [KMT10c], [KMT11b], [KMT12]. For details on Lie theory, we refer to [Hum72], [Bou06], [Bou81]. We will then apply some of our results to the study of Witten's zeta functions $\zeta_{g_2}(s)$, and $\zeta_{s_0(5)}(s)$, to deduce the explicit expression for the constant $C$ in Theorem 1.2.35, and an asymptotic formula for the number of representations $r_{g_2}(n)$ using Theorem 1.2.33.

### 6.1 Generalities on Lie algebras

We call $\mathfrak{g}$ a Lie algebra, a $C$-vector space of finite dimension equipped with a bilinear form $[,]$ that satisfies the following conditions

$$1. \forall x \in \mathfrak{g}, \quad [x, x] = 0$$

$$2. \forall x, y, z \in \mathfrak{g}, \quad [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0.$$

We call $[,]$ a Lie bracket associated with the Lie algebra $\mathfrak{g}$.

Let $\mathfrak{g}_1, \mathfrak{g}_2$ be two Lie algebras with two Lie brackets $[,]_1$ and $[,]_2$ respectively. Let $f: \mathfrak{g}_1 \to \mathfrak{g}_2$ be a linear map between the two vector spaces. We say that $f$ is a morphism of Lie algebra if $f$ preserves the Lie brackets, i.e:

$$\forall x, y \in \mathfrak{g}_1, \quad [f(x), f(y)]_2 = f([x, y]_1).$$

Moreover, if $f$ is also a linear isomorphism, we say that $f$ is an isomorphism of Lie algebras.

For $x \in \mathfrak{g}$, we note $ad_x$ the following derivation:

$$ad_x : \mathfrak{g} \to \mathfrak{g}, \quad y \mapsto [x, y].$$

**Definition 6.1.1.** Let $\mathfrak{g}$ a Lie algebra. Let $\mathfrak{a} \subset \mathfrak{g}$ be a subspace of $\mathfrak{g}$. We say that $\mathfrak{a}$ is a Lie subalgebra of $\mathfrak{g}$ if $\mathfrak{a}$ is stable under the Lie bracket, i.e., $[\mathfrak{a}, \mathfrak{a}] \subset \mathfrak{a}$.

Let $\mathfrak{a}_1$ and $\mathfrak{a}_2$ be subalgebras of a Lie algebra $\mathfrak{g}$, we define $[\mathfrak{a}_1, \mathfrak{a}_2]$ as the Lie subalgebra of $\mathfrak{g}$ generated by elements of the form $[x, y]$, where $x \in \mathfrak{a}_1$, and $y \in \mathfrak{a}_2$.

Let $\mathfrak{a}$ be a subalgebra of a Lie algebra $\mathfrak{g}$. We denote $N(\mathfrak{a})$ the subalgebra of $\mathfrak{g}$ consisting of elements $x$ such that $ad_x(\mathfrak{a}) \subset \mathfrak{a}$. If $N(\mathfrak{a}) = \mathfrak{g}$, then we say that $\mathfrak{a}$ is an ideal of $\mathfrak{g}$.

**Definition 6.1.2.** We define a sequence $(\mathfrak{g}^{(n)})_{n \in \mathbb{N}_0}$ such that

$$\mathfrak{g}^{(0)} = \mathfrak{g}, \quad \forall n \in \mathbb{N}_0, \quad \mathfrak{g}^{(n+1)} = [\mathfrak{g}^{(n)}, \mathfrak{g}^{(n)}].$$

We say that $\mathfrak{g}$ is solvable if there is an integer $n \in \mathbb{N}_0$ such that $\mathfrak{g}^{(n)} = 0$. We define $\mathrm{rad}(\mathfrak{g})$ as the (unique) maximal solvable ideal of $\mathfrak{g}$.
**Definition 6.1.3.** Let $\mathfrak{g}$ be a Lie algebra, we define the sequence $(\mathfrak{g}^n)_{n \in \mathbb{N}_0}$,

$$ \mathfrak{g}^0 = \mathfrak{g}, \quad \forall n \in \mathbb{N}_0, \quad \mathfrak{g}^{n+1} = [\mathfrak{g}, \mathfrak{g}^n]. $$

We say that $\mathfrak{g}$ is nilpotent if there's an integer $n \in \mathbb{N}_0$ such that $\mathfrak{g}^n = 0$.

**Definition 6.1.4.** Let $\mathfrak{g}$ be a Lie algebra. We say that $\mathfrak{g}$ is semi-simple if the only solvable ideal of $\mathfrak{g}$ is $\{0\}$. Moreover, if $\mathfrak{g}$ contains only a single ideal, we say that $\mathfrak{g}$ is simple.

We now provide some fundamental examples of semi-simple Lie algebras:

**Example 6.1.5.** Let $r \in \mathbb{N}$,

$$ \text{The type } A_r: A_r := \mathfrak{sl}_{r+1} = \{X \in \mathcal{M}_{r+1}(\mathbb{C}) \mid \text{tr}(X) = 0\}. $$

$$ \text{The type } B_r: B_r := \mathfrak{so}_{2r+1} = \{X \in \mathcal{M}_{2r+1}(\mathbb{C}) \mid {}^tX + X = 0\}. $$

$$ \text{The type } C_r: C_r := \mathfrak{sp}_r = \{X \in \mathcal{M}_{2r}(\mathbb{C}) \mid {}^tXJ_r + J_rX = 0\}, \text{ with} $$

$$ J_r := \begin{pmatrix} 0 & \mathrm{Id}_r \\ -\mathrm{Id}_r & 0 \end{pmatrix}. $$

$$ \text{The type } D_r: D_r := \mathfrak{so}_{2r} = \{X \in \mathcal{M}_{2r}(\mathbb{C}) \mid {}^tX + X = 0\}, \text{ for } r \ge 2. $$

In all these examples, the Lie bracket corresponds to $[x, y] := x \cdot y - y \cdot x$, with $x, y \in \mathcal{M}_r$ and where $\cdot$ means the usual matrix multiplication.

In the previous example, all these Lie algebras are simple, except for $D_2$. We also have the decomposition $D_2 = A_1 \oplus A_1$.

**Proposition 6.1.6.** A semi-simple Lie algebra is a direct sum of simple Lie algebras.

Thus, to classify semi-simple Lie algebras, it is sufficient to classify simple Lie algebras. We will see via the Cartan-Killing Theorem that the 4 types of algebras that we have described in the previous example almost completely classify the set of semi-simple Lie algebras.

**Definition 6.1.7.** A representation of a Lie algebra $\mathfrak{g}$ over a $\mathbb{C}$-vector space $E$ corresponds to a linear map $\varphi: \mathfrak{g} \to GL(E)$. We will call $\dim(\varphi)$ the dimension of this representation. We say that the representation $\varphi$ is irreducible if it does not admit any non-trivial vector subspace stable under the linear mapping $\varphi$.

**Example 6.1.8.** The derivation $\mathrm{ad}$ induces a representation of a Lie algebra $\mathrm{ad}(\mathfrak{g})$ on the vector space $\mathfrak{g}$:

$$ \mathrm{Ad}(\mathfrak{g}) : \mathfrak{g} \to GL(\mathfrak{g}), \quad x \mapsto \mathrm{ad}_x. $$

We can then define the Killing bilinear form on $\mathfrak{g} \times \mathfrak{g}$:

$$ <x, y> := \mathrm{Tr}(\mathrm{Ad}_x \circ \mathrm{Ad}_y). $$

**Theorem 6.1.9.** Let $\mathfrak{g}$ be a Lie algebra. The following conditions are equivalent:

a) $\mathfrak{g}$ is semi-simple.

b) $\mathrm{rad}(\mathfrak{g}) = 0$

c) The Killing form $<,>$ is non-degenerate.

The previous theorem allows us in particular to identify a semi-simple Lie algebra with its dual. Thus, for an element $x \in \mathfrak{g}$, we can canonically associate an element $x^* \in \mathfrak{g}^*$.

## 6.2 Root System of a Lie Algebra

Let $\mathfrak{g}$ be a Lie algebra. A Cartan subalgebra is defined as a nilpotent Lie subalgebra of $\mathfrak{g}$ such that $N(\mathfrak{h}) = \mathfrak{h}$. Cartan subalgebras are not unique, but they share the same dimension over $\mathbb{C}$. This dimension is then called the rank of $\mathfrak{g}$, denoted $rg(\mathfrak{g})$.

Throughout this section, we fix a semi-simple Lie algebra $\mathfrak{g}$ and a Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{g}$. The Killing form extends to the Cartan subalgebra $\mathfrak{h}$, and it is non-degenerate on $\mathfrak{h}$. Through the identification of $\mathfrak{h}$ with its dual, we can define the Killing form on $\mathfrak{h}^*$.
**Definition 6.2.1.** Let $\alpha \in \mathfrak{h}^*$ be an element of the dual of $\mathfrak{h}$. We say that $\alpha$ is a root of $\mathfrak{g}$ relative to $\mathfrak{h}$ when there exists a nonzero element $x \in \mathfrak{g} \setminus \{0\}$ such that

$$ \forall h \in \mathfrak{h}, \quad ad_h(x) = \alpha(h)x. \qquad (6.1) $$

We will denote $g_\alpha$ as the set of $x \in \mathfrak{g}$ satisfying 6.1. We will denote $\Delta(\mathfrak{g})$ as the set of roots of $\mathfrak{g}$ relative to $\mathfrak{h}$. We will call $\Delta(\mathfrak{g})$ the root system of the semi-simple Lie algebra $\mathfrak{g}$.

The root system of $\mathfrak{g}$ is a finite set. Moreover, the Lie algebras $\mathfrak{g}_\alpha$ have dimension 1, and they yield a decomposition of $\mathfrak{g}$:

$$ \mathfrak{g} = \mathfrak{h} \oplus \bigoplus_{\alpha \in \Delta} g_{\alpha}. $$

Through the identification between $\mathfrak{h}$ and its dual, we consider for every root $\alpha \in \mathfrak{h}^*$ the corresponding element $\alpha' \in \mathfrak{h}$. We set $\mathfrak{h}_0 := \mathrm{Vect}_{\mathbb{R}}((\alpha')_{\alpha \in \Delta(\mathfrak{g})})$ the $\mathbb{R}$-vector space generated by the elements $\alpha'$. We then define

$$ \alpha^{\vee} := \frac{2}{< \alpha', \alpha' >} \alpha'. $$

For every $\alpha, \beta \in \Delta(\mathfrak{g})$, we observe that

$$ \beta(\alpha^{\vee}) = \frac{2 < \beta, \alpha >}{< \alpha, \alpha >}. $$

We denote $a(\beta, \alpha)$ the above expression.

**Notation.** We denote $\mathfrak{h}_0$ as the $\mathbb{R}$-vector space generated by the roots $\Delta(\mathfrak{g})$, and $\mathfrak{h}_0^*$ as its dual.

**Fact 6.2.2.** Let $\alpha, \beta \in \Delta(\mathfrak{g})$. We have:

1) $\beta(\alpha^{\vee}) \in \mathbb{Z}$.

2) $\beta - a(\beta, \alpha)\alpha \in \Delta(\mathfrak{g})$.

3) If there exists $c \in \mathbb{C}$ such that $\beta = c\alpha$, then $c = \pm 1$.

These three properties characterize an "abstract" root system in the sense of [Ser66] and [Bou81]. Given two semi-simple Lie algebras, we know that there exists an isometry $(\mathrm{Vect}_{\mathbb{R}}(\Delta(\mathfrak{g}_1)), ||_1) \to (\mathrm{Vect}_{\mathbb{R}}(\Delta(\mathfrak{g}_2)), ||_2)$ mapping $\Delta(\mathfrak{g}_1)$ to $\Delta(\mathfrak{g}_2)$ if and only if $\mathfrak{g}_1$ is isomorphic to $\mathfrak{g}_2$ as a Lie algebra, denoting $||_1$ and $||_2$ as the Euclidean norms of the two vector spaces $\mathrm{Vect}_{\mathbb{R}}(\Delta(\mathfrak{g}_1))$ and $\mathrm{Vect}_{\mathbb{R}}(\Delta(\mathfrak{g}_2))$. This property, in particular, reduces the study of simple Lie algebras to the study of root systems. From this correspondence follows the subsequent theorem, allowing the classification of simple Lie algebras.

**Theorem 6.2.3.** [Bou81] [Cartan-Killing Theorem]

Any simple Lie algebra is isomorphic either to $A_r$, $B_r$, $C_r$, $D_r$, or to an exceptional Lie algebra $G_2$, $F_4$, $E_6$, $E_7$, $E_8$.

To see the construction of the exceptional Lie algebras $G_2$, $F_4$, $E_6$, $E_7$, or $E_8$, refer to [Bou81].

Let us define $H_{\alpha^{\vee}} := \{\beta \in \mathfrak{h}_0^* | \langle \beta, \alpha^{\vee} \rangle = 0\}$.

We note that, since the set of roots is finite, the set

$$ \mathfrak{h}_0^* \setminus \bigcup_{\alpha \in \Delta(\mathfrak{g})} H_{\alpha^{\vee}} $$

is non-empty. Let $\gamma$ be an element of this set. We then have the following decomposition:

$$ \Delta(\mathfrak{g}) = \{\alpha \in \Delta(\mathfrak{g}) | \langle \gamma, \alpha^{\vee} \rangle > 0\} \cup \{\alpha \in \Delta(\mathfrak{g}) | \langle \gamma, \alpha^{\vee} \rangle < 0\}. $$

We denote $\Delta_+(\mathfrak{g}, \gamma) = \Delta_+(\mathfrak{g}) := \{\alpha \in \Delta(\mathfrak{g}) | \langle \gamma, \alpha^{\vee} \rangle > 0\}$ as the set of positive roots. If a positive root $\alpha$ cannot be decomposed as a sum of two positive roots, we call $\alpha$ a fundamental (or simple) root.

**Definition 6.2.4.** Let $\alpha$ be a positive root. We say that $\alpha$ is a fundamental (or simple) root if $\alpha$ cannot be decomposed as a sum of two positive roots. We shall note $\psi(\mathfrak{g})$ as the set of fundamental roots.
**Fact 6.2.5.** The set of fundamental roots forms a basis for the $\mathbb{R}$-vector space $\mathfrak{h}_0^* = \mathrm{Vect}_{\mathbb{R}}((\alpha')_{\alpha \in \Delta(\mathfrak{g})})$. As a consequence, there are $r = \dim_{\mathbb{R}}(\mathfrak{h}_0)$ fundamental roots.

We now aim to introduce the concept of weight:

**Definition 6.2.6.** Let $\varphi : \mathfrak{g} \to GL(E)$ be a representation of the semi-simple Lie algebra $\mathfrak{g}$, with $E$ being of finite dimension. Let $u \in E$ be a vector. We say that $u$ is an eigenvector if there exists a linear form $\lambda \in \mathfrak{h}^*$ on $\mathfrak{h}$ such that

$$ \forall H \in \mathfrak{h}, \quad \varphi(H)u = \lambda(H)u. $$

We call $\lambda$ the weight of $u$.

If furthermore, we have $\lambda(\alpha^\vee) \ge 0$ for every simple root $\alpha$ of $\mathfrak{g}$, then we say that $\lambda$ is a dominant weight.

Let $\psi(\mathfrak{g}) = \{\alpha_1, \dots, \alpha_r\}$ be the set of simple roots of $\mathfrak{g}$. We then define $\lambda_j(\alpha_i^\vee) := \delta_{i,j}$ for all $1 \le i, j \le r$. In particular, the $\lambda_j$ are dominant weights. We call them fundamental weights (or simple weights).

**Fact 6.2.7.** Each dominant weight $\lambda$ has a unique decomposition in terms of the fundamental weights

$$ \lambda = n_1\lambda_1 + \dots + n_r\lambda_r \quad (n_1, \dots, n_r \in \mathbb{N}_0). $$

**Fact 6.2.8.** There is a bijective correspondence between the set of irreducible finite-dimensional representations (up to isomorphism), and the set of dominant weights.

**Theorem 6.2.9.** [of Weyl]

Let $\varphi$ be an irreducible representation of $\mathfrak{g}$. Then we have

$$ \dim(\varphi) = \prod_{\alpha \in \Delta^+(\mathfrak{g})} \frac{< \alpha^\vee, (n_1+1)\lambda_1 + \dots + (n_r+1)\lambda_r >}{< \alpha^\vee, \lambda_1 + \dots + \lambda_r >}. $$

**Remark 6.2.10.** At first glance, the result above depends on the choice of the Cartan subalgebra $\mathfrak{h}$, and the choice of $\gamma$. However, according to [KMT24, §2.3], these are independent of these choices. The reason being that, according to [Hum72, §16.2], two Cartan subalgebras are necessarily conjugate.

We note in passing that Fact 6.2.8 and Theorem 6.2.9 particularly allow us to determine an expression for $r_{\mathfrak{g}}$, the number of irreducible representations (up to isomorphism) of dimension $n$ of a semi-simple Lie algebra $\mathfrak{g}$.

**Definition 6.2.11.** [Zag94] We associate with a semi-simple Lie algebra $\mathfrak{g}$ a zeta function

$$ \zeta_{\mathfrak{g}}(s) := \sum_{\varphi} \frac{1}{\dim(\varphi)^s}, $$

where the sum is over the finite-dimensional irreducible representations $\varphi$ of $\mathfrak{g}$ (up to isomorphism).

This zeta function generalizes the series that appear in Witten's volume formulas in [Wit91].

From Fact 6.2.8 and Weyl's Theorem 6.2.9, it follows that Witten's zeta functions can be written as a series of products of linear forms in $r$ variables, $r$ being the number of fundamental roots. More precisely, we have the following expression:

$$ \zeta_{\mathfrak{g}}(s) := \prod_{\alpha \in \Delta^+(\mathfrak{g})} < \alpha^\vee, \lambda_1 + \dots + \lambda_r >^s \sum_{n_1, \dots, n_r \ge 0} \frac{1}{\prod_{\alpha \in \Delta^+(\mathfrak{g})} < \alpha^\vee, (n_1+1)\lambda_1 + \dots + (n_r+1)\lambda_r >^s}. $$

Matsumoto, Tsumura, and Komori introduced in [KMT10b] multizeta analogues of Witten's zeta functions, by considering Dirichlet series of the form

$$ \sum_{n_1, \dots, n_r \ge 0} \frac{1}{\prod_{\alpha \in \Delta^+(\mathfrak{g})} \langle \alpha^\vee, (n_1+1)\lambda_1 + \dots + (n_r+1)\lambda_r \rangle^{s_\alpha}}. $$

This allows for obtaining recursive formulas for these multizeta functions (see [KMT10b, Theorem 3.1]), and to derive several formulas for some positive integers
## 6.3 Some Results on $\zeta_{g_2}$ and $\zeta_{s_0(5)}$

Here, we aim to detail the formulas obtained in Theorem A and in Theorem D for the two Witten zeta functions $\zeta_{g_2}$ and $\zeta_{s_0(5)}$, primarily focusing on the values at $s = 0$. The special values $\zeta_{g_2}(0)$ and $\zeta_{s_0(5)}(0)$ are very simple to compute using Corollary A1. On the other hand, the calculations for the values $\zeta'_{g_2}(0)$ and $\zeta'_{s_0(5)}(0)$ will prove to be more challenging. We propose to establish a formula using Corollary D2 when $P = 2$, that is, when the Dirichlet series describing $Z(s, s')$ only involves two variables $n_1, n_2$. We furthermore fix the direction $(\mu, \mu') = (1, 1)$, and set $d_1 = d_2 = 1$. Let's start by examining the values of $Z_\Delta(s)$ at nonpositive integers using Corollary A2:

**Lemma 6.3.1.** Let $N \in \mathbb{N}_0$. We have

$$Z_{\Delta}(-N) = \frac{(-1)^{(1+Q)N}}{Q} N! \sum_{k=0}^{QN} Q_N^0(\emptyset, (k, QN-k)) \frac{\zeta(-N-k)}{k!} \frac{\zeta(-Q+1)N+k}{(QN-k)!} \\ + \frac{(-1)^{QN+1}}{Q+1} N!^2 (Q_N^0(\{1\}, (Q+1)N+1) + Q_N^0(\{2\}, (Q+1)N+1)) \frac{\zeta(-Q+2)N-1}{((Q+1)N+1)!},$$

with

$$Q_N^0(\emptyset, (k_1, k_2)) = (-1)^{N(Q-1)} N!^{Q-1} \sum_{\substack{j=1 \\ |\mathbf{w}_1|=k_1, |\mathbf{w}_2|=k_2 \\ \forall q, w_{1,q}+w_{2,q}=N}} \sum_{\mathbf{w}=(\mathbf{w}_p)_{p=1,2} \in (\mathbb{N}_0^Q)^2} \left(\binom{k_1}{\mathbf{w}_1}\binom{k_2}{\mathbf{w}_2}\right) \prod_{q=1}^Q c_{q,1}^{w_{1,q}} c_{q,2}^{w_{2,q}}, \quad (k_1+k_2=QN)$$

$$Q_N^0(\{1\}, (Q+1)N+1) = (-1)^{N(Q-1)} N!^{Q-1} \sum_{\substack{j=1 \\ |\mathbf{w}_2|=(Q+1)N+1 \\ w_{2,j} \ge 2N+1}} \sum_{\mathbf{w}_2 \in \mathbb{N}_0^Q} \frac{(-N-1)...(N+1-w_{2,j})}{\prod_{q \ne j} w_{2,q}!} c_{j,2}^{w_{2,j}} \prod_{\substack{q=1 \\ q \ne j}}^Q c_{q,1}^{N-w_{2,q}} c_{q,2}^{w_{2,q}},$$

$$Q_N^0(\{2\}, (Q+1)N+1) = (-1)^{N(Q-1)} N!^{Q-1} \sum_{\substack{j=1 \\ |\mathbf{w}_1|=(Q+1)N+1 \\ w_{1,j} \ge 2N+1}} \sum_{\mathbf{w}_1 \in \mathbb{N}_0^Q} \frac{(-N-1)...(N+1-w_{1,j})}{\prod_{q \ne j} w_{1,q}!} c_{j,1}^{w_{1,j}} \prod_{\substack{q=1 \\ q \ne j}}^Q c_{q,2}^{N-w_{1,q}} c_{q,1}^{w_{1,q}}.$$

*Proof.* By Corollary A2, we get

$$Z_{\Delta}(-N) = \sum_{\mathcal{P} \subseteq [1,2]} \frac{(-1)^{(|\mathcal{P}|+1+Q)N+|\mathcal{P}|}}{Q + |\mathcal{P}|} N!^{|\mathcal{P}|+1} \sum_{\substack{\mathbf{k}=(k_p)_{p \in \mathcal{P}} \in \mathbb{N}_0^{\mathcal{P}} \\ |\mathbf{k}|=(Q+|\mathcal{P}|)N+|\mathcal{P}|}} Q_N^0(\mathcal{P}, \mathbf{k}) \prod_{p \in \mathcal{P}^c} \frac{\zeta(-N-k_p)}{k!}, \\ = \frac{(-1)^{(1+Q)N}}{Q} N! \sum_{\substack{k_1,k_2 \ge 0 \\ k_1+k_2=QN}} Q_N^0(\emptyset, (k_1, k_2)) \prod_{p=1}^2 \frac{\zeta(-N-k_p)}{k_p!} \\ + \frac{(-1)^{QN+1}}{Q+1} N!^2 (Q_N^0(\{1\}, (Q+1)N+1) + Q_N^0(\{2\}, (Q+1)N+1)) \frac{\zeta(-Q+2)N-1}{((Q+1)N+1)!},$$

with

$$Q_N^0(\mathcal{P}, \mathbf{k}) = (-1)^{N(Q-1)} N!^{Q-1} \sum_{\substack{j=1 \\ |\mathbf{w}_1|=(\mathbf{w}_p)_{p \in \mathcal{P}} \in (\mathbb{N}_0^Q)^{\mathcal{P}} \\ \forall p \in \mathcal{P}^c |\mathbf{w}_p|=k_p}} \sum_{\substack{\mathbf{v}=(\mathbf{v}_p)_{p \in \mathcal{P}} \in (\mathbb{N}_0^{[1,Q]} \setminus \{j\})^\mathcal{P} \\ \forall q \ne j, \mathbf{v}(q)+\mathbf{w}(q)=N}} \left( \prod_{p \in \mathcal{P}^c} \binom{k_p}{\mathbf{w}_p} \prod_{q=1}^Q c_{q,p}^{w_{p,q}} \right) \left( \prod_{p \in \mathcal{P}} \binom{-N-1}{|\mathbf{v}_p|} \binom{|\mathbf{v}_p|}{\mathbf{v}_p} \prod_{\substack{q=1 \\ q \ne j}}^Q c_{q,p}^{v_{p,q}} \right).$$
• If $\mathcal{P} = \emptyset$ and $k_1 + k_2 = QN$, then we have

$$Q_N^0(\emptyset, \mathbf{k}) = (-1)^{N(Q-1)} N! Q^{-1} \sum_{\substack{j=1 \\ |\mathbf{w}_1|=k_1, |\mathbf{w}_2|=k_2 \\ \forall q \neq j, w_{1,q}+w_{2,q}=N}} \sum_{\mathbf{w}=(\mathbf{w}_p)_{p=1,2} \in (\mathbb{N}_0^Q)^2} \left(\binom{k_1}{\mathbf{w}_1}\binom{k_2}{\mathbf{w}_2}\right) \prod_{q=1}^Q c_{q,1}^{w_{1,q}} c_{q,2}^{w_{2,q}}.$$

Let's observe that

$$\begin{aligned} w_{1,j} + w_{2,j} &= \sum_{q=1}^{Q} w_{1,q} + w_{2,q} - \sum_{\substack{q=1 \\ q \neq j}}^{Q} w_{1,q} + w_{2,q}, \\ &= k_1 + k_2 - (Q-1)N, \\ &= N. \end{aligned}$$

Let $\mathbf{w} = (\mathbf{w}_p)_{p=1,2} \in (\mathbb{N}_0^Q)^2$. The condition $|\mathbf{w}_1| = k_1$, $|\mathbf{w}_2| = k_2$, and $w_{1,q} + w_{2,q} = N$ for all $q \neq j$, is equivalent to the condition $w_{1,q} + w_{2,q} = N$ for all $q$ and $|\mathbf{w}_1| = k_1$, $|\mathbf{w}_2| = k_2$. We then find that

$$Q_N^0(\emptyset, \mathbf{k}) = (-1)^{N(Q-1)} N! Q^{-1} \sum_{\substack{j=1 \\ |\mathbf{w}_1|=k_1, |\mathbf{w}_2|=k_2 \\ \forall q, w_{1,q}+w_{2,q}=N}} \sum_{\mathbf{w}=(\mathbf{w}_p)_{p=1,2} \in (\mathbb{N}_0^Q)^2} \left(\binom{k_1}{\mathbf{w}_1}\binom{k_2}{\mathbf{w}_2}\right) \prod_{q=1}^Q c_{q,1}^{w_{1,q}} c_{q,2}^{w_{2,q}}.$$

• If $\mathcal{P} = \{1\}$ and $k_2 = (Q+1)N+1$, then we have that

$$Q_N^0(\{1\}, (Q+1)N+1) = (-1)^{N(Q-1)} N! Q^{-1} \sum_{\substack{j=1 \\ |\mathbf{w}_2|=(Q+1)N+1 \\ \forall q \neq j, w_{2,q} \le N}} \sum_{\mathbf{w}_2 \in \mathbb{N}_0^Q} \left( \left( \binom{k_1}{w_{2,1}, \dots, w_{2,Q}} \right) \prod_{q=1}^Q c_{q,2}^{w_{2,q}} \right) \left( \left( \binom{-N-1}{\sum_{q \neq j} (N-w_{2,q})} \right) \left( \binom{\sum_{q \neq j} N-w_{2,q}}{(N-w_{2,q})_{q \neq j}} \right) \prod_{\substack{q=1 \\ q \neq j}}^Q c_{q,1}^{N-w_{2,q}} \right).$$

Let's note that

$$\begin{aligned} \sum_{\substack{q=1 \\ q \neq j}}^{Q} (N - w_{2,q}) &= (Q - 1)N - |\mathbf{w}_2| + w_{2,j} \\ &= -2N - 1 + w_{2,j}. \end{aligned}$$

If this term is negative, then the binomial coefficient $\left( \frac{-N-1}{\sum_{q \neq j} (N-w_{2,q})} \right)$ vanishes. Therefore we can assume the condition $w_{2,j} \ge 2N+1$ instead:

$$Q_N^0(\{1\}, k_2) = (-1)^{N(Q-1)} N! Q^{-1} \sum_{\substack{j=1 \\ |\mathbf{w}_2|=(Q+1)N+1 \\ w_{2,j} \ge 2N+1}} \sum_{\mathbf{w}_2 \in \mathbb{N}_0^Q} \frac{(-N-1) \dots (N+1-w_{2,j})}{\prod_{q \neq j} w_{2,q}!} c_{j,2}^{w_{2,j}} \prod_{\substack{q=1 \\ q \neq j}}^{Q} c_{q,1}^{N-w_{2,q}} c_{q,2}^{w_{2,q}}.$$

• If $\mathcal{P} = \{2\}$ and $k_1 = (Q+1)N+1$, therefore we get

$$Q_N^0(\{2\},) = (-1)^{N(Q-1)} N! Q^{-1} \sum_{\substack{j=1 \\ |\mathbf{w}_1|=(Q+1)N+1 \\ w_{1,j} \ge 2N+1}} \sum_{\mathbf{w}_1 \in \mathbb{N}_0^Q} \frac{(-N-1) \dots (N+1-w_{1,j})}{\prod_{q \neq j} w_{1,q}!} c_{j,1}^{w_{1,q}} \prod_{\substack{q=1 \\ q \neq j}}^{Q} c_{q,2}^{N-w_{1,q}} c_{q,1}^{w_{1,q}}.$$
<figure><img src="image_21.png" /></figure>

Via the previous lemma, and by using the fact that $\zeta(-n) = (-1)^{n+1} \frac{B_{n+1}}{n+1}$, we have the simplified version of the Lemma, which is easier to compute:

$$Z_{\Delta}(-N) = \frac{(-1)^{QN}}{Q} N!^Q \sum_{k=0}^{QN} \alpha_k(N) \frac{B_{N+k+1}}{(N+k+1)k!} \frac{B_{(Q+1)N-k+1}}{((Q+1)N-k+1)(QN-k)!} \quad (6.2)$$

with

$$\alpha_k(N) = \sum_{j=1}^{Q} \sum_{\substack{(w_q)_{q \in [1,Q]} \in [0,N]^Q \\ w_1 + \dots + w_Q = k}} \binom{k}{(w_1, \dots, w_Q)} \binom{QN-k}{(N-w_1, \dots, N-w_Q)} \prod_{q=1}^{Q} c_{q,1}^{w_q} c_{q,2}^{N-w_q},$$

$$\beta(N) = \sum_{j=1}^{Q} \sum_{\substack{(w_q)_{q \in [1,Q]} \in [0,(Q+1)N+1]^Q \\ w_1 + \dots + w_Q = (Q+1)N+1 \\ w_j \ge 2N+1}} \frac{(-N-1)\dots(N+1-w_j)}{\prod_{q \ne j} w_q!} \left( c_{j,2}^{w_j} \prod_{\substack{q=1 \\ q \ne j}}^{Q} c_{q,1}^{N-w_q} c_{q,2}^{w_q} + c_{j,1}^{w_j} \prod_{\substack{q=1 \\ q \ne j}}^{Q} c_{q,2}^{N-w_q} c_{q,1}^{w_q} \right).$$

**Lemma 6.3.2.** We have

$$Z'_{\Delta}(0) = \\ Q\zeta(0)^2\gamma - \zeta(-1) \left(\sum_{j=1}^{Q} (c_{j,1} + c_{j,2})\right)\gamma \\ - \frac{1}{Q+1}\zeta(-1) \sum_{j=1}^{Q} \sum_{\substack{f=1 \\ f \ne j}}^{Q} \left(\frac{1}{c_{j,2}}(c_{f,1} - c_{j,1})\ln\left(1 + \frac{c_{f,2}}{c_{j,2}}\right) + \frac{1}{c_{j,1}}(c_{f,2} - c_{j,2})\ln\left(1 + \frac{c_{f,1}}{c_{j,1}}\right)\right) \\ + \frac{\zeta'(0)\zeta(0)}{Q} - \frac{\zeta'(-1)}{Q+1} \sum_{j=1}^{Q} (c_{j,1} + c_{j,2}) - \gamma\zeta(0)^2 \sum_{j=1}^{Q} (c_{j,1}c_{j,2})^{-1} + \gamma\zeta(-1) \sum_{j=1}^{Q} (c_{j,1}^{-1}c_{j,2} + c_{j,2}^{-1}c_{j,1}) \\ + \sum_{j=1}^{Q} \partial_s \zeta^B(s, d'_j|c_{j,1}, c_{j,2})|_{s=0}.$$

*Proof.* Let's recall that $Z'_{\Delta}(0) = Z'(-(0,0))$ We set $P=2$ and we fixed the directions $(\mu, \mu') = (1,1)$.

Therefore, by Corollary D2, we have:

$$Z'(-(0,0)) = \\ \sum_{\mathcal{P} \subseteq [1,2]} (-1)^{|\mathcal{P}|} \frac{1}{Q + |\mathcal{P}|} \sum_{\mathbf{k}=(k_p)_{p \in \mathcal{P}^c} \in \mathbb{N}_0^{\mathcal{P}^c} \atop |\mathbf{k}| = |\mathcal{P}|} \prod_{p \in \mathcal{P}^c} \frac{\zeta(-k_p)}{k_p!} \sum_{j=1}^{Q} \left( \left( \prod_{p \in \mathcal{P}^c} c_{j,p}^{k_p} \right) \gamma (1 + |\mathcal{P}|) + Q^1(j, \mathcal{P}, \mathbf{k}) \right) \\ + \sum_{\mathcal{P} \subseteq [1,2]} \sum_{j=1}^{Q} (-1)^{|\mathcal{P}|} \frac{1}{Q + |\mathcal{P}|} \sum_{\mathbf{k}=(k_p)_{p \in \mathcal{P}^c} \in \mathbb{N}_0^{\mathcal{P}^c} \atop |\mathbf{k}| = |\mathcal{P}|} \prod_{p \in \mathcal{P}^c} c_{j,p}^{k_p} \left( \sum_{i \in \mathcal{P}^c} \frac{\zeta'(-k_i)}{k_i!} \prod_{p \in \mathcal{P}^c \setminus \{i\}} \frac{\zeta(-k_p)}{k_p!} \right) \\ + \sum_{j=1}^{Q} \partial_s \zeta^B(s, d'_j|c_{j,1}, c_{j,2})|_{s=0} - \sum_{j=1}^{Q} \gamma \sum_{\mathcal{P} \subseteq [1,2]} (-1)^{|\mathcal{P}|} \prod_{p \in \mathcal{P}} c_{j,p}^{-1} \sum_{\mathbf{k}=(k_p)_{p \in \mathcal{P}^c} \in \mathbb{N}_0^{\mathcal{P}^c} \atop |\mathbf{k}| = |\mathcal{P}|} \prod_{p \in \mathcal{P}^c} c_{j,p}^{k_p} \frac{\zeta(-k_p)}{k_p!},$$
with

$$Q^1(j, \mathcal{P}, \mathbf{k}) = \sum_{\substack{f=1 \\ f \neq j}}^{Q} \sum_{\mathbf{w}' = (w'_p)_{p \in \mathcal{P}^c} \in \prod_{p \in \mathcal{P}^c} [0, k_p]} F_{\mathcal{P}, j, f, \mathbf{0}, (w'_p \mathbf{e}_f + (k_p - w'_p) \mathbf{e}_j)_{p \in \mathcal{P}^c}} (\mathbf{0}, \mathbf{0})) \left( \prod_{p \in \mathcal{P}^c} \binom{k_p}{w'_p} c_{f, p}^{w'_p} c_{j, p}^{k_p - w'_p} \right) + (Q-1)\gamma \prod_{p \in \mathcal{P}^c} c_{j, p}^{k_p},$$

where we recall that $\mathbf{e}_n = (0, \dots, 1, \dots, 0) \in \mathbb{R}^Q$ is the vector with a 1 in the n-th component, and 0 elsewhere.

Developing the sum in the formula of $Z'(-(\mathbf{0}, \mathbf{0}))$, we get

$$Z'(-(\mathbf{0}, \mathbf{0})) = \zeta(0)^2 \gamma - \frac{2\zeta(-1)}{Q+1}\gamma \left(\sum_{j=1}^{Q} (c_{j,1} + c_{j,2})\right) + \frac{1}{Q}\zeta(0)^2 \sum_{j=1}^{Q} Q^1(j, \emptyset, (0,0)) \\ - \frac{1}{Q+1}\zeta(-1) \sum_{j=1}^{Q} (Q^1(j, \{1\}, 1) + Q^1(j, \{2\}, 1)) + \frac{\zeta'(0)\zeta(0)}{Q} - \frac{\zeta'(-1)}{Q+1} \sum_{j=1}^{Q} (c_{j,1} + c_{j,2}) \\ + \sum_{j=1}^{Q} \partial_s \zeta^B(s, d'_j | c_{j,1}, c_{j,2})_{s=0} - \gamma\zeta(0)^2 \sum_{j=1}^{Q} (c_{j,1}c_{j,2})^{-1} + \gamma\zeta(-1) \sum_{j=1}^{Q} (c_{j,1}^{-1}c_{j,2} + c_{j,2}^{-1}c_{j,1}).$$

Let's now study the values of $Q^1(j, \mathcal{P}, \mathbf{k})$:

* If $\mathcal{P} = \emptyset$ and $\mathbf{k} = (0,0)$, we have

$$Q^1(j, \emptyset, (0,0)) = \sum_{\substack{f=1 \\ f \neq j}}^{Q} F_{\emptyset, j, f, \mathbf{0}}(\mathbf{0}, \mathbf{0})) + (Q-1)\gamma.$$

By the Example 1.3.4, we have that $F_{\emptyset, j, f, \mathbf{0}}(\mathbf{0}, \mathbf{0})) = 0$ therefore $Q^1(j, \emptyset, (0,0)) = (Q-1)\gamma$.

* If $\mathcal{P} = \{1\}$ and $k_2 = 1$, we have that

$$Q^1(j, \{1\}, 1) = (Q-1)\gamma c_{j,2} + \sum_{\substack{f=1 \\ f \neq j}}^{Q} \sum_{w'=0}^{1} F_{\{1\}, j, f, \mathbf{0}, w' \mathbf{e}_f + (1-w') \mathbf{e}_j}(\mathbf{0}, \mathbf{0})) \left( \binom{1}{w'} c_{f,2}^{w'} c_{j,2}^{1-w'} \right) \\ = (Q-1)\gamma c_{j,2} + \sum_{\substack{f=1 \\ f \neq j}}^{Q} (F_{\{1\}, j, f, \mathbf{0}, \mathbf{e}_f}(\mathbf{0}, \mathbf{0})) c_{f,2} + F_{\{1\}, j, f, \mathbf{0}, \mathbf{e}_j}(\mathbf{0}, \mathbf{0}) c_{j,2}).$$

We then find by Example 1.3.6 that

$$F_{\{1\}, j, f, \mathbf{0}, \mathbf{e}_j}(\mathbf{0}, \mathbf{0}) = -\frac{1}{c_{j,1}} \ln \left( 1 + \frac{c_{f,1}}{c_{j,1}} \right),$$

and thanks to Example 1.3.5, we have that

$$F_{\{1\}, j, f, \mathbf{0}, \mathbf{e}_f}(\mathbf{0}, \mathbf{0}) = \frac{1}{c_{f,1}} \ln \left( 1 + \frac{c_{f,1}}{c_{j,1}} \right),$$
therefore we obtain

$$Q^1(j, \{1\}, 1) = (Q-1)\gamma c_{j,2} + \sum_{\substack{f=1 \\ f \neq j}}^{Q} \left( \frac{c_{f,2}}{c_{f,1}} - \frac{c_{j,2}}{c_{j,1}} \right) \ln \left( 1 + \frac{c_{f,1}}{c_{j,1}} \right)$$

* If $\mathcal{P} = \{1\}$ and $k_1 = 1$, mutatis mutandis, on get

$$Q^1(j, \{2\}, 1) = (Q-1)\gamma c_{j,1} + \sum_{\substack{f=1 \\ f \neq j}}^{Q} \left( \frac{c_{f,1}}{c_{f,2}} - \frac{c_{j,1}}{c_{j,2}} \right) \ln \left( 1 + \frac{c_{f,2}}{c_{j,2}} \right).$$

From the last three points, we get

$$\begin{aligned} Z'(-(\mathbf{0}, \mathbf{0})) = & \sum_{j=1}^{Q} \left( \frac{c_{j,1} + c_{j,2}}{c_{j,1} + c_{j,2}} \right) \gamma \\ & - \frac{Q-1}{Q+1} \zeta(-1) \left( \sum_{j=1}^{Q} (c_{j,1} + c_{j,2}) \right) \gamma \\ & - \frac{1}{Q+1} \zeta(-1) \sum_{\substack{j=1 \\ f \neq j}}^{Q} \left( \left( \frac{c_{f,1}}{c_{f,2}} - \frac{c_{j,1}}{c_{j,2}} \right) \ln \left( 1 + \frac{c_{f,2}}{c_{j,2}} \right) + \left( \frac{c_{f,2}}{c_{f,1}} - \frac{c_{j,2}}{c_{j,1}} \right) \ln \left( 1 + \frac{c_{f,1}}{c_{j,1}} \right) \right) \\ & + \frac{\zeta'(0)\zeta(0)}{Q} - \frac{\zeta'(-1)}{Q+1} \sum_{j=1}^{Q} (c_{j,1} + c_{j,2}) - \gamma\zeta(0)^2 \sum_{j=1}^{Q} (c_{j,1}c_{j,2})^{-1} + \gamma\zeta(-1) \sum_{j=1}^{Q} (c_{j,1}^{-1}c_{j,2} + c_{j,2}^{-1}c_{j,1}) \\ & + \sum_{j=1}^{Q} \partial_s \zeta^B(s, d'_j|c_{j,1}, c_{j,2})|_{s=0}. \end{aligned}$$

□

**Remark 6.3.3.** Under the assumptions of the previous lemma, by giving an integer $Q \ge 1$ and the coefficients $c = (c_{q,p})_{q \in [1,Q], p \in [1,2]}$, the SAGE code 7.1 shall compute the following expression

$$\begin{aligned} & Q\zeta(0)^2\gamma - \frac{Q-1}{Q+1}\zeta(-1)\left(\sum_{j=1}^{Q}(c_{j,1} + c_{j,2})\right)\gamma \\ & - \frac{1}{Q+1}\zeta(-1)\sum_{\substack{j=1 \\ f \neq j}}^{Q}\sum_{\substack{f=1 \\ f \neq j}}^{Q}\left(\left(\frac{c_{f,1}}{c_{f,2}} - \frac{c_{j,1}}{c_{j,2}}\right)\ln\left(1 + \frac{c_{f,2}}{c_{j,2}}\right) + \left(\frac{c_{f,2}}{c_{f,1}} - \frac{c_{j,2}}{c_{j,1}}\right)\ln\left(1 + \frac{c_{f,1}}{c_{j,1}}\right)\right) \\ & + \frac{\zeta'(0)\zeta(0)}{Q} - \frac{\zeta'(-1)}{Q+1}\sum_{j=1}^{Q}(c_{j,1} + c_{j,2}) - \gamma\zeta(0)^2\sum_{j=1}^{Q}(c_{j,1}c_{j,2})^{-1} + \gamma\zeta(-1)\sum_{j=1}^{Q}(c_{j,1}^{-1}c_{j,2} + c_{j,2}^{-1}c_{j,1}). \end{aligned}$$

Let's note that the result of such computations are not rounded up, but give an expression using Euler's constant $\gamma$, $\zeta(0)$, $\zeta(-1)$, $\zeta'(0)$, $\zeta'(-1)$ and some logarithm.

Even though we have a formula to compute at nonpositive integers of the derivative with respect to $s$ of Barnes' zeta function $\partial_s\zeta^B(s, d'_j|c_{j,1}, c_{j,2})|_{s=0}$ via Corollary B1, by setting $N = 0$ and $\mathbf{R} = \mathbf{0}$, we will instead use Theorem 1.2.11 proved by Aoki and Sakane. The formula in this theorem provides more easily exploitable expressions. In particular, we will use the formulas obtained in Example 1.2.12 to simplify the expressions for $\zeta'_{g_2}(0)$ and $\zeta'_{s_0(5)}(0)$. We will use the formulas (1.4), (1.3) and (1.2) in order to get
a more suitable expression of the derivative of the values at nonpositive integers of the derivative of the
Barnes zeta function:

$$
\begin{align}
\zeta^{B'}(0, 2|1, 1) &= -\zeta'(0) + \zeta'(-1), \tag{6.3} \\
&= \frac{1}{2} \ln(2\pi) + \frac{1}{12} - \ln(A) \nonumber
\end{align}
$$

$$
\begin{align}
\zeta^{B'}(0, 3|1, 2) &= -\frac{5}{8}\ln(2) - \frac{1}{4}\ln(\pi) - \frac{3}{2}\zeta'(0) + \frac{1}{2}\zeta'(-1), \tag{6.4} \\
&= \frac{1}{8}\ln(2) + \frac{1}{2}\ln(\pi) + \frac{1}{24} - \frac{\ln(A)}{2} \nonumber
\end{align}
$$

$$
\zeta^{B'}(0, 4|1, 3) = -\frac{19 \ln(3)}{36} - \zeta'(0) - \frac{2}{3} \zeta'\left(0, \frac{5}{3}\right) - \frac{1}{3} \zeta'\left(0, \frac{4}{3}\right) + \zeta'(-1) + \zeta'\left(-1, \frac{5}{3}\right) + \zeta'\left(-1, \frac{4}{3}\right), \quad (6.5)
$$

$$
\begin{equation}
\begin{split}
\zeta^{B'}(0, 5|2, 3) = & -\frac{31 \ln(6)}{72} - \frac{\ln(2)}{6} - \frac{1}{4} \ln(\pi) - \frac{3}{2} \zeta'(0) - \frac{2}{3} \zeta'\left(0, \frac{5}{3}\right) - \frac{1}{6} \zeta'\left(0, \frac{7}{6}\right) + \frac{1}{6} \zeta'\left(0, \frac{5}{6}\right) \\
& \qquad + \frac{1}{2} \zeta'(-1) + \zeta'\left(-1, \frac{5}{3}\right) + \zeta'\left(-1, \frac{4}{3}\right) + \zeta'\left(-1, \frac{7}{6}\right) + \zeta'\left(-1, \frac{5}{6}\right).
\end{split}
\tag{6.6}
\end{equation}
$$

We hereby used the following facts:

$$
\begin{align*}
\zeta'\left(0, \frac{3}{2}\right) &= \zeta'(0) - \ln(2) + \ln\Gamma\left(\frac{1}{2}\right), \\
&= \zeta'(0) + \ln(\sqrt{\pi}) - \ln(2),
\end{align*}
$$

and

$$
\begin{align*}
\zeta'\left(-1, \frac{3}{2}\right) &= \zeta'\left(-1, \frac{1}{2}\right) - \frac{1}{2}\ln(2) \\
&= -\frac{\ln(2)}{6} - \frac{1}{2}\zeta'(-1) - \frac{1}{2}\ln(2) \\
&= -\frac{2}{3}\ln(2) - \frac{1}{2}\zeta'(-1).
\end{align*}
$$

6.3.1 Poles and residues of $\zeta_{g_2}$

Via[KMT11a], we've seen in 1.6 that $\zeta_{g_2}$ has poles of the form $s = \frac{1}{3}$ and $s = \frac{1-k}{5}$, with $k \ge 0$ and
$k \ne 1 \pmod 5$. We shall give an expression of the residues of $\zeta_{g_2}$ at $s = \frac{1}{5}$ and at $s = \frac{1}{3}$. We set

$$
\begin{align*}
l_1(x_1, x_2) &= x_1 + x_2, & l_2(x_1, x_2) &= x_1 + 2x_2, & l_3(x_1, x_2) &= x_1 + 3x_2, & l_4(x_1, x_2) &= 2x_1 + 3x_2 \\
d_1 &= 1, & d_2 &= 1, & d'_1 &= 2, & d'_2 &= 3, & d'_3 &= 4, & d'_4 &= 5.
\end{align*}
$$

Via Proposition 2.2.4, we find that

$$
\zeta_{g_2}(s) = K(\theta, s) + J(\theta, s),
$$

with

$$
K(\theta, s) = 120^{-s} \sum_{\emptyset \neq Q \subseteq [1, 4]} \sum_{A_Q \subset Q^c} (-1)^{|Q^c| - |A_Q|} \frac{1}{\Gamma(s)^{4 - |A_Q|}} \left( \sum_{\mathbf{n} \in \mathbb{N}_0^2} (n_1 + 1)^{-s} (n_2 + 1)^{-s} \cdot \prod_{q \in A_Q} (l_q(\mathbf{n} + 1))^{-s} \prod_{q \in [1, 4] \setminus A_Q} \Gamma(s, \theta, l_q(\mathbf{n} + 1)) \right),
$$
and

$$J(\theta, s) = 120^s \sum_{\mathcal{P} \subsetneq [1,2]} \Gamma(1-s)^{|\mathcal{P}|} \sum_{\mathbf{k}=(k_p)_{p \in \mathcal{P}^c} \in \mathbb{N}_0^{\mathcal{P}^c}} (-1)^{|\mathbf{k}|} \left( \sum_{j=1}^4 h_{\mathcal{P},j,\mathbf{k}}(s) \right) \frac{\theta^{(4+|\mathcal{P}|)s-|\mathcal{P}|+|\mathbf{k}|}}{\Gamma(s)((4+|\mathcal{P}|)s-|\mathcal{P}|+|\mathbf{k}|)} \prod_{p \in \mathcal{P}^c} \frac{\zeta(s-k_p)}{k_p!} \\ + 120^s \Gamma(1-s)^2 \left( \sum_{j=1}^4 h_{[1,2],j}(s) \right) \frac{\theta^{6s-2}}{\Gamma(s)(6s-2)}, \qquad (6.7)$$

with

$$h_{\mathcal{P},1,\mathbf{k}}(s) = \frac{1}{\Gamma(s)^3} \int_{[0,1]^3} (x_2 x_3 x_4)^{s-1} \prod_{p \in \mathcal{P}} l_p^*(1, x_2, x_3, x_4)^{s-1} \prod_{p \in \mathcal{P}^c} l_p^*(1, x_2, x_3, x_4)^{k_p} dx_2 dx_3 dx_4,$$

$$h_{\mathcal{P},2,\mathbf{k}}(s) = \frac{1}{\Gamma(s)^3} \int_{[0,1]^3} (x_1 x_3 x_4)^{s-1} \prod_{p \in \mathcal{P}} l_p^*(x_1, 1, x_3, x_4)^{s-1} \prod_{p \in \mathcal{P}^c} l_p^*(x_1, 1, x_3, x_4)^{k_p} dx_1 dx_3 dx_4,$$

$$h_{\mathcal{P},3,\mathbf{k}}(s) = \frac{1}{\Gamma(s)^3} \int_{[0,1]^3} (x_1 x_2 x_4)^{s-1} \prod_{p \in \mathcal{P}} l_p^*(x_1, x_2, 1, x_4)^{s-1} \prod_{p \in \mathcal{P}^c} l_p^*(x_1, x_2, 1, x_4)^{k_p} dx_1 dx_2 dx_4,$$

$$h_{\mathcal{P},4,\mathbf{k}}(s) = \frac{1}{\Gamma(s)^3} \int_{[0,1]^3} (x_1 x_2 x_3)^{s-1} \prod_{p \in \mathcal{P}} l_p^*(x_1, x_2, x_3, 1)^{s-1} \prod_{p \in \mathcal{P}^c} l_p^*(x_1, x_2, x_3, 1)^{k_p} dx_1 dx_2 dx_3,$$

$$l_1^*(x_1, x_2) = x_1 + x_2 + x_3 + 2x_4,$$

$$l_2^*(x_1, x_2) = x_1 + 2x_2 + 3x_3 + 4x_4,$$

and

$$\sum_{j=1}^{4} h_{\llbracket 1,2 \rrbracket, j}(s) = \frac{1}{\Gamma(s)^3} \int_{[0,1]^3} (x_1 x_2 x_3)^{s-1} \prod_{p=1}^{2} l_p^*(x_1, x_2, x_3, 1)^{s-1} dx_1 dx_2 dx_3, \\ + \frac{1}{\Gamma(s)^3} \int_{[0,1]^3} (x_1 x_2 x_4)^{s-1} \prod_{p=1}^{2} l_p^*(x_1, x_2, 1, x_4)^{s-1} dx_1 dx_2 dx_4 \\ + \frac{1}{\Gamma(s)^3} \int_{[0,1]^3} (x_1 x_3 x_4)^{s-1} \prod_{p=1}^{2} l_p^*(x_1, 1, x_3, x_4)^{s-1} dx_1 dx_3 dx_4 \\ + \frac{1}{\Gamma(s)^3} \int_{[0,1]^3} (x_2 x_3 x_4)^{s-1} \prod_{p=1}^{2} l_p^*(1, x_2, x_3, x_4)^{s-1} dx_2 dx_3 dx_4.$$

Let's recall that $K(\theta, s)$ is holomorphic on $\mathbb{C}$, thus this term doesn't contribute to the residues. We just have to study the denominators of the terms inside of $J(\theta, s)$, in the formula 6.7.

* Let $\mathcal{P} \subsetneq [1,2]$, and $\mathbf{k} = (k_p)_{p \in \mathcal{P}^c} \in \mathbb{N}_0^{\mathcal{P}^c}$. For $s = \frac{1}{5}$, we have

$$(4 + \mathcal{P})s - |\mathcal{P}| + |\mathbf{k}| = 0 \text{ if and only if } |\mathcal{P}| = 1, \mathbf{k} = 0.$$
We then find that

$$
\begin{align*}
\mathrm{Res}_{s=\frac{1}{5}}(\zeta_{\mathfrak{g}_2}(s)) &= 120 \frac{\Gamma(\frac{4}{5})}{\Gamma(\frac{1}{5})} \left( \sum_{j=1}^{4} h_{\{1\},j,0} \left(\frac{1}{5}\right) + h_{\{2\},j,0} \left(\frac{1}{5}\right) \right) \zeta\left(\frac{1}{5}\right), \\
&= 120 \frac{\Gamma(\frac{4}{5})}{\Gamma(\frac{1}{5})^4} \zeta\left(\frac{1}{5}\right) \left( \int_{[0,1]^3} (x_2 x_3 x_4)^{-\frac{4}{5}} (l_1^*(1, x_2, x_3, x_4)^{-\frac{4}{5}} + l_2^*(1, x_2, x_3, x_4)^{-\frac{4}{5}}) dx_2 dx_3 dx_4 \right. \\
&\qquad \left. + \int_{[0,1]^3} (x_1 x_3 x_4)^{-\frac{4}{5}} (l_1^*(x_1, 1, x_3, x_4)^{-\frac{4}{5}} + l_2^*(x_1, 1, x_3, x_4)^{-\frac{4}{5}}) dx_1 dx_3 dx_4 \right. \\
&\qquad \left. + \int_{[0,1]^3} (x_1 x_2 x_4)^{-\frac{4}{5}} (l_1^*(x_1, x_2, 1, x_4)^{-\frac{4}{5}} + l_2^*(x_1 1, x_2, 1, x_4)^{-\frac{4}{5}}) dx_1 dx_2 dx_4 \right. \\
&\qquad \left. + \int_{[0,1]^3} (x_1 x_2 x_3)^{-\frac{4}{5}} (l_1^*(x_1, x_2, x_3, 1)^{-\frac{4}{5}} + l_2^*(x_1, x_2, x_3, 1)^{-\frac{4}{5}}) dx_1 dx_2 dx_3 \right).
\end{align*}
$$

• Via a similar argument, we get that

$$
\begin{align*}
\mathrm{Res}_{s=\frac{1}{3}}(\zeta_{\mathfrak{g}_2}(s)) &= 120 \frac{\Gamma(\frac{2}{3})^2}{\Gamma(\frac{1}{3})^4} \left( \sum_{j=1}^{4} h_{\{1,2\},j} \left(\frac{1}{3}\right) + h_{\{1,2\},j} \left(\frac{1}{5}\right) \right) \zeta\left(\frac{1}{3}\right), \\
&= 120 \frac{\Gamma(\frac{2}{3})^2}{\Gamma(\frac{1}{3})} \left( \int_{[0,1]^3} (x_1 x_2 x_3)^{-\frac{2}{3}} \prod_{p=1}^{2} l_p^*(x_1, x_2, x_3, 1)^{-\frac{2}{3}} dx_1 dx_2 dx_3 \right. \\
&\qquad \left. + \int_{[0,1]^3} (x_1 x_2 x_4)^{-\frac{2}{3}} \prod_{p=1}^{2} l_p^*(x_1, x_2, 1, x_4)^{-\frac{2}{3}} dx_1 dx_2 dx_4 \right. \\
&\qquad \left. + \int_{[0,1]^3} (x_1 x_3 x_4)^{-\frac{2}{3}} \prod_{p=1}^{2} l_p^*(x_1, 1, x_3, x_4)^{-\frac{2}{3}} dx_1 dx_3 dx_4 \right. \\
&\qquad \left. + \int_{[0,1]^3} (x_2 x_3 x_4)^{-\frac{2}{3}} \prod_{p=1}^{2} l_p^*(1, x_2, x_3, x_4)^{-\frac{2}{3}} dx_2 dx_3 dx_4 \right).
\end{align*}
$$

6.3.2 Special values of ζ<sub>g2</sub> and of ζ'<sub>g2</sub>

Let's take the same notations as in the previous sub-section:

$$
\begin{align*}
l_1(x_1, x_2) &= x_1 + x_2, \quad l_2(x_1, x_2) = x_1 + 2x_2, \quad l_3(x_1, x_2) = x_1 + 3x_2, \quad l_4(x_1, x_2) = 2x_1 + 3x_2 \\
d_1 &= 1, \quad d_2 = 1, \quad d'_1 = 2, \quad d'_2 = 3, \quad d'_3 = 4, \quad d'_4 = 5.
\end{align*}
$$

The zeta function ζ<sub>g2</sub> is defined as

$$
\zeta_{\mathfrak{g}_2}(s) = 120^s \sum_{n_1, n_2 \ge 1} \frac{1}{n_1^s n_2^s (n_1 + n_2)^s (n_1 + 2n_2)^s (n_1 + 3n_3)^s (2n_1 + 3n_2)^s} = 120^s Z_{\Delta}(s)
$$

Therefore, we have

$$
\zeta_{\mathfrak{g}_2}(-N) = 120^{-N} Z(-(N,N), (N,N,N,N)), \quad \zeta'_{\mathfrak{g}_2}(-N) = 120^{-N} Z'(-(N,N), (N,N,N,N)) + \ln(120) \zeta_{\mathfrak{g}_2}(-N).
\begin{array}{l}
(1,1), (1,1,1,1) \\
(1,1), (1,1,1,1)
\end{array}
$$

Via Corollary A1, we then get that

$$
\zeta_{\mathfrak{g}_2}(0) = \sum_{\mathcal{P} \subseteq [1,2]} \frac{(-1)^{|\mathcal{P}|}}{4 + |\mathcal{P}|} \sum_{\mathbf{k}=(k_p)_{p \in \mathcal{P}} \in \mathbb{N}_0^{\mathcal{P}}} \left( \sum_{j=1}^{4} \prod_{p \in \mathcal{P}} c_{j,p}^{k_p} \right) \frac{\zeta(-k_p)}{k_p!}.
$$
1) If $\mathcal{P} = \emptyset$ and $|\mathbf{k}| = 0$, then $\mathbf{k} = (0, 0)$, and

$$ \sum_{j=1}^{4} \prod_{p=1}^{2} c_{j,p}^{0} = 4. $$

2) If $\mathcal{P} = \{1\}$ and $|\mathbf{k}| = 1$, then $\mathbf{k} = k_2 = 1$, and

$$ \sum_{j=1}^{4} c_{j,2}^{1} = 9. $$

3) If $\mathcal{P} = \{2\}$ and $|\mathbf{k}| = 1$, then $\mathbf{k} = k_1 = 1$, and

$$ \sum_{j=1}^{4} c_{j,1}^{1} = 5. $$

Using the fact that $\zeta(0) = -\frac{1}{2}$, and that $\zeta(-1) = -\frac{1}{12}$, we get the value of $\zeta_{g_2}$ at $s = 0$:

$$ \begin{aligned} \zeta_{g_2}(0) &= \zeta(0)^2 - \frac{14}{5}\zeta(-1) \\ &= \frac{29}{60}. \end{aligned} $$

We also find thanks to Corollary A2 the following formula

$$ \zeta_{g_2}(-N) = 120^{-N} \sum_{\mathcal{P} \subseteq [1,2]} \frac{(-1)^{(|\mathcal{P}|+5)N+|\mathcal{P}|}}{4+|\mathcal{P}|} N!^{|\mathcal{P}|+1} \sum_{\substack{\mathbf{k}=(k_p)_{p \in \mathcal{P}^c} \in \mathbb{N}_0^{\mathcal{P}^c} \\ |\mathbf{k}|=(4+|\mathcal{P}|)N+|\mathcal{P}|}} Q_N^0(\mathcal{P}, \mathbf{k}) \prod_{p \in \mathcal{P}^c} \frac{\zeta(-N_p - k_p)}{k_p!}, $$

with

$$ Q_N^0(\mathcal{P}, \mathbf{k}) = (-1)^{3N} N!^3 \sum_{j=1}^4 \sum_{\substack{\mathbf{w}=(\mathbf{w}_p)_{p \in \mathcal{P}^c} \in (\mathbb{N}_0^4)^{\mathcal{P}^c} \\ \forall p \in \mathcal{P}^c | \mathbf{w}_p| = k_p \\ \mathbf{v}=(\mathbf{v}_p)_{p \in \mathcal{P}^c} \in (\mathbb{N}_0^{[1,4] \setminus \{j\}})^{\mathcal{P}^c} \\ \forall q \neq j, \mathbf{v}(q) + \mathbf{w}(q) = N}} \left( \prod_{p \in \mathcal{P}^c} \binom{k_p}{\mathbf{w}_p} \prod_{q=1}^4 c_{q,p}^{w_{p,q}} \right) \left( \prod_{p \in \mathcal{P}^c} \binom{-N-1}{|\mathbf{v}_p|} \binom{|\mathbf{v}_p|}{|\mathbf{w}_p|} \prod_{\substack{q=1 \\ q \neq j}}^4 c_{q,p}^{v_{p,q}} \right). $$

In order to compute $\zeta'_{g_2}(0)$, we see that

$$ \zeta'_{g_2}(0) = \ln(120)\zeta_{g_2}(0) + Z'_{\Delta}(0). $$

By applying Lemma 6.3.2, we see by using the SAGE code 7.1 that

$$ \begin{aligned} \zeta'_{g_2}(0) = & \ln(120)\zeta_{g_2}(0) + 2\gamma\zeta(0)^2 + \frac{1}{4}\zeta(0)\zeta'(0) - 4\gamma\zeta(-1) + \frac{1}{5}\zeta(-1)(\ln(3) + 2\ln(2)) - \frac{14}{5}\zeta'(-1) \\ & + \zeta^{B'}(0, 2|1, 1) + \zeta^{B'}(0, 3|1, 2) + \zeta^{B'}(0, 4|1, 3) + \zeta^{B'}(0, 5|2, 3). \end{aligned} $$
By the formulas 6.3, 6.4, 6.5 and 6.6, we find that

$$
\begin{align*}
\zeta^{B'}(0, 2|1, 1) + \zeta^{B'}(0, 3|1, 2) + \zeta^{B'}(0, 4|1, 3) + \zeta^{B'}(0, 5|2, 3) = & \\
& - \zeta'(0) + \zeta'(-1) - \frac{5}{8} \ln(2) - \frac{1}{4} \ln(\pi) - \frac{3}{2} \zeta'(0) + \frac{1}{2} \zeta'(-1) \\
& - \frac{19 \ln(3)}{36} - \zeta'(0) - \frac{2}{3} \zeta'\left(0, \frac{5}{3}\right) - \frac{1}{3} \zeta'\left(0, \frac{4}{3}\right) + \zeta'(-1) + \zeta'\left(-1, \frac{5}{3}\right) + \zeta'\left(-1, \frac{4}{3}\right) \\
& - \frac{31 \ln(6)}{72} - \frac{\ln(2)}{6} - \frac{1}{4} \ln(\pi) - \frac{3}{2} \zeta'(0) - \frac{2}{3} \zeta'\left(0, \frac{5}{3}\right) - \frac{1}{6} \zeta'\left(0, \frac{7}{6}\right) + \frac{1}{6} \zeta'\left(0, \frac{5}{6}\right) \\
& + \frac{1}{2} \zeta'(-1) + \zeta'\left(-1, \frac{5}{3}\right) + \zeta'\left(-1, \frac{4}{3}\right) + \zeta'\left(-1, \frac{7}{6}\right) + \zeta'\left(-1, \frac{5}{6}\right). &
\end{align*}
$$

Simplifying those terms, we get:

$$
\begin{align*}
\zeta'_{g_2}(0) ={}& \ln(120)\zeta_{g_2}(0) + 2\gamma\zeta(0)^2 + \frac{1}{4}\zeta(0)\zeta'(0) - 4\gamma\zeta(-1) + \frac{1}{5}\zeta(-1)(\ln(3) + 2\ln(2)) - \frac{14}{5}\zeta'(-1) \\
& - \frac{11}{9}\ln(2) - \frac{23}{24}\ln(3) - \frac{1}{2}\ln(\pi) - 5\zeta'(0) + 3\zeta'(-1) - \frac{4}{3}\zeta'\left(0, \frac{5}{3}\right) - \frac{1}{3}\zeta'\left(0, \frac{4}{3}\right) \\
& + \zeta'\left(-1, \frac{5}{3}\right) + \zeta'\left(-1, \frac{4}{3}\right) - \frac{1}{6}\zeta'\left(0, \frac{7}{6}\right) + \frac{1}{6}\zeta'\left(0, \frac{5}{6}\right).
\end{align*}
$$

6.3.3 Poles and residues of ζ_{s0(5)}

Via [KMT10d], we saw in 1.5 that $\zeta_{s0(5)}$ has poles of the form $s = \frac{1}{2}$ and $s = \frac{1-k}{3}$, for $k \ge 0$ an
arbitrary integer such that $k \ne 1 \pmod{3}$. We aim here to obtain expressions for the residues at $s = \frac{1}{3}$
and $s = \frac{1}{2}$ for $\zeta_{s0(5)}(s)$. We note that Bridges, Brindle, Bringmann, and Franke have already obtained a
very simple expression for these two residues in [BBBF23, Proposition 5.16] using different techniques.

We know that, for $\theta > 0$ sufficiently small, the zeta function $\zeta_{s0(5)}$ admits the decomposition $\zeta_{s0(5)}(s) = K(\theta, s) + J(\theta, s)$, where $K(\theta, s)$ is a holomorphic function in $s$ over all $\mathbb{C}$, and $J(\theta, s)$ is a meromorphic function in $s$ over all $\mathbb{C}$, having the same poles as $\zeta_{s0(5)}(s)$. By Proposition 2.2.4, we have

$$
\begin{align}
J(\theta, s) ={}& 6^s \sum_{\mathcal{P} \subseteq [1, 2]} \Gamma(1-s)^{|\mathcal{P}|} \sum_{\mathbf{k}=(k_p)_{p \in \mathcal{P}^c} \in \mathbb{N}_0^{\mathcal{P}^c}} (-1)^{|\mathbf{k}|} \left( \sum_{j=1}^2 h_{\mathcal{P},j,\mathbf{k}}(s) \right) \frac{\theta^{(2+|\mathcal{P}|)s-|\mathcal{P}|+|\mathbf{k}|}}{\Gamma(s)((2+|\mathcal{P}|)s-|\mathcal{P}|+|\mathbf{k}|)} \prod_{p \in \mathcal{P}^c} \frac{\zeta(s-k_p)}{k_p!} \nonumber \\
& + 6^s \Gamma(1-s)^2 \left( \sum_{j=1}^2 h_{[1,2],j,0}(s) \right) \frac{\theta^{4s-2}}{\Gamma(s)(4s-2)}, \tag{6.8}
\end{align}
$$

with

$$
h_{\mathcal{P},1,\mathbf{k}}(s) = \frac{1}{\Gamma(s)} \int_0^1 x_2^{s-1} \prod_{p \in \mathcal{P}} l_p^*(1, x_2)^{s-1} \prod_{p \in \mathcal{P}^c} l_p^*(1, x_2)^{k_p} dx_2,
$$

$$
h_{\mathcal{P},2,\mathbf{k}}(s) = \frac{1}{\Gamma(s)} \int_0^1 x_1^{s-1} \prod_{p \in \mathcal{P}} l_p^*(x_1, 1)^{s-1} \prod_{p \in \mathcal{P}^c} l_p^*(x_1, 1)^{k_p} dx_1,
$$

$$
l_1^*(x_1, x_2) = x_1 + x_2,
$$

$$
l_2^*(x_1, x_2) = x_1 + 2x_2.
$$

As $K(\theta, s)$ is holomorphic in $s$ on all $\mathbb{C}$, we can get the residues of $\zeta_{s0(5)}(s)$ by studying the expression of $J(\theta, s)$ above. At $s = 1/2$, the only contributing term of the residue of $\zeta_{s0(5)}(s)$ is

$$
\Gamma(1-s)^2 \left( \sum_{j=1}^{2} h_{[1,2],j,0}(s) \right) \frac{\theta^{4s-2}}{\Gamma(s)(4s-2)}.
$$
We then get the following residue at $s = 1/2$:

$$
\begin{align*}
\mathrm{Res}_{s=1/2}(\zeta_{s_0(5)}(s)) &= \sqrt{6} \frac{\Gamma(3/2)^2}{4\Gamma(1/2)} \left( \sum_{j=1}^{2} h_{[1,2],j,0}(1/2) \right), \\
&= \sqrt{6} \frac{\Gamma(3/2)^2}{4\Gamma(1/2)} \frac{1}{\Gamma(1/2)} \left( \int_0^1 x^{-1/2}(x+1)^{-1/2}(x+2)^{-1/2} dx + \int_0^1 x^{-1/2}(1+x)^{-1/2}(1+2x)^{-1/2} dx \right) \\
&= \sqrt{6} \frac{\Gamma(3/2)^2}{4\Gamma(1/2)} \frac{1}{\Gamma(1/2)} \left( \int_0^1 x^{-1/2}(x+1)^{-1/2}(x+2)^{-1/2} dx + \int_0^1 x^{-1/2}(1+x)^{-1/2}(1+2x)^{-1/2} dx \right).
\end{align*}
$$

For the poles of the form $\frac{n}{3}$, with $n \le 1$ being an integer such that $n \ne 0 (\mathrm{mod}~3)$, the only contributing
term to the residues at $s = \frac{n}{3}$ are of the form

$$
-\Gamma(1 - n/3)^{|\mathcal{P}|} (-1)^{|\mathbf{k}|} \left( \sum_{j=1}^{2} h_{\mathcal{P},j,\mathbf{k}}(1 - n/3) \right) \frac{1}{\Gamma(n/3)((2 + |\mathcal{P}|))} \prod_{p \in \mathcal{P}^c} \frac{\zeta(n/3 - k_p)}{k_p!},
$$

with $\mathcal{P} \subsetneq [1,2]$, $\mathbf{k} \in \mathbb{N}^{\mathcal{P}^c}$ such that $(2 + |\mathcal{P}|)\frac{n}{3} - |\mathcal{P}| + |\mathbf{k}| = 0$. This condition implies that $\mathcal{P} = \{i\}$ is a singleton. We set $\mathcal{P}^c = \{p\}$, we then have $|k| = k_p = 1 - n \in \mathbb{N}_0$. We found

$$
\mathrm{Res}_{s=n/3}(\zeta_{s_0(5)}(s)) = (-1)^{1-n} \Gamma\left(\frac{3-n}{3}\right) \left(\sum_{i=1}^{2} \sum_{j=1}^{2} h_{\{i\},j,k}\left(1-\frac{n}{3}\right)\right) \frac{1}{3\Gamma\left(\frac{n}{3}\right)} \frac{\zeta\left(\frac{4n}{3}-1\right)}{(1-n)!},
$$

Let's note that, when considering the general terms of the series present in 6.8 with $\mathcal{P} = \emptyset$, we observe
terms of the form $\frac{\theta^{2s+|k|}}{\Gamma(s)(2s+|k|)}$, where $\mathbf{k} = (k_1, k_2) \in \mathbb{N}_0$. The zeros of $2s + |k|$ do not constitute poles since
these poles are canceled by the zeros of the terms of the form $\sum_{j=1}^2 h_{\emptyset, j, \mathbf{k}}$. For example, at $s = -1/2$,
we find that the term $2s + |k|$ is zero if and only if $|k| = k_1 + k_2 = 1$, that is, when $\mathbf{k} \in \{(1,0), (0,1)\}$.
However, we note that

$$
\begin{align*}
h_{1,\emptyset,(1,0)}(s) + h_{2,\emptyset,(1,0)}(s) + h_{1,\emptyset,(0,1)}(s) + h_{1,\emptyset,(0,1)}(s) &= \frac{1}{\Gamma(s)} \int_0^1 (x^{s-1}(1+2x) + 2x^{s-1}(1+x) + x^{s-1}(2+x))dx \\
&= \frac{5}{\Gamma(s)} \left( \frac{1}{s+1} + \frac{1}{s} \right),
\end{align*}
$$

and at $s = -\frac{1}{2}$, this term vanish. This matches the list of poles that we obtained from Komori, Matsumoto and Tsumura in [KMT10d, Theorem 6.2].

**6.3.4 Special values of ζ<sub>so(5)</sub> and ζ'<sub>so(5)</sub>**

Let

$$
l_1(x_1, x_2) = x_1 + x_2, \quad l_2(x_1, x_2) = x_1 + 2x_2,
$$

$$
d_1 = 1, d_2 = 1, d'_1 = 2, d'_2 = 3,
$$

then $\zeta_{s_0(5)}(s)$ is of the form

$$
\zeta_{s_0(5)}(s) = 6^s \sum_{\mathbf{n} \in \mathbb{N}_0^2} \frac{1}{n_1^s n_2^s (n_1 + n_2)^s (n_1 + 2n_2)^s} = 6^s Z_{\Delta}(s)
$$

we then get that

$$
\zeta_{s_0(5)}(-N) = 6^{-N} Z(-(N,N), (N,N)), \quad \zeta'_{g_2}(-N) = 120^{-N} Z'(-(N,N), (N,N)) + \ln(6)\zeta_{g_2}(-N).
$$
At $N = 0$, we find by Corollary A1:

$$
\begin{align*}
\zeta_{so}(5)(0) &= \sum_{\mathcal{P} \subseteq [1,2]} \frac{(-1)^{|\mathcal{P}|}}{2 + |\mathcal{P}|} \sum_{\mathbf{k}=(k_p)_{p \in \mathcal{P}^c} \in \mathbb{N}_0^{\mathcal{P}^c} \atop |\mathcal{P}| = |\mathbf{k}|} \left( \sum_{j=1}^2 \prod_{p \in \mathcal{P}^c} c_{j,p}^{k_p} \right) \prod_{p \in \mathcal{P}^c} \frac{\zeta(-k_p)}{k_p!} \\
&= \zeta(0)^2 - \frac{5}{3}\zeta(-1) \\
&= \frac{7}{18}.
\end{align*}
$$

We can also have an expression of the values at nonpositive integers using Corollary A2,

$$
\zeta_{so}(5)(-N) = 6^{-N} \sum_{\mathcal{P} \subseteq [1,2]} \frac{(-1)^{(|\mathcal{P}|+3)N+|\mathcal{P}|}}{3+|\mathcal{P}|} N!^{|\mathcal{P}|+1} \sum_{\substack{\mathbf{k}=(k_p)_{p \in \mathcal{P}^c} \in \mathbb{N}_0^{\mathcal{P}^c} \\ |\mathbf{k}|=(3+|\mathcal{P}|)N+|\mathcal{P}|}} Q_N^0(\mathcal{P}, \mathbf{k}) \prod_{p \in \mathcal{P}^c} \frac{\zeta(-N-k_p)}{k_p!},
$$

with

$$
Q_N^0(\mathcal{P}, \mathbf{k}) = N! \sum_{j=1}^2 \sum_{\substack{\mathbf{w}=(\mathbf{w}_p)_{p \in \mathcal{P}^c} \in (\mathbb{N}_0^2)^{\mathcal{P}^c} \\ \forall p \in \mathcal{P}^c | \mathbf{w}_p|=k_p \\ \mathbf{v}=(\mathbf{v}_p)_{p \in \mathcal{P}} \in (\mathbb{N}_0^{[1,2]} \setminus \{j\})^\mathcal{P} \\ \forall q \neq j, \mathbf{v}(q)+\mathbf{w}(q)=N}} \left( \prod_{p \in \mathcal{P}^c} \binom{k_p}{\mathbf{w}_p} \prod_{q=1}^2 c_{q,p}^{w_p,q} \right) \left( \prod_{p \in \mathcal{P}} \binom{-N-1}{|\mathbf{v}_p|} \binom{|\mathbf{v}_p|}{\mathbf{v}_p} \prod_{\substack{q=1 \\ q \neq j}}^2 c_{q,p}^{v_p,q} \right).
$$

Then by Lemma 6.3.2, using SAGE code 7.1 we have:

$$
\zeta'_{so(5)}(0) = \ln(6)\zeta_{so(5)}(0) + \frac{1}{2}\gamma\zeta(0)^2 + \frac{1}{2}\zeta(0)\zeta'(0) - \frac{1}{2}\gamma\zeta(-1) - \frac{5}{3}\zeta'(-1) + \frac{1}{6}\ln(2) \\
+ \partial_s\zeta^B(s, 2|1, 1)|_{s=0} + \partial_s\zeta^B(s, 3|1, 2)|_{s=0},
$$

and with the formulas (6.3) and (6.4), we get:

$$
\zeta'_{so(5)}(0) = \ln(6)\zeta_{so(5)}(0) + \frac{1}{2}\gamma\zeta(0)^2 + \frac{1}{2}\zeta(0)\zeta'(0) - \frac{1}{2}\gamma\zeta(-1) - \frac{5}{3}\zeta'(-1) + \frac{1}{6}\ln(2) \\
- \zeta'(0) + \zeta'(-1) - \frac{5}{8}\ln(2) - \frac{1}{4}\ln(\pi) - \frac{3}{2}\zeta'(0) + \frac{1}{2}\zeta'(-1).
$$

By injecting these last two formulas into the one obtained for $\zeta'_{so(5)}(0)$, we get:

$$
\zeta'_{so(5)}(0) = \frac{7}{18} \ln(6) - \frac{11}{24} \ln(2) - \frac{1}{4} \ln(\pi) + \left( \frac{1}{2} \zeta(0)^2 - \frac{1}{2} \zeta(-1) \right) \gamma - \frac{11}{4} \zeta'(0) - \frac{13}{6} \zeta'(-1).
$$

This last expression allows us to have an explicit expression of the constant $C$ in Theorem 1.2.35, which allow us to get an explicit asymptotic formula for the number of representation $r_{so(5)}$.

6.4 Application to the study of the asymptotic behavior of $r_{\mathfrak{g}_2}(n)$

Let's recall that the number of representations of dimension $n$ of the exceptional Lie algebra $\mathfrak{g}_2$ is

$$
r_{\mathfrak{g}_2}(n) = \left| \left\{ (k_{i,j})_{i,j \ge 1} \in (\mathbb{N}_0)^{\mathbb{N}} \middle| \sum_{i,j \ge 1} k_{i,j} \frac{ij(i+j)(i+2j)(i+3j)(2i+3j)}{5!} = n \right\} \right|.
$$
We set $P(i, j) := \frac{ij(i+j)(i+2j)(i+3j)(2i+3j)}{5!}$. We have that

$$ \prod_{i,j \ge 1} \frac{1}{1 - q^{P(i,j)}} = \prod_{i,j \ge 1} \sum_{k_{i,j} \ge 1} q^{k_{(i,j)} P(i,j)}, $$

Developing the infinite product, and using the expression of the set $r_{g_2}(n)$, we get:

$$ \sum_{n=1}^{+\infty} r_{g_2}(n)q^n = \prod_{i,j \ge 1} \frac{1}{1 - q^{P(i,j)}}. $$

We set $f(n) = |\{(i, j) \in \mathbb{N}^2 | P(i, j) = n\}|$, we then get that

$$ \sum_{n=1}^{+\infty} r_{g_2}(n)q^n = \prod_{n=1}^{+\infty} \frac{1}{(1 - q^n)f(n)}. $$

Let's define $L_f(s) = \zeta_{g_2}(s)$, and $L_f^*(s) = \zeta_{g_2}(s)\Gamma(s)\zeta(s+1)$. We see that $L_f^*(s)$ has only two poles in $H_0$, at $s = \frac{1}{3}$ and $s = \frac{1}{5}$. Moreover, $L_f^*(s)$ has a double pole at $s = 0$. We also note that, thanks to the list of poles obtained in 1.6, $L_f^*$ has simple poles at $s = \frac{1}{3}$ and $s = \frac{k}{5}$, $k \in \mathbb{Z}_{\le 1}$, with $k \ne 0 \bmod 5$.

We want to apply Theorem 1.2.33. First, let's verify if, with the data above, the conditions of the theorem are valid. Let $\Lambda := \mathbb{N}_0 \setminus f^{-1}(\{0\})$. We note that $\frac{3}{2} \cdot \frac{1}{5} < \frac{1}{3} < \frac{2}{1} < \frac{1}{5}$. Thus, the integer $l = 2$ satisfies the hypothesis of Theorem 1.2.33. Now, let's check whether the conditions (P1), (P2), and (P3) are met:

(P1) Let $p \ge 2$ be a prime number. We have that $|\Lambda \setminus (p\mathbb{N}_0 \cap \Lambda)| = +\infty$. Indeed, the sequences

$$ u(2) := (P(8k + 1, 1))_{k \in \mathbb{N}_0}, $$

$$ u(3) := (P(9k + 1, 1))_{k \in \mathbb{N}_0}, $$

$$ u(5) := (P(25k + 1, 1))_{k \in \mathbb{N}_0}, $$

$$ u(p) := (P(kp + 1, 1))_{k \ge 1} \quad \text{if } p \ge 7. $$

are strictly increasing, and belong to $\Lambda$ by definition. Moreover, we observe that the sequence $u(p)$ belongs in $\Lambda \setminus (p\mathbb{N}_0 \cap \Lambda)$. We then conclude that any real number $L \ge \frac{1}{6}$ satisfies condition (P1).

(P2) We have already seen previously that $L_f^*$ has two poles (who are simple) at $s = \frac{1}{3}$ and $s = \frac{1}{5}$, a double pole at $s = 0$, and simple poles at $s = \frac{k}{5}$, $k \in \mathbb{Z}_{\le 1}$, with $k \ne 0 \bmod 5$. Therefore, for any real $R \in \mathbb{R}_+$ such that $R \ne \frac{k}{5}$ with $k \ne 0 \bmod 5$, $R$ satisfies the condition of the second hypothesis.

(P3) We note that the polynomial $P$ satisfies the H0S condition of the article [Ess97]. In particular, we know from [Ess97, Theorem 3] that the zeta function associated with this polynomial, (i.e., $\zeta_{g_2}$) admits a polynomial bound in the imaginary part of $s = \sigma + i\tau$ on each vertical strip $\sigma_1 \le \sigma \le \sigma_2$. We then conclude that condition (P3) is met.

**Remark 6.4.1.** In condition (P2), $R$ is arbitrarily large, and in condition (P1), $L$ is also arbitrarily large.

Let's note $\alpha = \frac{1}{3}$, $\beta = \frac{1}{5}$, $\omega_\alpha := \text{Res}_{s=\frac{1}{3}} \zeta_{g_2}(s)$ and $\omega_\beta := \text{Res}_{s=\frac{1}{5}} \zeta_{g_2}(s)$. Applying Theorem 1.2.33, we get that

$$ r_{g_2}(n) = \sum_{n \to +\infty} \frac{C}{n^b} \exp \left( A_1 n^{\frac{1}{4}} + A_2 n^{\frac{3}{20}} + A_3 n^{\frac{1}{20}} \right) \left( 1 + \sum_{j=2}^{N} \frac{B_j}{n^{\nu_j}} + O_{L,R} \left( n^{-\min\left\{\frac{2L-\alpha}{2(\alpha+1)}, \frac{R}{\alpha+1}\right\}} \right) \right), $$
with

$$C = \frac{e^{\zeta'_{g_2}(0)} (\omega_\alpha \Gamma(\frac{4}{3}) \zeta(\frac{4}{3}))^{\frac{1-6\zeta_{g_2}(0)}{8}} \sqrt{3}}{\sqrt{8\pi}}, \quad b = \frac{6-6\zeta_{g_2}(0)+1}{8},$$

and $A_1 := 4 \left(\omega_\alpha \Gamma\left(\frac{4}{3}\right) \zeta\left(\frac{4}{3}\right)\right)^{\frac{3}{4}}$, $A_2 := \frac{\omega_\beta \Gamma\left(\frac{1}{5}\right) \zeta\left(\frac{6}{5}\right)}{\left(\omega_\alpha \Gamma\left(\frac{1}{3}\right) \zeta\left(\frac{4}{3}\right)\right)^{\frac{3}{20}}}$, and

$$A_3 := K_3 + 3 \left( \omega_\alpha \Gamma\left(\frac{4}{3}\right) \zeta\left(\frac{4}{3}\right) \right)^{\frac{3}{4}} \left( \begin{aligned} & \left(-\frac{1}{3}\right) \frac{K_3}{\left(\omega_\alpha \Gamma\left(\frac{4}{3}\right) \zeta\left(\frac{4}{3}\right)\right)^{\frac{3}{4}}} + \left(-\frac{1}{3}\right) \frac{K_2^2}{\left(\omega_\alpha \Gamma\left(\frac{4}{3}\right) \zeta\left(\frac{4}{3}\right)\right)^{\frac{3}{2}}} \\ & + \frac{5 \left(\omega_\beta \Gamma\left(\frac{6}{5}\right) \zeta\left(\frac{6}{5}\right)\right)}{\left(\omega_\alpha \Gamma\left(\frac{4}{3}\right) \zeta\left(\frac{4}{3}\right)\right)^{\frac{3}{20}}} \left(-\frac{1}{5}\right) \frac{K_2}{\left(\omega_\alpha \Gamma\left(\frac{4}{3}\right) \zeta\left(\frac{4}{3}\right)\right)^{\frac{3}{4}}} \end{aligned} \right),$$

with $0 < \nu_2 < ...$ being the positive elements of the set $\mathcal{N} + \mathcal{M}$.

By [BBBF23, Lemma 4.3], we get

$$K_2 = \frac{3}{4} \cdot \frac{\omega_\beta \Gamma\left(\frac{6}{5}\right) \zeta\left(\frac{6}{5}\right)}{\left(\omega_\alpha \Gamma\left(\frac{4}{3}\right) \zeta\left(\frac{4}{3}\right)\right)^{\frac{3}{20}}}$$

$$K_3 = - \frac{3}{160} \cdot \frac{\left(\omega_\beta \Gamma\left(\frac{6}{5}\right) \zeta\left(\frac{6}{5}\right)\right)^2}{\left(\omega_\alpha \Gamma\left(\frac{4}{3}\right) \zeta\left(\frac{4}{3}\right)\right)^{\frac{21}{20}}}.$$

From the definition of the sets (1.8), (1.9), and (1.10) in [BBBF23], we found that the sequence $\nu_j$ in Theorem 1.2.33 give, with our assumptions, $\nu_2 = \frac{1}{20}, \nu_3 = \frac{2}{20}, \nu_4 = \frac{3}{20}, ...$
Chapter 7

Appendix

7.1 SAGE codes

Listing 7.1: SAGE Code to compute Z'_Δ(0) with P = 2 and d₁ = d₂ = 1

from sage.all import *

c_g2 =[[1,1],[1,2],[1,3],[2,3]]
c.so5 =[[1,1],[1,2]]

gamma = eulergamma
zeta0 = var('zeta0')
zeta_m1 = var('zeta_m1')
zeta0_prime = var('zeta0_prime')
zeta_m1_prime = var('zeta_m1_prime')

def Z_delta_prime_zero(c_matrix):
    Q =len(c_matrix)
    sum1 = sum(c[j][0] + c[j][1] for j in range(Q))
    sum2 = sum(sum((c[f][0]/c[f][1] - c[j][0]/c[j][1])
                  *log(1 + c[f][1]/c[j][1])+(c[f][1]/c[f][0] - c[j][1]/c[j][0])
                  *log(1 + c[f][0]/c[j][0]))
    for f in range(Q) if f != j) for j in range(Q))
    sum3 = sum((c[j][0] * c[j][1])**-1 for j in range(Q))
    sum4 = sum(c[j][0]**-1 * c[j][1] + c[j][1]**-1 * c[j][0] for j in range(Q))
    result = Q*zeta0**2*gamma - zeta_m1*sum1*gamma - 1/(Q+1)*zeta_m1*sum2
    +zeta0_prime*zeta0/Q - zeta_m1_prime/(Q+1)*sum1 - gamma*zeta0**2*sum3
    + gamma*zeta_m1*sum4
    return(result)
print(result)

Listing 7.2: SAGE Code to compute Z_Δ(-N) with P = 2

from math import comb
from itertools import product

def alpha_k_N(Q, N, k, c_matrix):
    alpha = 0
    for j in range(1, Q + 1):
        for w in product(range(N + 1), repeat=Q):
if sum(w) == k:

coef1 = multinomial(w)

coef2 = multinomial([N - w_i for w_i in w])

prod = 1

for q in range(Q):

prod *= (c_matrix[q][0] ** w[q]) * (c_matrix[q][1] ** (N - w[q]))

alpha += coef1 * coef2 * prod

return alpha

def beta_N(Q, N, c_matrix):

beta = 0

for j in range(1, Q + 1):

for w in product(range((Q + 1) * N + 2), repeat=Q):

if sum(w) == (Q + 1) * N + 1 and w[j - 1] >= 2 * N + 1:

numerator = 1

for i in range(N + 1 - w[j - 1], -N - 1):

numerator *= i

denominator = 1

for q in range(Q):

if q != j - 1:

denominator *= factorial(w[q])

term1 = c_matrix[j - 1][1] ** w[j - 1]

term2 = c_matrix[j - 1][0] ** w[j - 1]

for q in range(Q):

if q != j - 1:

term1 *= (c_matrix[q][0] ** (N - w[q]))

*(c_matrix[q][1] ** w[q])

term2 *= (c_matrix[q][1] ** (N - w[q]))

*(c_matrix[q][0] ** w[q])

beta += (numerator / denominator) * (term1 + term2)

return beta

def Z_Delta_minus_N(N, c_matrix):

Q = len(c_matrix)

sum1 = 0

N_fact_Q = factorial(N) ** Q

for k in range(Q * N + 1):

alpha = alpha_k_N(Q, N, k, c_matrix)

term1 = zeta(-N - k) / factorial(k)

term2 = zeta(-(Q + 1) * N + k) / factorial(Q * N - k)

sum1 += alpha * term1 * term2

beta = beta_N(Q, N, c_matrix)

sum2 = ((-1) ** (N + 1)) * (N_fact_Q * factorial(N)) * beta

*zeta(-(Q + 2) * N - 1) / factorial((Q + 1) * N + 1)

Z = (1 / Q) * N_fact_Q * sum1 + (1 / (Q + 1)) * sum2

return Z
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