On single-variable Witten zeta functions of rank 2 root systems

Kam Cheong Au University of Cologne
Department of Mathematics and Computer Science
Weyertal 86-90, 50931 Cologne, Germany [email protected]

Abstract.

We use a Mellin-transform-based method to simultaneously analyse single-variable Witten zeta functions of all rank $2$ irreducible root systems. We obtain satisfactory details regarding their pole locations, values of residues and special values. We also mention a connection between Witten zeta function to Eisenstein series.

Key words and phrases:

Witten zeta functions, analytic continuation, residues, asymptotic formula

2020 Mathematics Subject Classification:

Primary: 11M32, 11M35. Secondary: 11P82

Introduction

Let $\mathfrak{g}$ be a simple Lie algebra over $\mathbb{C}$ ,

\zeta_{\mathfrak{g}}(s):=\sum_{\rho}\frac{1}{(\text{dim }\rho)^{s}},\quad\rho\in\{\text{finite-dimensional irreducible representations of $\mathfrak{g}$}\},

be its Witten zeta function. If $\Phi$ is the root system associated to $\mathfrak{g}$ , we abuse notation by writing $\zeta_{\mathfrak{g}}(s)$ as $\zeta_{\Phi}(s)$ .

There are vast literatures on Witten zeta function ([12], [14], [17], [10], [11], [9], [16]), especially on functional relations of their multi-variable generalizations. Finer analytic properties of single-variable $\zeta_{\Phi}(s)$ , however, received less attention and seem not easily derivable from their multi-variable versions. In this article, we look at following properties of $\zeta_{\Phi}(s),\Phi=A_{2},B_{2},G_{2}$ :

•

values of residue at abscissa of convergence;
•

values at $0$ and negative integers;
•

derivative at zero;
•

location of all poles.

Answers to them are nice yet non-trivial, this leads one to suspect $\zeta_{\Phi}(s)$ underlies deep analytic and arithmetic properties that were up-to-now overlooked. We denote

	$\displaystyle\omega_{A}(s):=2^{-s}\zeta_{A_{2}}(s)$	$\displaystyle=\sum_{n,m\geq 1}\frac{1}{m^{s}n^{s}(m+n)^{s}};$
	$\displaystyle\omega_{B}(s):=6^{-s}\zeta_{B_{2}}(s)$	$\displaystyle=\sum_{n,m\geq 1}\frac{1}{m^{s}n^{s}(m+n)^{s}(m+2n)^{s}};$
	$\displaystyle\omega_{G}(s):=120^{-s}\zeta_{G_{2}}(s)$	$\displaystyle=\sum_{n,m\geq 1}\frac{1}{m^{s}n^{s}(m+n)^{s}(m+2n)^{s}(m+3n)^{s}(2m+3n)^{s}}.$

Major properties of these functions are tabulated below.

\omega_{\bullet}(s)

A

B

G

Converges when

\Re(s)>\frac{2}{3}

\Re(s)>\frac{1}{2}

\Re(s)>\frac{1}{3}

Residue at abscissa

of convergence

\dfrac{\Gamma(\frac{1}{3})^{3}}{2\sqrt{3}\pi}

\dfrac{\Gamma(\frac{1}{4})^{2}}{8\sqrt{2\pi}}

\dfrac{\Gamma(\frac{1}{3})^{3}}{2^{8/3}3^{3/2}\pi}

\omega(0)

\frac{1}{3}

\frac{3}{8}

\frac{5}{12}

\omega(-n)

Vanishes for

n\in\mathbb{N}

Vanishes for

n\in 2\mathbb{N}

\omega^{\prime}(0)

\log 2+\log\pi

\frac{5}{4}\log 2+\frac{3}{2}\log\pi

2\log 2-\frac{1}{2}\log 3+\frac{5}{2}\log\pi

Location of

other poles

s=\frac{k}{2}

k\leq 1,k\equiv 1\pmod{2}

s=\frac{k}{3}

k\leq 1,k\equiv 1,5\pmod{6}

s=\frac{k}{5}

k\leq 1,k\equiv 1,3,7,9\pmod{10}

Table 1. Major properties of

\omega_{A}(s),\omega_{B}(s)

and

\omega_{G}(s)

All facts concerning $\omega_{A}(s)$ proved in this article are known, thanks to a plethora of literatures on Tornheim’s double zeta function ([17], [3], [19], [15]), which specializes to $\omega_{A}(s)$ . Main novelty are thus corresponding facts on $\omega_{B}(s)$ and $\omega_{G}(s)$ , which received much less investigation. Bringmann et. al in [4] proved first three rows for $\omega_{B}(s)$ ; Rutard [20] establishes the rows $\omega(0),\omega^{\prime}(0)$ .

This article proves all (and re-proves some) entries of table above. We analyse $\omega_{A}(s),\omega_{B}(s),\omega_{G}(s)$ in a uniform manner, this is done by introducing a more general series

\zeta_{f}(s):=\sum_{n,m\geq 1}\left(nm^{d+1}f(\frac{n}{m})\right)^{-s},

where $f(x)$ is a degree $d$ polynomial with real coefficient such that $f(\mathbb{R}^{\geq 0})\subset\mathbb{R}^{>0}$ . We recover $\omega_{A}(s),\omega_{B}(s)$ and $\omega_{G}(s)$ by letting $f$ to be

(0.1)

f_{A}(x):=1+x,\quad f_{B}(x):=(1+x)(1+2x),\quad f_{G}(x):=(1+x)(1+2x)(1+3x)(2+3x).

Theorem 0.1.

Let $f$ be a degree $d$ polynomial with real coefficients such that $f(\mathbb{R}^{\geq 0})\subset\mathbb{R}^{>0}$ , then $\zeta_{f}(s)$ has meromorphic continuation to $\mathbb{C}$ , also

(1)

$\zeta_{f}(s)$ has abscissa of convergence $s=\frac{2}{d+2}$ , it has a simple pole at this point with residue

$\frac{1}{d+2}\int_{0}^{\infty}(xf(x))^{-2/(d+2)}dx;$
(2)

$\zeta_{f}(s)$ is of moderate increase along imaginary direction, that is, for any compact subset $K\subset\mathbb{R}$ , there exists $M>0$ such that

$|\zeta_{f}(\sigma+it)|=O(|t|^{M}),\qquad\sigma\in K,|t|\gg 1;$
(3)

$\zeta_{f}(s)$ is analytic at $s=0,-1,-2,\cdots$ , with an explicit formula to calculate them (Theorem 3.1);
(4)

Apart from the pole at $s=\frac{2}{d+2}$ , all other poles are at $s\in\frac{1}{1+d}\mathbb{Z}^{\leq 1}$ , they are at most simple (Theorem 4.1);
(5)

If roots of $f$ are rational numbers, then $\zeta^{\prime}_{f}(0)$ is a linear combination of $\gamma$ (Euler’s constant), $\zeta^{\prime}(-1)$ and $\log\Gamma(r)$ for various $r\in\mathbb{Q}$ .

We will see $\omega_{\bullet}(s)$ manifest unusual properties not found in generic $\zeta_{f}(s)$ . Here we give two examples: (1) $\zeta_{f}(-2n)\neq 0$ for generic $f$ , whereas $\omega_{\bullet}(-2n)=0$ ; (2) for generic $f$ of degree $4$ , $\zeta_{f}(s)$ has poles at $s=k/5$ for all $k\leq 1$ such that $k/5\notin\mathbb{Z}$ , whereas for $\omega_{G}(s)$ , only those $k$ which are odd give poles. These properties indicate Witten zeta functions are nice objects to study and provide motivation to investigate the corresponding situation at higher rank.

We structure the article as follows: In §1, we develop prerequisite materials on our Mellin transform setup, which is needed in §2 for meromorphic continuation of $\zeta_{f}(s)$ . §3 looks at values of $\zeta_{f}$ at non-positive integers. §4 classifies all poles of $\zeta_{f}$ . §5 calculates $\zeta_{f}^{\prime}(0)$ . §6 applies our previously gained knowledge on $\zeta_{G_{2}}(s)$ to derive an asymptotic formula for number of representation of Lie algebra $\mathfrak{g}_{2}$ .

Acknowledgment

The author thanks Prof. Kathrin Bringmann valuable discussions and feedbacks. The author has received funding from the European Research Council (ERC) under the European Union’s Horizon 2020 research and innovation programme (grant agreement No. 101001179).

1. An integration kernel

In this section, we assemble materials that we will need for our investigation of $\zeta_{f}(s)$ . Let $f(x)$ be a degree $d$ polynomial with real coefficient such that $f(\mathbb{R}^{\geq 0})\subset\mathbb{R}^{>0}$ , define

F_{f}(a,s)=\int_{0}^{\infty}(f(x))^{-a}x^{s}\frac{dx}{x},\qquad d\Re(a)>\Re(s)>0.

It is absolutely convergent on the region $d\Re(a)>\Re(s)>0$ , so defines an analytic function there.

Theorem 1.1.

The function $F_{f}(a,s)$ has meromorphic continuation to $\mathbb{C}^{2}$ , and

K_{f}(a,s)=\frac{\Gamma(a)}{\Gamma(s)\Gamma(da-s)}F_{f}(a,s)

is an entire function.

Proof.

Let $C$ be a contour on complex plane that starts at $+\infty$ with small negative imaginary part, goes around $0$ , wraps around positive real axis and ends at $+\infty$ with small positive imaginary part.

Then, by agreeing $0\leq\arg(x)<2\pi$ , we have

(1.1)

F_{f}(a,s)=\frac{1}{e^{2\pi is}-1}\int_{C}f(x)^{-a}x^{s}\frac{dx}{x}

now since the path of integral no longer passes through $x=0$ , only rapid decrease at $x=\infty$ (i.e. $d\Re(a)>\Re(s)$ ) is needed to ensure $\int_{C}f(x)^{-a}x^{s}\frac{dx}{x}$ being analytic, thus we obtain a continuation on a larger region with the condition $\Re(s)>0$ removed. From the poles of $1/(e^{2\pi is}-1)$ , we see that $F_{f}(a,s)$ has simple poles at $s=0,-1,-2,\cdots$ , they are cancelled by zeroes of $1/\Gamma(s)$ , so

\frac{F_{f}(a,s)}{\Gamma(s)}\text{ is analytic on }d\Re(a)>\Re(s).

Let $g(x)=x^{d}f(1/x)\in\mathbb{R}[x]$ , then substitution $x\mapsto 1/x$ in the original integral definition of $F_{f}(a,s)$ shows $F_{f}(a,s)=F_{g}(a,ad-s)$ . Apply this conclusion to $g$ :

\frac{F_{g}(a,ad-s)}{\Gamma(ad-s)}\text{ is analytic on }d\Re(a)>\Re(ad-s),

that is $F_{f}(a,s)/\Gamma(ad-s)$ is analytic on $\Re(s)>0$ . From integral representation (1.1) again: only poles on $\Re(s)\leq 0$ are simple poles at $s=0,-1,-2,\cdots$ , so $\frac{F_{f}(a,s)}{\Gamma(s)\Gamma(ad-s)}$ is an entire function in $a$ and $s$ .

To prove $K_{f}(a,s)$ is still entire, we need to show $F_{f}(-n,s)=0$ when $n=0,1,2,\cdots$ . By analytic continuation, we can assume $s$ is sufficiently negative. By representation (1.1), we need to prove

\int_{C}f(x)^{n}x^{s}\frac{dx}{x}=0

Let $C(R)$ be truncation of $C$ at large real number $R$ , let $C^{\prime}(R)$ be a circle of large radius $R$ as in figure.

Since $n\geq 0$ , $f(x)^{n}x^{s}\frac{1}{x}$ is analytic inside the contour, so

\int_{C(R)}f(x)^{n}x^{s}\frac{dx}{x}+\int_{C^{\prime}(R)}f(x)^{n}x^{s}\frac{dx}{x}=0

and when $s$ is sufficiently negative, the integral along $C^{\prime}(R)$ tends to $0$ as $R\to\infty$ , so integral along $C$ is indeed zero. ∎

Remark 1.2.

For $d=1,2$ , $F_{f}(a,s)$ and hence $K_{f}(a,s)$ can be expressed in terms of well-known special functions:

(1.2)		$\displaystyle K_{f}(a,s)$	$\displaystyle=\alpha_{1}^{-s}$	$\displaystyle f(x)=1+\alpha_{1}x$
(1.3)		$\displaystyle K_{f}(a,s)$	$\displaystyle=\frac{\Gamma(a)}{\Gamma(2a)}\alpha_{1}^{-s}{}_{2}F_{1}{\left[\genfrac{.}{.}{0.0pt}{}{a{\mathchar 44\relax}\mskip 8.0mus}{2a};1-\frac{\alpha_{2}}{\alpha_{1}}\right]}\qquad\qquad$	$\displaystyle f(x)=(1+\alpha_{1}x)(1+\alpha_{2}x)$

with ${}_{2}F_{1}$ being Gauss hypergeometric function, they follow from the general formula:

\int_{0}^{\infty}\frac{x^{s-1}}{(1+\alpha x)^{a}}dx=\alpha^{-s}\frac{\Gamma(s)\Gamma(a-s)}{\Gamma(a)}

(1.4)

\int_{0}^{\infty}x^{s-1}(1+x)^{-a}(1+\lambda x)^{-b}dx=\frac{\Gamma(s)\Gamma(a+b-s)}{\Gamma(a+b)}\times{}_{2}F_{1}{\left[\genfrac{.}{.}{0.0pt}{}{b{\mathchar 44\relax}\mskip 8.0mus}{a+b};1-\lambda\right]}

For higher $d$ , an explicit formula is in general not possible. Nonetheless, most of the properties of $K_{f}(a,s)$ we need can be derived without relying on an explicit representation using special functions.

Proposition 1.3.

When $n$ is a non-negative integer,

K_{f}(a,-n)=(-1)^{n}n!\frac{\Gamma(a)}{\Gamma(ad+n)}[f(x)^{-a}][x^{n}]

where $[f(x)^{-a}][x^{n}]$ is the coefficient of $x^{n}$ in the series expansion of $f(x)^{-a}$ around $x=0$ .

Proof.

By analytic continuation, we only need to prove this when $a$ is sufficiently large. In this case

K_{f}(a,s)=\frac{1}{e^{2\pi is}-1}\frac{\Gamma(a)}{\Gamma(s)\Gamma(da-s)}\int_{C}f(x)^{-a}x^{s}\frac{dx}{x}

here $C$ is the contour used in proof of previous theorem. When $s$ tends to $-n$ , we have

K_{f}(a,-n)=\frac{(-1)^{n}n!\Gamma(a)}{\Gamma(da+n)}(\frac{1}{2\pi i})\int_{C}f(x)^{-a}x^{-n}\frac{dx}{x}

Now the integrand remains single-valued across the positive real axis, the only pole is at $x=0$ , its residue there is $[f(x)^{-a}][x^{n}]$ . ∎

Proposition 1.4.

When $n,n$ are non-negative integers,

K_{f}(-m,-n)=\begin{cases}(-1)^{(d+1)m}d\frac{n!(dm-n)!}{m!}[f(x)^{m}][x^{n}]&\quad n\leq dm\\ (-1)^{n+m}\frac{n!}{m!(n-dm-1)!}[-f(x)^{m}\log f(x)][x^{n}]&\quad n>dm\end{cases}.

In particular, if $f(x)\in\mathbb{Q}[x]$ , then

K_{f}(-n,m)\in\mathbb{Q}\qquad n\in\mathbb{Z}^{\geq 0},m\in\mathbb{Z}.

Proof.

When $n\leq dm$ , we simply let $a\to-m$ , here we note that both $\Gamma(a)$ and $\Gamma(ad+n)$ has simple pole at $a=-m$ , this gives the formula in the case $n\leq dm$ . When $n>dm$ , $\Gamma(a)$ has a pole at $a=-m$ , $\Gamma(ad+n)$ is regular with value $(dm-n-1)!$ , and $[f(x)^{-a}][x^{n}]$ has a zero at $a=-m$ ,¹¹1because $f(x)^{m}$ is a polynomial of degree $dm$ and we are extracting coefficient higher than the degree.. By differentiating with respect to $a$ , we obtain its first order term:

[f(x)^{-a}][x^{n}]=[-f(x)^{m}\log f(x)](a+m)+O((a+m)^{2})\qquad a\to-m

therefore the formula in the case $n>dm$ .

Regarding the last claim of rationality, this is evident from the formula when $m\leq 0$ . For $m>0$ , let $g(x)=x^{d}f(1/x)$ , as in the proof of previous theorem, $K_{f}(a,s)=K_{g}(a,ad-s)$ , thus $K_{f}(-n,m)=K_{g}(-n,-nd-m)$ with $-nd-m<0$ , so we can apply the formula again with $f$ replaced by $g$ and produces the same conclusion. ∎

Proposition 1.5.

Let $f(x)=c_{0}(1+\alpha_{1}x)\cdots(1+\alpha_{d}x)$ ,²²2note that our restriction on $f$ implies $c_{0}>0$ and each $\alpha_{i}\in\mathbb{C}-(-\infty,0]$ , then

K_{f}(0,s)=\sum_{i=1}^{d}\alpha_{i}^{-s}

Proof.

By analytic continuation, it suffices to prove this when $s$ is sufficiently negative, so the following representation is valid:

K_{f}(a,s)=\frac{1}{e^{2\pi is}-1}\frac{\Gamma(a)}{\Gamma(s)\Gamma(ad-s)}\int_{C}f(x)^{-a}x^{s}\frac{dx}{x}

with $C$ same contour as in proof of previous theorem. Letting $a\to 0$ gives

K_{f}(0,s)=-\frac{1}{e^{2\pi is}-1}\frac{1}{\Gamma(s)\Gamma(-s)}\int_{C}x^{s}\log f(x)\frac{dx}{x}:=h(s)\int_{C}x^{s}\log f(x)\frac{dx}{x}

We know from remark above that $K_{f}(0,s)=\alpha^{-s}$ when $f=1+\alpha x$ , so

\int_{C}x^{s}\log(1+\alpha x)\frac{dx}{x}=\alpha^{-s}h(s)^{-1}

So when $f(x)=c_{0}(1+\alpha_{1}x)\cdots(1+\alpha_{d}x)$ ,

\begin{aligned} K_{f}(0,s)&=h(s)\int_{C}x^{s}\log[c(1+\alpha_{1}x)\cdots(1+\alpha_{d}x)]\frac{dx}{x}\\ &=h(s)(\log c_{0})\underbrace{\int_{C}x^{s}\frac{dx}{x}}_{=0}+\sum_{i=1}^{d}h(s)\int_{C}x^{s}\log(1+\alpha_{i}x)\frac{dx}{x}\\ &=h(s)\sum_{i=1}^{d}h(s)^{-1}\alpha_{i}^{-s}\end{aligned}.

as desired. ∎

Proposition 1.6.

Write

f(x)=c_{0}(1+\alpha_{1}x)\cdots(1+\alpha_{d}x)=c_{0}+c_{1}x+\cdots+c_{d}x^{d},

we have

K_{f}(s,1-s)=\frac{c_{d-1}}{c_{d}}+s\left(\frac{c_{d-1}}{c_{d}}((d-1)\gamma-\log c_{d})+(d+1)\sum_{i}\frac{\log\alpha_{i}}{\alpha_{i}}\right)+O(s^{2}),\qquad s\to 0

where $\gamma$ is Euler’s constant.

We will need this result during evaluation of $\zeta^{\prime}_{f}(0)$ .

Proof.

The constant term is $K_{f}(0,1)=\sum_{i}\alpha^{-1}=c_{d-1}/c_{d}$ , coefficient of $s$ is

\left.\frac{d}{ds}\right|_{s=0}K_{f}(s,1-s)=\left.\frac{d}{ds}\right|_{s=0}K_{f}(s,1)-\left.\frac{d}{ds}\right|_{s=1}K_{f}(0,s)

where chain rule has been used. The second term is easy to evaluate by the previous proposition: $\left.\frac{d}{ds}\right|_{s=1}K_{f}(0,s)=-\sum_{i}(\log\alpha_{i})/\alpha_{i}$ . For first term, let $g(x)=x^{d}f(x)$ ,

\left.\frac{d}{ds}\right|_{s=0}K_{f}(s,1)=\left.\frac{d}{ds}\right|_{s=0}K_{g}(s,sd-1)=\left.\frac{d}{ds}\right|_{s=0}K_{g}(s,-1)+d\left.\frac{d}{ds}\right|_{s=-1}K_{g}(0,s)

Since $K_{g}(0,s)=\sum_{i}\alpha^{s}$ , the second term derivative on RHS equals $\sum_{i}(\log\alpha_{i})/\alpha_{i}$ . For first term, by Proposition 1.3,

K_{g}(s,-1)=-\frac{\Gamma(s)}{\Gamma(sd+1)}[f(x)^{-s}][x^{1}]=\frac{\Gamma(s+1)}{\Gamma(sd+1)}c_{d}^{-s-1}c_{d-1}

Evaluating derivatives of the gamma functions gives the claim. ∎

Following two technical lemmas concern growth of $F_{f}(a,s)$ when arguments $a,s$ have large imaginary part.

Lemma 1.7.

(a) There exists a positive integer $N$ and polynomials $p_{0},\cdots,p_{N}\in\mathbb{R}[a,s]$ such that $F_{f}(a,s)$ satisfies a recurrence of form

\sum_{i=0}^{N}p_{i}(a,s)F_{f}(a,s+i)=0

(b) Denote $s=\sigma+it$ to be its real and imaginary part. There exists a positive number $\delta>0$ , depending only on $f$ , such that for any $\varepsilon>0$ ,

F_{f}(a,\sigma+it)=O_{K,a}(e^{-(\delta-\varepsilon)|t|})\qquad|t|>1,\sigma\in K\textit{ a compact subset of }\mathbb{R}

We can choose $\delta=\pi$ if all roots of $f$ are real.

Proof.

(a) This is a direct consequence of creative telescoping, which guarantees existence of $p_{i}$ and a rational function $q(a,s,x)\in\mathbb{R}(a,s,x)$ such that

\sum_{i=0}^{N}p_{i}(a,s)(f(x))^{-a}x^{s+i}\frac{1}{x}=\frac{d}{dx}\left(q(a,s,x)(f(x))^{-a}x^{s+i}\frac{1}{x}\right)

Then integrate above expression with respect to $x$ along contour $C$ mentioned in proof of Theorem 1.1, integral of RHS is zero by fundamental theorem of calculus.

(b) For fixed $a$ , we only need to prove the statement when $\sup K<d\Re(a)$ : the recurrence in (a) allows us to extend this estimate to higher $\Re(s)=\sigma$ . Moreover, since $f$ is assumed to be in $\mathbb{R}[x]$ , $\overline{F_{f}(a,s)}=F_{f}(\overline{a},\overline{s})$ , we can assume $t>0$ . Let $\beta_{1},\cdots,\beta_{d}$ be roots of $f(x)$ , I claim we can let $\delta$ to be

0<\delta=\min_{1\leq i\leq d}\{\arg\beta_{i}\}\qquad 0<\arg\beta_{i}<2\pi

From (1.1), we have

F_{f}(a,s)=\frac{1}{e^{2\pi is}-1}\int_{C}f(x)^{-a}x^{s}\frac{dx}{x},\quad d\Re(a)>\Re(s)

with $C$ the same contour as in the proof there. Fix $r>0$ such that all roots of $f$ have absolute value $>r$ . For given small $\varepsilon>0$ , write $\delta^{\prime}=\delta-\varepsilon$ . We deform the contour $C$ into another contour that consists of three parts:

(1)

a ray from $e^{\delta^{\prime}i}\infty$ to $e^{\delta^{\prime}i}r$
(2)

a circular arc from $re^{\delta^{\prime}i}$ to $re^{-\delta^{\prime}i}$
(3)

a ray from $e^{-\delta^{\prime}i}r$ to $e^{-\delta^{\prime}i}\infty$ .

This is allowed by our choice of $\delta$ : $f(x)^{-a}$ is analytic on the sector $|\arg x|<\delta^{\prime}$ . The integral along (1) is

-e^{\delta^{\prime}i(\sigma+it)}\int_{r}^{\infty}f(e^{\delta^{\prime}i}x)^{-a}x^{\sigma+it}\frac{dx}{x}

whose absolute value is upper bounded by

e^{-\delta^{\prime}t}\left|\int_{r}^{\infty}f(e^{\delta^{\prime}i}x)^{-a}x^{\sigma}\frac{dx}{x}\right|=O_{K,a}(e^{-\delta^{\prime}t})

Similarly, integral along (3) is $O_{K,a}(e^{-(2\pi-\delta^{\prime})t})$ . For integral along (2): its absolute value is

\left|\int_{\delta^{\prime}}^{2\pi-\delta^{\prime}}f(re^{i\theta})^{-a}(re^{i\theta})^{\sigma+it}d\theta\right|=O_{K,a}\left(\int_{\delta^{\prime}}^{2\pi-\delta^{\prime}}e^{-t\theta}d\theta\right)=O_{K,a}(e^{-\delta^{\prime}t})+O_{K,a}(e^{-(2\pi-\delta^{\prime})t})

therefore

F_{f}(a,\sigma+it)=\frac{O_{K,a}(e^{-\delta^{\prime}t})+O_{K,a}(e^{-(2\pi-\delta^{\prime})t})}{e^{2\pi i(\sigma+it)}-1}

since $\delta^{\prime}<\pi$ and $t>0$ , above is $O_{K,a}(e^{-\delta^{\prime}t})$ , as desired. ∎

We say a meromorphic function $f(s)$ has moderate increase along imaginary direction if for every compact subset $K\subset\mathbb{R}$ , there exists $M>0$ such that

|f(\sigma+it)|=O(|t|^{M})\qquad\sigma\in K,\quad|t|\gg 1

Lemma 1.8.

(a) $F_{f}(s,1-s)$ has moderate increase along imaginary direction.

(b) For fixed $z$ , $F_{f}(s,z)$ has moderate increase along imaginary direction in variable $s$ .

Proof.

(a) By creative telescoping, $f(s)=F_{f}(s,1-s)$ satisfies a recurrence of some order with coefficient rational functions of $s$ . So it suffices to prove moderate growth when $\Re(s)$ is large, formula (1.1) is valid and

f(s)=\frac{1}{e^{-2\pi is}-1}\int_{C}(xf(x))^{-s}dx.

For small $r>0$ that we will fix later, deform contour $C$ into three pieces:

•

straight line path from $\infty$ to $r$ above real axis;
•

full counter-clockwise circle from $r$ to $r$ ;
•

straight line path from $r$ to $\infty$ below real axis.

Giving

(1.5)

f(s)=\int_{r}^{\infty}(xf(x))^{-s}dx+\frac{1}{e^{-2\pi is}-1}\int_{0}^{2\pi}[re^{i\theta}f(re^{i\theta})]^{-s}(ire^{i\theta})d\theta

Writing $s=\sigma+it$ , first term is of moderate increase:

\left|\int_{r}^{\infty}(xf(x))^{-s}dx\right|\leq\int_{r}^{\infty}(xf(x))^{-\sigma}dx=O(r^{-\sigma+1})

Since $c:=f(0)\neq 0$ and $c>0$ , we have

f(re^{i\theta})^{-s}=c^{-s}(1+O(re^{i\theta}))^{-s}=c^{-s}\exp[-s\log(1+O(re^{i\theta}))]=c^{-s}\exp[O(rse^{i\theta})](1+O(-r^{2}s)),

this term is $O(1)$ if we choose $r=|t|^{-1}$ , thus second term in (1.5) is bounded above by

\frac{r^{1-\sigma}c^{-\sigma}}{|e^{-2\pi is}-1|}O\left(\int_{0}^{2\pi}e^{(1-s)i\theta}d\theta\right)=\frac{r^{1-\sigma}c^{-\sigma}}{|e^{-2\pi is}-1|}O\left(\int_{0}^{2\pi}e^{t\theta}d\theta\right)=O(1)

Combining both terms $f(s)=O(r^{-\sigma+1})+O(1)=O(|t|^{\sigma-1})$ , so is of moderate growth.

(b) For fixed $z$ , we only need to do this for $\Re(s)$ sufficiently large (again by creative telescoping),

F_{f}(s,z)=\int_{r}^{\infty}f(x)^{-s}x^{z}\frac{dx}{x}+\frac{ir}{e^{2\pi iz}-1}\int_{0}^{2\pi}f(re^{i\theta})^{-s}(re^{i\theta})^{z}d\theta

choose $r=|t|^{-1}$ and same argument as above shows the claim. ∎

2. Meromorphic continuation and residue at abscissa of convergence

We retain all notations from previous section: $f(x)\in\mathbb{R}[x]$ is a polynomial which is positive on $[0,\infty)$ , $d=\deg f$ ,

F_{f}(a,s)=\int_{0}^{\infty}(f(x))^{a}x^{s-1}dx,\qquad K_{f}(a,s)=\frac{\Gamma(a)}{\Gamma(s)\Gamma(ad-s)}F_{f}(a,s)

Proposition 2.1.

The series

\zeta_{f}(s)=\sum_{n,m\geq 1}\left(nm^{d+1}f(\frac{n}{m})\right)^{-s}

converges absolutely when $\Re(s)>\frac{2}{d+2}$ and defines an analytic function in this region.

Proof.

Let $g(x)=x^{d}f(1/x)$ , then

\zeta_{f}(s)=\sum_{n\geq m}(n^{d+1}mg(\frac{m}{n}))^{-s}+\sum_{m>n}(nm^{d+1}f(\frac{n}{m}))^{-s}

Since $c_{1}<f(x)<c_{2}$ for positive constants $c_{i}$ on $x\in[0,1]$ . Above two sums converge absolutely if and only if

\sum_{n\geq m}(n^{d+1}m)^{-s},\qquad\sum_{m>n}(nm^{d+1})^{-s}

both converge. From $\sum_{1\leq n<m}n^{-s}=O(m^{1-s})$ , we see this happens if and only if $\Re(s)>2/(d+2)$ . ∎

We remark that abscissa of convergence for series much more general than $\zeta_{f}(s)$ is known, see for example [8].

From our definition of $F_{f}$ and inverse Mellin transform, we have

f(x)^{-a}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}F_{f}(a,z)x^{-z}dz\quad d\Re(a)>c>0,x>0.

Therefore

	$\displaystyle\zeta_{f}(s)$	$\displaystyle=\frac{1}{2\pi i}\sum_{n,m\geq 1}\int_{c-i\infty}^{c+i\infty}n^{-s}m^{-(d+1)s}F_{f}(s,z)(\frac{n}{m})^{-z}dz$
		$\displaystyle=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}F_{f}(s,z)\zeta(s+z)\zeta((d+1)s-z)dz$
		$\displaystyle=\frac{1}{2\pi i}\frac{1}{\Gamma(s)}\int_{c-i\infty}^{c+i\infty}\Gamma(z)\Gamma(sd-z)K_{f}(s,z)\zeta(s+z)\zeta((d+1)s-z)dz,$

the exchange of summation and integral holds when the series is absolute convergent, that is

\Re(s)+c>1\quad\Re((d+1)s)-c>1\quad d\Re(s)>c>0,

in particular when $\Re(s)$ is sufficiently large. We shall use the above integral representation of $\zeta_{f}(s)$ to extend it meromorphically to all $\mathbb{C}$ .

Due to the poles of $\Gamma$ and $\zeta$ , it is best to visualize this situation using a "singularity diagram". By Theorem 1.1, the integrand

\Gamma(z)\Gamma(sd-z)K_{f}(s,z)\zeta(s+z)\zeta((d+1)s-z)

has poles³³3here we highlight importance of Theorem 1.1, which says $K_{f}(s,z)$ is an entire function. at

z=-n,\quad z=ds+n,\quad z=1-s,\quad z=(d+1)s-1\text{ with }n\in\mathbb{Z}^{\geq 0}

The diagram is constructed by plotting these lines with respect to coordinates $(\Re(s),\Re(z))=(\Re(s),c)$ . Note that his diagram depends only on $d=\deg f$ , not on particular appearance of $f$ .

Refer to caption — Figure 1. Singularity diagram when $d=2$ . Readers can convince themselves this diagram remain essentially same for other $d$ .

These lines cut the plane into different connected components. The component of point $A$ corresponds to the choice of $(\Re(s),c)$ for which our original integral representation

\zeta_{f}(s)=\frac{1}{2\pi i}\frac{1}{\Gamma(s)}\int_{c-i\infty}^{c+i\infty}\Gamma(z)\Gamma(sd-z)K_{f}(s,z)\zeta(s+z)\zeta((d+1)s-z)dz:=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}H_{f}(s,z)dz

is valid. Since

H_{f}(s,z)=F_{f}(s,z)\zeta(s+z)\zeta((d+1)s-z),

Lemma 1.7 tells us the integrand is exponentially decreasing along path of integration, we can shift the contour of integration with $\Re(z)=c$ to an arbitrary $c^{\prime}$ , picking up the residue in between:

\int_{c-i\infty}^{c+i\infty}H_{f}(s,z)dz=\int_{c^{\prime}-i\infty}^{c^{\prime}+i\infty}H_{f}(s,z)dz\pm\left(\text{Sum of residues at poles between }c\text{ and }c^{\prime}\right)

here $+$ is taken if $c^{\prime}<c$ and $-$ if $c^{\prime}>c$ .

For example, we can cross the line $\Re(z)=1-\Re(s)$ to reach point $B$ , then

	$\displaystyle\zeta_{f}(s)$	$\displaystyle=\text{Res}_{z=1-s}[H_{f}(s,z)]+\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}H_{f}(s,z)dz$
		$\displaystyle=\frac{\Gamma(1-s)\Gamma(ds+s-1)}{\Gamma(s)}\zeta((d+2)s-1)K_{f}(s,1-s)+\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}H_{f}(s,z)dz$

where now $(\Re(s),c)$ in last integral should be in the connected component of point $B$ , it is analytic as a function of $s$ for whatever $\Re(s)$ in this region. In particular, it is analytic near $s=2/(d+2)$ , the term in front has a simple pole at this point, so we established

Theorem 2.2.

$\zeta_{f}(s)$ has a simple pole at $s=2/(d+2)$ , with residue

\frac{1}{d+2}F_{f}(\frac{2}{d+2},\frac{d}{d+2})=\frac{1}{d+2}\int_{0}^{\infty}(xf(x))^{-2/(d+2)}dx

For general $f$ , one does not expect this integral to simplify further. However, for $f$ arising from rank $2$ root systems, they all can be evaluated nicely:

Theorem 2.3.

Residue of $\omega_{A}(s)$ at $s=2/3$ is

\frac{\Gamma(\frac{1}{3})^{3}}{2\sqrt{3}\pi}.

Residue of $\omega_{B}(s)$ at $s=1/2$ is

\frac{\Gamma\left(\frac{1}{4}\right)^{2}}{8\sqrt{2\pi}}.

Residue of $\omega_{G}(s)$ at $s=1/3$ is

\frac{\Gamma\left(\frac{1}{3}\right)^{3}}{2^{8/3}3^{3/2}\pi}.

Proof.

We skip proofs for $\omega_{A},\omega_{B}$ , partly because evaluation of integrals $\int_{0}^{\infty}(xf(x))^{-2/(d+2)}$ are trivial in these cases, partly because they are known result ([19] for $\omega_{A}$ and [4] for $\omega_{B}$ ). We concentrate on $\omega_{G}$ , the claim is equivalent to non-trivial evaluation

\int_{0}^{\infty}\frac{dx}{\sqrt[3]{x(1+x)(1+2x)(1+3x)(2+3x)}}=\frac{1}{2^{5/3}3^{1/2}\pi}\Gamma\left(\frac{1}{3}\right)^{3}.

In order not to interrupt our discussion, we put its proof in Appendix. ∎

Back to meromorphic continuation of $\zeta_{f}(s)$ . In the diagram, we can continue from point $B$ to $C$ , to $D$ and so on, picking up residues at $z=0,-1,-2,\cdots$ in the process, so we obtain

(2.1)

\zeta_{f}(s)=\frac{\Gamma(1-s)\Gamma(ds+s-1)}{\Gamma(s)}\zeta((d+2)s-1)K_{f}(s,1-s)+\sum_{i=0}^{M}\frac{(-1)^{i}}{i!}\frac{\Gamma(sd+i)}{\Gamma(s)}K_{f}(s,-i)\zeta(s-i)\zeta((d+1)s+i)\\ +\frac{1}{\Gamma(s)}\frac{1}{2\pi i}\int_{-M-1/2-i\infty}^{-M-1/2+i\infty}\Gamma(z)\Gamma(sd-z)K_{f}(s,z)\zeta(s+z)\zeta((d+1)s-z)dz

for any non-negative integer $M$ . From the diagram, we see that as $M$ gets sufficiently large, the latter integral becomes analytic at any point $s\in\mathbb{C}$ , so we proved the meromorphic continuation of $\zeta_{f}(s)$ .

Proposition 2.4.

$\zeta_{f}(s)$ is moderately increasing along imaginary direction, that is, for any compact subset $K\subset\mathbb{R}$ , there exists $M>0$ such that

|\zeta_{f}(\sigma+it)|=O(|t|^{M}),\qquad\sigma\in K,|t|\gg 1

Proof.

Rewrite equation 2.1 as

\zeta_{f}(s)=\zeta((d+2)s-1)F_{f}(s,1-s)+\sum_{i=0}^{M}\zeta(s-i)\zeta((d+1)s+i)[f(x)^{-s}][x^{i}]\\ +\frac{1}{2\pi i}\int_{-M-1/2-i\infty}^{-M-1/2+i\infty}\zeta(s+z)\zeta((d+1)s-z)F_{f}(s,z)dz

Recall $\zeta(s)$ is moderately increasing, $F_{f}(s,1-s)$ is too (Lemma 1.8), so first two terms are moderately increasing. As for the integral, same lemma implies $F_{f}(s,z)$ is moderately increasing in $s$ when $\Re(z)$ is fixed. ∎

Actually, a much more general moderate growth result on zeta functions of hyperplane arrangement is known, see [7].

3. Values at non-positive integers

In the representation (2.1), when let $s$ tends to $-n$ , where $n$ is a non-negative integer, the last term involving the integral is $0$ because of $1/\Gamma(s)$ at the front. As for the sum,

\sum_{i=0}^{M}\frac{(-1)^{i}}{i!}\frac{\Gamma(sd+i)}{\Gamma(s)}K_{f}(s,-i)\zeta(s-i)\zeta((d+1)s+i)

when $s\to-n$ , only terms that are non-zero come from $i$ such that $0\leq i\leq nd$ and $i=(d+1)n+1$ . Now it is easy to write down the formula of $\zeta_{f}(-n)$ :

\zeta_{f}(-n)=\frac{(n!)^{2}}{d+1}\frac{(-1)^{dn+1}}{((d+1)n+1)!}\zeta(-(d+2)n-1)\left[K_{f}(-n,1+n)+K_{f}(-n,-1-(d+1)n)\right]\\ +\sum_{i=0}^{nd}\frac{(-1)^{n(d+1)}n!}{d\times i!(nd-i)!}K_{f}(-n,-i)\zeta(-n-i)\zeta(-(d+1)n+i)

Note that $K_{f}(-n,1+n)=K_{g}(-n,-1-(d+1)n)$ with $g(x)=x^{d}f(1/x)$ . Some simplifications using Proposition 1.4 yields the following form:

Theorem 3.1.

$\zeta_{f}(s)$ is analytic at $s=0,-1,-2,\cdots$ and

\zeta_{f}(-n)=\frac{\zeta(-(d+2)n-1)}{d+1}\left([-f(x)^{n}\log f(x)][x^{1+(1+d)n}]+[-g(x)^{n}\log g(x)][x^{1+(1+d)n}]\right)\\ +\sum_{i=0}^{nd}[f(x)^{n}][x^{i}]\zeta(-n-i)\zeta(-(d+1)n+i),

where $g(x)=x^{d}f(1/x)$ .

Corollary 3.2.

When $d=\deg f$ is odd and $n$ is a non-negative integer, we have

\zeta_{f}(-2n-1)=0

Proof.

Conditions on parity of $d$ and $n$ ensure every term in above expression is zero because Riemann’s zeta at negative even integers are $0$ . ∎

Example 3.3.

Let $f(x)=(1+x)(1+3x)$ ,

\zeta_{f}(s)=\sum_{n,m\geq 1}\frac{1}{n^{s}m^{s}(n+m)^{s}(n+3m)^{s}}.

$\zeta_{f}(0),\zeta_{f}(-1),\cdots,\zeta_{f}(-4)$ equal respectively

\frac{43}{108},\quad-\frac{23}{87480},\quad\frac{809}{72171},\quad\frac{10828231}{51963120},\quad\frac{2383543667}{157837977}.

The values of $\zeta_{f}(-n)$ with $f$ coming from root system, i.e., $\omega_{A}(s),\omega_{B}(s)$ and $\omega_{G}(s)$ are much more interesting.

Example 3.4.

For $\omega_{A}(s)$ , one calculates, using above formula, that $\omega_{A}(0)=1/3$ and one quickly notices $\omega_{A}(-n)=0$ for first few $n$ .

This is indeed true in general. For odd $n$ , this is explained by above corollary; the vanishing of $\omega_{A}$ at even $n$ is a much deeper fact (see below).

Example 3.5.

For $\omega_{B}(s)$ , its values at $s=0,-1,\cdots,-6$ are

\frac{3}{8},\quad-\frac{11}{4480},\quad 0,\quad-\frac{4581}{1576960},\quad 0,\quad-\frac{287820799443}{256502005760},\quad 0.

For $\omega_{G}(s)$ , its values at $s=0,-1,\cdots,-6$ are

\frac{5}{12},\quad\frac{33205}{2612736},\quad 0,\quad\frac{313071820474581425}{1580471680499712},\quad 0,\quad\frac{34067192614177690045793856810696875}{2320107746327814532497408},\quad 0.

These values have already been reported in [20]. The obvious pattern $\omega_{B}(-2n)=\omega_{G}(-2n)=0$ for $n\geq 1$ is indeed true. It is special case of the following deep result.

Theorem 3.6 ([1]).

Let $\Phi$ be a root system, $m$ be a negative even integer, the order of vanishing of $\zeta_{\Phi}(s)$ at $s=m$ is at least the rank of $\Phi$ , that is ⁴⁴4Here $\text{ord}_{s=s_{0}}f(s)$ for a meromorphic function is order of zero or pole at point $s=s_{0}$ .

\text{ord}_{s=m}(\zeta_{\Phi}(s))\geq\text{ Rank of $\Phi$}.

Its proof relies on root system’s special symmetry and employs technique of different nature. It follows $\omega_{A}(s),\omega_{B}(s),\omega_{G}(s)$ not only vanish at $s=-2n$ , but also these are at least double zeroes. For $\omega_{A}(s)$ , this is already shown in [17].

3.1. A connection to Eisenstein series

We make some interesting observations on vanishing of $\omega_{B}(-2n),\omega_{G}(-2n)$ .

First we write out, using Theorem 3.1, an explicit expression that is equivalent to vanishing of $\omega_{\bullet}(-2n)$ with $n\geq 1$ .

For $\omega_{A}$ , it is better to directly use the fact that $K_{f}(a,s)\equiv 1$ for $f(x)=1+x$ , arriving at:

(3.1)

\omega_{A}(-2n)=0\iff\frac{(2n)!}{(4n+1)!}\zeta(-6n-1)=\sum_{i=0}^{2n}\frac{1}{i!(2n-i)!}\zeta(-i-2n)\zeta(i-4n)

For $\omega_{B}$ and $\omega_{G}$ , denote $f_{B}(x)=(1+x)(1+2x)$ and $f_{G}(x)=(1+x)(1+2x)(1+3x)(2+3x)$ ,

\omega_{B}(-2n)=0\iff\frac{\zeta(-8n-1)}{3}(1+2^{-1-4n})[f_{B}(x)^{2n}\log f_{B}(x)][x^{1+6n}]=\sum_{i=0}^{4n}[f_{B}(x)^{2n}][x^{i}]\zeta(-i-2n)\zeta(i-6n)

\omega_{G}(-2n)=0\iff\frac{\zeta(-12n-1)}{5}(1+3^{-1-6n})[f_{G}(x)^{2n}\log f_{G}(x)][x^{1+10n}]=\sum_{i=0}^{8n}[f_{G}(x)^{2n}][x^{i}]\zeta(-i-2n)\zeta(i-10n)

Recall following Eisenstein series, for $\tau$ in upper-half plane,

E_{2k}(\tau)=\frac{1}{2\zeta(2k)}\sum_{\begin{subarray}{c}(m,n)\in\mathbb{Z}^{2}\\ (m,n)\neq(0,0)\end{subarray}}\frac{1}{(m+n\tau)^{2k}}\qquad k\geq 2

We also agree that $E_{k}(\tau)=0$ for $k\geq 3$ odd. Note that $\lim_{\Im(\tau)\to\infty}E_{2k}(\tau)=1$ .

Romik ([19]) observed that, in equation (3.1), if we replace every occurrence of $\zeta(1-2k)$ by $\zeta(1-2k)E_{2k}(\tau)$ , we still obtain a valid equality. That is

\frac{(2n)!}{(4n+1)!}\zeta(-6n-1)E_{6n+2}(\tau)=\sum_{i=0}^{2n}\frac{1}{i!(2n-i)!}\zeta(-i-2n)\zeta(i-4n)E_{2n+i+1}(\tau)E_{4n-i+1}(\tau)

the original equality (3.1) is recovered by letting $\Im(\tau)\to\infty$ . Romik gave a proof that applies to both (3.1) and the Eisenstein series version.

Surprisingly, when one performs the same replacement to the equalities coming from $\omega_{B},\omega_{G}$ , they seem still hold⁵⁵5They have been checked for $n\leq 10$ . We formulate them as conjectures.

Conjecture 3.7.

For positive integers $n$ , the following two equalities hold:

\frac{\zeta(-8n-1)E_{8n+2}(\tau)}{3}(1+2^{-1-4n})\times[f_{B}(x)^{2n}\log f_{B}(x)][x^{1+6n}]\\ =\sum_{i=0}^{4n}[f_{B}(x)^{2n}][x^{i}]\zeta(-i-2n)\zeta(i-6n)E_{2n+i+1}(\tau)E_{6n-i+1}(\tau)

\frac{\zeta(-12n-1)E_{12n+2}(\tau)}{5}(1+3^{-1-6n})\times[f_{G}(x)^{2n}\log f_{G}(x)][x^{1+10n}]\\ =\sum_{i=0}^{8n}[f_{G}(x)^{2n}][x^{i}]\zeta(-i-2n)\zeta(i-10n)E_{2n+i+1}(\tau)E_{10n-i+1}(\tau)

4. Poles of $\zeta_{f}(s)$

Apart from the pole of $\zeta_{f}(s)$ at abscissa of convergence $s=2/(d+2)$ , it also has a family of simple poles at other $s$ , they are easily described.

Theorem 4.1.

Let $s_{0}\neq 2/(d+2)$ such that $\zeta_{f}(s)$ have a pole at $s=s_{0}$ . Then this pole is (at most) simple, $s_{0}\notin\mathbb{Z}$ , $s_{0}(1+d)-1=-n\in\mathbb{Z}^{\leq 0}$ , the residue at this point is⁶⁶6recall $[h(x)][x^{m}]$ means coefficient of $x^{m}$ in series expansion of $h(x)$ around $x=0$ .

\frac{\zeta((d+2)s_{0}-1)}{d+1}\left([f(x)^{-s_{0}}][x^{n}]+[g(x)^{-s_{0}}][x^{n}]\right)

where $g(x)=x^{d}f(1/x)$ .

Proof.

We have seen that $\zeta_{f}(s)$ is analytic at integer $s$ , so $s_{0}$ is not an integer. From (2.1),

\zeta_{f}(s)=\frac{\Gamma(1-s)\Gamma((d+1)s-1)}{\Gamma(s)}\zeta((d+2)s-1)K_{f}(s,1-s)+\sum_{i=0}^{M}\frac{(-1)^{i}}{i!}\frac{\Gamma(sd+i)}{\Gamma(s)}K_{f}(s,-i)\zeta(s-i)\zeta((d+1)s+i)\\ +\frac{1}{\Gamma(s)}\frac{1}{2\pi i}\int_{-M-1/2-i\infty}^{-M-1/2+i\infty}\Gamma(z)\Gamma(sd-z)K_{f}(s,z)\zeta(s+z)\zeta((d+1)s-z)dz

For $s\notin\mathbb{Z}$ , the only possible pole that could arise are those coming from

\zeta((d+2)s-1)\text{ or }\Gamma((d+1)s-1)\text{ or }\Gamma(sd+i).

The first choice gives $s=2/(d+2)$ , which we already excluded. The third case cannot give a pole either: let $s\to-n_{0}/d$ where $n_{0}\geq\mathbb{Z}^{\geq 0}$ , then residue of the sum $\sum_{i=0}^{M}$ term is

\sum_{i=0}^{n_{0}}\frac{(-1)^{i}}{i!\Gamma(-n_{0}/d)}\frac{(-1)^{n_{0}-i}}{d(n_{0}-i)!}K_{f}(-\frac{n_{0}}{d},-i)\zeta(-\frac{n_{0}}{d}-i)\zeta(-\frac{n_{0}}{d}+i-n_{0})

for this range of $i$ , Proposition 1.3, together with the assumption $-n_{0}/d\notin\mathbb{Z}$ , implies $K_{f}(-\frac{n_{0}}{d},-i)=0$ , so above sum is zero.

So only the second case remains: $s\to s_{0}$ such that $s_{0}(1+d)-1=-n_{0}$ with $n\in\mathbb{Z}^{\geq 0}$ . At this point $(d+2)s-1,sd+i\notin\mathbb{Z}$ , so only contribution of residue comes from the term $\frac{\Gamma(1-s)\Gamma((d+1)s-1)}{\Gamma(s)}\zeta((d+2)s-1)K_{f}(s,1-s)$ and inside summation with $i=-(d+1)s_{0}+1=n$ . A straightforward calculation says residue is

\frac{(-1)^{n}}{n!}\frac{\Gamma(1-s_{0})}{\Gamma(s_{0})}\zeta((d+2)s_{0}-1)\frac{1}{1+d}\left[K_{f}(s_{0},1-s_{0})+K_{f}(s_{0},-n)\right]

Note that $K_{f}(s_{0},1-s_{0})=K_{g}(s_{0},-n)$ and Proposition 1.3 completes the proof. ∎

Corollary 4.2.

Let $f(x)=c_{0}+c_{1}x+\cdots+c_{d}x^{d}$ . Apart from $s=2/(2+d)$ , the only other pole of $\zeta_{f}(s)$ with $\Re(s)\geq 0$ is $s=1/(1+d)$ , it is simple and residue there is

\frac{c_{0}^{-1/(d+1)}+c_{d}^{-1/(d+1)}}{d+1}\zeta(\frac{1}{d+1})

Proof.

This is special case $s_{0}=1/(d+1),n=0$ in above theorem. ∎

Example 4.3.

For $\omega_{A}(s)$ , we have $d=1$ , and $n$ has to be even. Letting $k=2n$ , residue at $s=1/2-k$ with $k\geq 0$ is

\zeta(\frac{1}{2}-3k)\binom{k-1/2}{2k}

a result already obtained by Romik [19].

Proposition 4.4.

(a) Let $n$ be a non-negative integer, residue of $\omega_{B}(s)$ at $s=(1-n)/3$ is

\frac{1}{3}\zeta(\frac{1-4n}{3})(1+2^{(-1-2n)/3})b_{n}

with $b_{n}\in\mathbb{Q}$ equals the coefficient of $x^{n}$ in power series expansion of $\left((1+x)(1+2x)\right)^{(n-1)/3}$ . Moreover, $b_{n}$ satisfies the recurrence

27(n+4)(n+6)b_{n+6}-4(n+2)(n+5)b_{n}=0

(b) Let $n$ a non-negative integer, residue of $\omega_{G}(s)$ at $s=(1-n)/5$ is

\frac{1}{5}\zeta(\frac{1-6n}{5})(1+3^{(-3n-2)/5})2^{(n-1)/5}g_{n}

with $g_{n}\in\mathbb{Q}$ equals the coefficient of $x^{n}$ in power series expansion of $\left((1+x)(1+2x)(1+3x)(1+\frac{3}{2}x)\right)^{(n-1)/5}$ . Moreover, $g_{n}$ satisfies the recurrence

729(n+4)(n+9)(n+14)(n+19)(6n+103)g_{n}-1875(n+14)(n+19)\left(216n^{3}+7596n^{2}+83694n+290603\right)g_{n+10}\\ +50000(n+12)(n+16)(n+18)(n+20)(6n+43)g_{n+20}=0

Proof.

Firstly, note that $b_{n}=0$ when $n\equiv 1\pmod{3}$ and $g_{n}=0$ when $n\equiv 1\pmod{5}$ , because $\left((1+x)(1+2x)\right)^{(n-1)/3}$ and $\left((1+x)(1+2x)(1+3x)(1+\frac{3}{2}x)\right)^{(n-1)/5}$ become polynomials, this is consistent with the fact that $\omega_{B}(s),\omega_{G}(s)$ being analytic at integral $s$ .

The assertion on expression of residue follows immediately from Theorem 4.1. Here one notes that, for

f_{B}(x)=(1+x)(1+2x),\quad g_{B}(x)=x^{2}f_{B}(1/x),

$g_{B}(2x)=2f_{B}(x)\implies[g_{B}(x)^{-s}][x^{n}]=2^{-s-n}[f_{B}(x)^{s}][x^{n}]$ ; and similarly

f_{G}(x)=(1+x)(1+2x)(1+3x)(2+3x),\quad g_{G}(x)=x^{4}f_{G}(1/x),

$g_{G}(3x)=9f_{G}(x)\implies[g_{G}(x)^{-s}][x^{n}]=3^{-2s-n}[f_{G}(x)^{s}][x^{n}]$ . It remains to prove the recurrences satisfied by $b_{n}$ and $g_{n}$ , this is follows from creative telescoping: we have

b_{n}=\frac{1}{2\pi i}\int_{C(\varepsilon)}(f_{B}(x))^{(n-1)/3}x^{-n-1}dx,\qquad g_{n}=\frac{1}{2\pi i}\int_{C(\varepsilon)}(f_{G}(x))^{(n-1)/5}x^{-n-1}dx,

where $C(\varepsilon)$ is a small counter-clockwise circle around $x=0$ . Let $F(x,n)=(f_{B}(x))^{(n-1)/3}x^{-n-1}$ , then one checks,

27(4+n)(6+n)F(x,n+6)-4(2+n)(5+n)F(x,n)\\ =\frac{d}{dx}\left(\frac{3(x+1)(2x+1)\left(2nx^{4}-6nx^{3}-39nx^{2}-36nx-9n+10x^{4}-12x^{3}-150x^{2}-144x-36\right)}{x^{5}}F(x,n)\right).

Applying $\int_{C(\varepsilon)}dx$ on both sides shows the recurrence for $b_{n}$ . Analogue method works for $g_{n}$ (see remark below). ∎

Remark 4.5.

In the above proof, certificate of creative telescoping⁷⁷7it refers to the complicated rational function inside $\frac{d}{dx}\left(\cdots\cdots F(x,n)\right)$ . is obtained with help of Mathematica package HolonomicFunctions developed by C. Koutschan, [13]. One simply executes the following command:

CreativeTelescoping[((1+x)(1+2x))^((n-1)/3)*x^(-n-1),Der[x],{S[n]}] // Factor

For $g_{n}$ , one make corresponding modification, this certificate is too long to be displayed here.

For a generic $f(x)$ of degree $d$ , residue of $\zeta_{f}(s)$ at $s=s_{0}$ allowed by Theorem 4.1 is in general non-zero. Witten zeta functions display here another unexpected property: residues at many more points turn out to be zero, $\zeta_{f}(s)$ is thus analytic there.

Theorem 4.6.

(a) $\omega_{B}(s)$ has a pole at $s=1/2$ and also at $s=\frac{k}{3}$ with $k\equiv 1,5\pmod{6}$ and $k\leq 1$ .

(b) $\omega_{G}(s)$ has a pole at $s=1/3$ and also (possibly) at $s=\frac{k}{5}$ with $k\equiv 1,3,7,9\pmod{10}$ and $k\leq 1$ .

In both cases, all poles are simple, these are only poles of $\omega_{B}(s)$ and $\omega_{G}(s)$ .

Proof.

Poles of $\omega_{B}(s)$ at $s=1/2$ and of $\omega_{G}(s)$ at $s=1/3$ comes from the abscissa of convergence.

For remaining poles, we concentrate on $\omega_{G}$ since the following argument for $\omega_{B}$ is similar. For $k\equiv 0,5\pmod{10}$ , $\omega_{G}(s)$ is analytic at $s=k/5$ because $k/5$ is an integer. To show $\omega_{G}(s)$ is also analytic at $s=k/5$ for $k\equiv 2,4,6,8\pmod{10}$ , it suffices to show $g_{n}=0$ for $n\equiv 3,5,7,9\pmod{10}$ . This can be easily checked from the recurrence: for example, one computes $g_{3}=g_{13}=0$ , the recurrence implies $g_{3+10n}=0$ . ∎

$\omega_{B}(s)$ has a pole of exact order $1$ at $s=\frac{k}{3}$ for $k\leq 1,k\equiv 1,5\pmod{6}$ : this follows from the non-vanishing of $b_{n}$ at $n\equiv 0,2\pmod{6}$ . For $\omega_{G}$ , non-vanishing of $g_{n}$ at $n\equiv 0,2,4,8\pmod{10}$ implies we can remove the word "possibly" in theorem above. This should not be difficult to prove by deriving an asymptotic of corresponding second order recurrence (Poincaré-Perron theorem [18]).

Conjecture 4.7.

Consider above defined sequence $\{g_{n}\}_{n\geq 0}$ , then $g_{n}\neq 0$ when $n\equiv 0,2,4,8\pmod{10}$ .

Results of this section concerning $\omega_{B}(s)$ are already obtained by Bringmann et al. in [4]. $\omega_{G}(s)$ having possible poles at $s=k/5$ was already known in [9], however, the fact that half of such $k$ actually does not give poles seems unknown until now.

5. Derivative at $s=0$

In this section, we show how to evaluate $\zeta_{f}^{\prime}(0)$ when all roots of $f$ are rational. We will be brief because $\zeta_{f}^{\prime}(0)$ was already evaluated in [20], using a different approach.

As a preliminary, we need to evaluate the following integral

\mathcal{I}(\alpha)=\frac{1}{2\pi i}\int_{-3/2-i\infty}^{-3/2+i\infty}\Gamma(z)\zeta(z)\alpha^{-z}\Gamma(-z)\zeta(-z)dz\qquad\alpha>0

Proposition 5.1.

Let $\alpha=p/q$ , with $p,q\in\mathbb{N}$ coprime. Then $\mathcal{I}(\alpha)$ is a $\mathbb{Q}$ -linear combination of following constants:

1,\quad\gamma,\quad\zeta^{\prime}(0),\quad\zeta^{\prime}(-1),\quad\log p,\quad\log q,\quad\log\Gamma(\frac{i}{p})\text{ with }1\leq i\leq p-1,\quad\log\Gamma(\frac{i}{q})\text{ with }1\leq i\leq q-1

Here we point out that $\zeta^{\prime}(0)=-\log(2\pi)/2$ which a well-known conclusion. We now prove the Proposition. Denote

(5.1)

\mathcal{I}(s,\alpha)=\frac{1}{\Gamma(s)}\frac{1}{2\pi i}\int_{3/2-i\infty}^{3/2+i\infty}\Gamma(z)\zeta(s)\alpha^{-z}\Gamma(s-z)\zeta(s-z)dz\qquad\Re(s)>5/2

Lemma 5.2.

When $\Re(s)>5/2$ , we have

\mathcal{I}(s,\alpha)=\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{x^{s-1}}{(e^{\alpha x}-1)(e^{x}-1)}dx

Proof.

This is a simple consequence of Parseval identity: if

F_{i}(s)=\int_{0}^{\infty}x^{s-1}f_{i}(x)dx\qquad\alpha_{i}<\Re(s)<\beta_{i}

then

\int_{0}^{\infty}x^{s-1}f_{1}(x)f_{2}(x)dx=\frac{1}{2\pi i}\int_{c-\infty}^{c+\infty}F_{1}(z)F_{2}(s-z)dz\qquad\alpha_{2}+c<\Re(s)<\beta_{2}+c,\alpha_{1}<c<\beta_{1}

Specialize this to $f_{1}(x)=\frac{1}{e^{x}-1},f_{2}(x)=\frac{1}{e^{\alpha x}-1}$ , with corresponding $F_{1}(s)=\Gamma(s)\zeta(s)$ and $F_{2}(s)=\alpha^{-s}\Gamma(s)\zeta(s)$ . ∎

Recall polylogarithm and Hurwitz zeta function:

\operatorname{Li}_{s}(z)=\sum_{n\geq 1}\frac{z^{n}}{n^{s}},\qquad\zeta(s,a)=\sum_{n\geq 0}\frac{1}{(n+a)^{s}}

Lemma 5.3.

When $\mu\neq 1$ is an $N$ -th root of unity,

\left.\frac{d}{ds}\right|_{s=0}\operatorname{Li}_{s}(\mu)=\frac{\mu}{\mu-1}\log N+\sum_{j=1}^{N-1}\mu^{j}\log\Gamma(\frac{j}{N})

Proof.

For $\Re(s)>1$ , we have

\operatorname{Li}_{s}(\mu)=\sum_{n\geq 1}\frac{\mu^{n}}{n^{s}}=\sum_{i=0}^{\infty}\sum_{j=1}^{N}\frac{\mu^{j}}{(j+iN)^{s}}=N^{-s}\sum_{j=1}^{N}\mu^{j}\zeta(s,\frac{j}{N}),

Recall following two classical identities: ([21])

\zeta(0,a)=\frac{1}{2}-a,\qquad\left.\frac{d}{ds}\right|_{s=0}\zeta(s,a)=\log\Gamma(a)-\frac{\log(2\pi)}{2}.

Differentiate initial displayed equation at $s=0$ and plug in these two identities on $\zeta(s,a)$ completes the proof. ∎

Let $\alpha=p/q$ be a positive rational number indicated in Proposition, and let

q^{-s}\mathcal{I}(s,\alpha)=c_{0}(p,q)+c_{1}(p,q)s+O(s^{2})\qquad s\to 0

Lemma 5.4.

We have

c_{0}(p,q)=\frac{1}{4}+\frac{1}{12}(\frac{p}{q}+\frac{q}{p}).

and

c_{1}(p,q)=\frac{\zeta^{\prime}(-1)}{pq}-\frac{p+q}{2pq}\zeta^{\prime}(0)+\frac{\log q}{q}r_{q}(p)+\frac{\log p}{p}r_{p}(q)+\frac{1}{q}\sum_{j=1}^{q}s_{q,p}(j)\log\Gamma\left(\frac{j}{q}\right)+\frac{1}{p}\sum_{j=1}^{p}s_{p,q}(j)\log\Gamma\left(\frac{j}{p}\right)

where

s_{q,p}(j)=-\frac{q-1}{2}+\left(\text{the unique }0\leq i\leq q-1\text{ such that }ip+j\equiv 0\pmod{q}\right)

and

r_{q}(p)=\frac{1}{q}\sum_{i=1}^{q-1}is_{q,p}(i)

Proof.

q^{-s}\mathcal{I}(s,\alpha)=\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{x^{s-1}}{(e^{px}-1)(e^{qx}-1)}dx

$c_{0}(p,q)$ is simply residue of $\frac{1}{x(e^{px}-1)(e^{qx}-1)}$ at $x=0$ . $c_{1}(p,q)$ requires a more elaborate computation. Partial fraction decomposition

\frac{1}{(x^{p}-1)(x^{q}-1)}=\frac{1}{pq}\frac{1}{(x-1)^{2}}+\frac{2-p-q}{2pq}\frac{1}{(x-1)}+\sum_{\nu\neq 1,\nu^{q}=1}\frac{\nu}{q(\nu^{p}-1)}\frac{1}{x-\nu}+\sum_{\mu:\mu\neq 1,\mu^{p}=1}\frac{\mu}{p(\mu^{q}-1)}\frac{1}{x-\mu}

together with Mellin transforms

\int_{0}^{\infty}\frac{x^{s-1}dx}{(e^{x}-1)^{2}}=(\zeta(s-1)-\zeta(s))\Gamma(s),\quad\int_{0}^{\infty}\frac{x^{s-1}dx}{e^{x}-a}=\frac{\Gamma(s)\operatorname{Li}_{s}(a)}{a}

produce

q^{-s}\mathcal{I}(s,\alpha)=\frac{1}{pq}(\zeta(s-1)-\zeta(s))+\frac{2-p-q}{2pq}\zeta(s)+\sum_{\nu\neq 1,\nu^{q}=1}\frac{\operatorname{Li}_{s}(\nu)}{q(\nu^{p}-1)}+\sum_{\mu:\mu\neq 1,\mu^{p}=1}\frac{\operatorname{Li}_{s}(\mu)}{p(\mu^{q}-1)}

From Lemma above, we obtain the expression of $c_{1}(p,q)$ as in statement with

r_{q}(p)=\sum_{\mu\neq 1,\mu^{q}=1}\frac{\mu}{(\mu-1)(\mu^{p}-1)},\qquad s_{q,p}(j)=\sum_{\mu\neq 1,\mu^{q}=1}\frac{\mu^{j}}{\mu^{p}-1}

here the sum is over all $q$ -th roots of unity that is not $1$ . Simplification of these two trigonometric sums gives the expression in statement. ∎

Therefore

(5.2)

\mathcal{I}(s,\alpha)=c_{0}(p,q)+[c_{0}(p,q)\log q+c_{1}(p,q)]s+O(s^{2}).

The original representation $\mathcal{I}(s,\alpha)$ in (5.1) is valid only for $\Re(s)>5/2$ , in order to get a representation that is valid at $s=0$ , we shift the contour from $\Re(z)=3/2$ to $\Re(z)=-3/2$ , picking up residues at $z=1,0,-1$ to yield:

\mathcal{I}(s,\alpha)=\frac{\zeta(s-1)}{\alpha(s-1)}+\frac{1}{12}(\alpha s\zeta(s+1)-6\zeta(s))+\frac{1}{\Gamma(s)}\frac{1}{2\pi i}\int_{-3/2-i\infty}^{-3/2+i\infty}\Gamma(z)\zeta(s)\alpha^{-z}\Gamma(s-z)\zeta(s-z)dz.

This is now valid at $s=0$ , its $s$ coefficient is

\frac{\gamma\alpha^{2}+3\alpha\log(2\pi)+1-12\zeta^{\prime}(-1)}{12\alpha}+\mathcal{I}(\alpha)

Comparing with (5.2) finally gives an explicit expression of $\mathcal{I}(\alpha)$ :

\mathcal{I}(\alpha)=c_{0}(p,q)\log q+c_{1}(p,q)-\frac{\gamma\alpha^{2}+3\alpha\log(2\pi)+1-12\zeta^{\prime}(-1)}{12\alpha}

here $c_{i}(p,q)$ are computed according to formula in Lemma above.

For example:

	$\displaystyle\mathcal{I}(1)$	$\displaystyle=2\zeta^{\prime}(-1)-\frac{\gamma}{12}-\frac{1}{12}+\frac{1}{4}\log(2\pi);$
	$\displaystyle\mathcal{I}(\frac{1}{2})$	$\displaystyle=\frac{5\zeta^{\prime}(-1)}{2}-\frac{1}{6}-\frac{\gamma}{24}+\frac{11\log 2}{24}+\frac{\log\pi}{4};$
	$\displaystyle\mathcal{I}(\frac{1}{3})$	$\displaystyle=\frac{10}{3}\zeta^{\prime}(-1)-\frac{1}{4}-\frac{\gamma}{36}+\frac{11\log 3}{36}+\frac{1}{12}\log(2\pi)+\frac{1}{3}\log\Gamma\left(\frac{1}{3}\right);$
	$\displaystyle\mathcal{I}(\frac{3}{2})$	$\displaystyle=\frac{5}{6}\zeta^{\prime}(-1)-\frac{\gamma}{8}-\frac{1}{18}+\frac{19\log 2}{72}+\frac{\log\pi}{12}-\frac{\log 3}{9}+\frac{1}{3}\log\Gamma\left(\frac{2}{3}\right).$

Now we go back to the problem of evaluating $\zeta_{f}^{\prime}(0)$ . Let

f(x)=c_{0}(1+\alpha_{1}x)\cdots(1+\alpha_{d}x)=c_{0}+c_{1}x+\cdots+c_{d}x^{d},

recall that $K_{f}(0,z)=\sum_{i}\alpha_{i}^{-z}$ (Proposition 1.5). From (2.1), the following representation is valid at a neighbourhood of $s=0$ :

\zeta_{f}(s)=\frac{\Gamma(1-s)\Gamma(ds+s-1)}{\Gamma(s)}\zeta((d+2)s-1)K_{f}(s,1-s)+\sum_{i=0}^{1}\zeta(s-i)\zeta((d+1)s+i)[f(x)^{-s}][x^{i}]\\ +\frac{1}{\Gamma(s)}\frac{1}{2\pi i}\int_{-3/2-i\infty}^{-3/2+i\infty}\Gamma(z)\Gamma(sd-z)K_{f}(s,z)\zeta(s+z)\zeta((d+1)s-z)dz

Let $A_{f}$ be sum of coefficients of $s$ in Taylor expansion of first three terms in the front, then

	$\displaystyle\zeta_{f}^{\prime}(0)$	$\displaystyle=A_{f}+\frac{1}{2\pi i}\int_{-3/2-i\infty}^{-3/2+i\infty}\Gamma(z)\Gamma(sd-z)K_{f}(0,z)\zeta(s+z)\zeta((d+1)s-z)dz$
		$\displaystyle=A_{f}+\sum_{i=1}^{d}\mathcal{I}(\alpha_{i}),$

with same $\mathcal{I}(\alpha)$ as before. Observe

\sum_{i=0}^{1}\zeta(s-i)\zeta((d+1)s+i)[f(x)^{-s}][x^{i}]=\zeta(s)\zeta((d+1)s)c_{0}^{-s}+\zeta(s-1)\zeta((d+1)s+1)\left(-sc_{0}^{-s-1}c_{1}\right);

Proposition 1.6 enables us to expand $K_{f}(s,1-s)$ upto $O(s^{2})$ , so $A_{f}$ can be found. When all $\alpha_{i}\in\mathbb{Q}$ , we showed previously how to evaluate $\mathcal{I}(\alpha_{i})$ . This completes our description on how to evaluate $\zeta^{\prime}_{f}(0)$ .

Example 5.5.

Let $f(x)=(1+x)(1+3x)$ ,

\zeta_{f}(s)=\sum_{n,m\geq 1}\frac{1}{n^{s}m^{s}(n+m)^{s}(n+3m)^{s}}

an explicit calculation gives

\zeta_{f}^{\prime}(0)=-\frac{4}{9}\zeta^{\prime}(-1)-\frac{25}{108}\log 3+\frac{4}{3}\log(2\pi)+\frac{1}{3}\log\Gamma(\frac{1}{3})

In view of occurrence of $\zeta^{\prime}(-1)$ and $\log\Gamma(1/3)$ in this generic example, simplicity of following derivatives might be surprising:

Theorem 5.6.

	$\displaystyle\omega_{A}^{\prime}(0)$	$\displaystyle=\log 2+\log\pi$
	$\displaystyle\omega_{B}^{\prime}(0)$	$\displaystyle=\frac{5}{4}\log 2+\frac{3}{2}\log\pi$
	$\displaystyle\omega_{G}^{\prime}(0)$	$\displaystyle=2\log 2-\frac{1}{2}\log 3+\frac{5}{2}\log\pi$

The author does not know a conceptual explanation for their simplicities, the terms $\zeta^{\prime}(-1)$ or $\log\Gamma(r)$ end up cancelling each other during the computation⁸⁸8For $\omega_{G}$ , one also needs relation $\Gamma(1/3)\Gamma(2/3)=2\pi/\sqrt{3}$ ., analogous miraculous cancellation also takes place when one uses the approach in [20]. The case of $\omega_{A}$ were also proved in [2], [3] and [17].

6. Number of representations for Lie algebra $G_{2}$

Let $\mathfrak{g}$ be a finite dimensional simple Lie algebra over $\mathbb{C}$ , denote $r_{\mathfrak{g}}(n)$ to be number of non-isomorphic representation of $\mathfrak{g}$ with dimension $n$ . We abuse notation to write $r_{\mathfrak{g}}(n)$ as $r_{\Phi}(n)$ if $\Phi$ is corresponding root system. For $\Phi=A_{2},B_{2},G_{2}$ ,

	$\displaystyle\sum_{n\geq 0}r_{A_{2}}(n)q^{n}$	$\displaystyle=\prod_{n,m\geq 1}(1-q^{mn(m+n)/2})^{-1};$
	$\displaystyle\sum_{n\geq 0}r_{B_{2}}(n)q^{n}$	$\displaystyle=\prod_{n,m\geq 1}(1-q^{mn(m+n)(m+2n)/6})^{-1};$
	$\displaystyle\sum_{n\geq 0}r_{G_{2}}(n)q^{n}$	$\displaystyle=\prod_{n,m\geq 1}(1-q^{mn(m+n)(m+2n)(m+3n)(2m+3n)/120})^{-1}.$

Circle method, adapted in [6] and [4], can be used to find asymptotic of $r_{\mathfrak{g}}(n)$ provided we have enough information on poles location and certain special values of $\zeta_{\mathfrak{g}}(s)$ .

Asymptotic of $r_{A_{2}}(n),r_{B_{2}}(n)$ has been found previously ([19], [5], [4]):

	$\displaystyle r_{A_{2}}(n)$	$\displaystyle\sim\frac{C_{1}}{n^{3/5}}\exp\left(A_{1}n^{2/5}+A_{2}n^{3/10}+A_{3}n^{1/5}+A_{4}n^{1/10}\right)$
	$\displaystyle r_{B_{2}}(n)$	$\displaystyle\sim\frac{C_{2}}{n^{7/12}}\exp\left(B_{1}n^{1/3}+B_{2}n^{2/9}+B_{3}n^{1/9}+B_{4}\right)$

for explicit constants $C_{1},C_{2},A_{i},B_{i}$ . Our gathered information on $\zeta_{G_{2}}(s)$ enables us to do the same for $G_{2}$ by applying a formula given in Theorem 4.4 of [4]. Some technical hypotheses have to be satisfied (e.g. poles being simple and moderate growth along imaginary direction), they are met for $\zeta_{A_{2}}(s),\zeta_{B_{2}}(s),\zeta_{G_{2}}(s)$ (Theorems 1.8, 4.1).

For $\zeta_{G_{2}}(s)$ , these hypotheses have been checked in Rutard [20], and we have

r_{G_{2}}(n)\sim\frac{C_{3}}{n^{9/16}}\exp\left(G_{1}n^{1/4}+G_{2}n^{3/20}+G_{3}n^{1/20}\right),\qquad n\to\infty,

for some constants $C_{3},G_{1},G_{2},G_{3}$ . However, in [20], residues at $s=1/5,1/3$ remains unevaluated, plugging in these residues, which we figured out in this article, we obtain values of these constants:

Theorem 6.1.

r_{G_{2}}(n)\sim\frac{C_{3}}{n^{9/16}}\exp\left(G_{1}n^{1/4}+G_{2}n^{3/20}+G_{3}n^{1/20}\right),\qquad n\to\infty

with

C_{3}=2^{79/48}3^{9/32}5^{7/16}\pi^{31/16}\zeta\left(\frac{4}{3}\right)^{1/16}\Gamma\left(\frac{1}{3}\right)^{1/4},\qquad G_{1}=\frac{5^{1/4}\Gamma\left(\frac{1}{3}\right)^{3}}{3^{13/8}}\left(\frac{2\zeta(4/3)}{\pi}\right)^{3/4},

G_{2}=\frac{2^{13/20}3^{1/8}(1+3^{2/5})\pi^{3/20}\zeta(\frac{1}{5})\zeta(\frac{6}{5})\Gamma\left(\frac{1}{5}\right)}{5^{17/20}\zeta(\frac{4}{3})^{3/20}\Gamma\left(\frac{1}{3}\right)^{3/5}},\qquad G_{3}=\frac{3^{23/8}(1+3^{2/5})^{2}\zeta\left(\frac{1}{5}\right)^{2}\zeta\left(\frac{6}{5}\right)^{2}\Gamma\left(\frac{1}{5}\right)^{2}}{2^{9/20}5^{79/20}\Gamma\left(\frac{1}{3}\right)^{21/5}}\left(\frac{\pi}{\zeta\left(\frac{4}{3}\right)}\right)^{21/20}

7. Appendix: evaluation of a definite integral

Here we prove the evaluation

\int_{0}^{\infty}\frac{dx}{\sqrt[3]{x(1+x)(1+2x)(1+3x)(2+3x)}}=\frac{1}{2^{5/3}3^{1/2}\pi}\Gamma\left(\frac{1}{3}\right)^{3},

which is related to residue of $\omega_{G}(s)$ at the point $s=1/3$ . We remark this evaluation seems highly non-trivial: algebraic curve $y^{3}=x(1+x)(1+2x)(1+3x)(2+3x)$ being genus $4$ . The fact it can be expressed in terms of gamma function at all indicates deep properties of $\omega_{G}(s)$ .

We shall give an indirect proof, it is following Lemma specialized at $u=-1/6$ .

Lemma 7.1.

For $-1/2<\Re(u)<1/6$ , we have

\int_{0}^{\infty}(x(1+x)(1+2x))^{2u}((1+3x)(2+3x))^{-2u-2/3}dx=\frac{3^{-3u-3/2}\Gamma\left(\frac{1}{6}-u\right)\Gamma\left(u+\frac{1}{2}\right)}{2^{2/3}\Gamma\left(\frac{2}{3}\right)}.

Proof.

The key idea is to do a creative telescoping. Let $F(x,u)$ be the integrand,

I_{1}(u):=\int_{0}^{\infty}F(x,u)dx,\qquad I_{2}(u):=\int_{C}F(x,u)dx=(e^{4\pi iu}-1)I_{1}(u),

here $C$ is the contour wrapping around positive real axis mentioned in the proof of Theorem 1.1. Note that $I_{1}(u)$ is analytic on $-1/2<\Re(u)<1/6$ and $I_{2}(u)$ on $\Re(u)<1/6$ . Creative telescoping produces the following relation

(2u+1)F(x,u)+9(6u+5)F(x,u+1)=\frac{\partial}{\partial x}\left(\frac{2x(x+1)(2x+1)\left(3x^{2}+3x+1\right)}{(3x+1)(3x+2)}F(x,u)\right).

Integrate with respect to $x$ along $C$ , fundamental theorem of calculus says integral on the RHS equals $0$ , so we have

(2u+1)I_{2}(u)+9(6u+5)I_{2}(u+1)=0.

This recurrence amounts to say that

(7.1)

J(u)=\frac{3^{3u}}{\Gamma\left(\frac{1}{6}-u\right)\Gamma\left(\frac{1}{2}+u\right)}(1-e^{4\pi iu})I_{1}(u)

is a function of period $1$ : $J(u+1)=J(u)$ .

Since $(1-e^{4\pi iu})I_{1}(u)=I_{2}(u)$ is analytic on $\Re(u)<1/6$ , $J(u)$ is entire. Note that for any $\varepsilon>0$ , $y$ a large positive real number, $|\Gamma(a\pm iy)|\geq O(e^{(-\pi/2-\varepsilon)|y|})$ and $|I_{1}(\pm iy)|=O(e^{\varepsilon|y|})$ . We have

J(iy)=O(e^{(\pi+\varepsilon)y}),\qquad J(-iy)=O(e^{(5\pi+\varepsilon)y}),\qquad y\to+\infty.

These two growth conditions, together with $J(u+1)=J(u)$ imply $J(u)=a+be^{2\pi iu}+ce^{4\pi iu}$ for some $a,b,c\in\mathbb{C}$ .

In (7.1), set $u=0$ gives $J(0)=0\iff a+b+c=0$ . Also, since $I_{1}(u)$ has a simple pole at $u=-1/2$ , and $\frac{1-e^{4\pi iu}}{\Gamma(\frac{1}{2}+u)}$ has double zero at $u=-1/2$ , $J(-1/2)=0\iff a-b+c=0$ . Therefore $J(u)=a(1-e^{4\pi iu})$ . So

I_{1}(u)=3^{-3u}\Gamma\left(\frac{1}{6}-u\right)\Gamma\left(\frac{1}{2}+u\right)a

for some constant $a$ independent of $u$ . Thus it suffices to evaluate $I_{1}(u)$ at any point, we will find $I_{1}(-1/3)$ , from equation (1.4),

I_{1}(-1/3)=\int_{0}^{\infty}(x(1+x)(1+2x))^{-2/3}dx=3\times{{}_{2}}F_{1}(1/3,2/3;4/3;-1),

and its evaluation follows from Kummer’s formula

{{}_{2}}F_{1}(a,b;1+a-b;-1)=\frac{\Gamma(1+a-b)\Gamma(1+a/2)}{\Gamma(1+a)\Gamma(1+a/2-b)}.

∎

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On single-variable Witten zeta functions of rank 2 root systems

Abstract.

Key words and phrases:

2020 Mathematics Subject Classification:

Introduction

Theorem 0.1.

Acknowledgment

1. An integration kernel

Theorem 1.1.

Proof.

Remark 1.2.

Proposition 1.3.

Proof.

Proposition 1.4.

Proof.

Proposition 1.5.

Proof.

Proposition 1.6.

Proof.

Lemma 1.7.

Proof.

Lemma 1.8.

Proof.

2. Meromorphic continuation and residue at abscissa of convergence

Proposition 2.1.

Proof.

Theorem 2.2.

Theorem 2.3.

Proof.

Proposition 2.4.

Proof.

3. Values at non-positive integers

Theorem 3.1.

Corollary 3.2.

Proof.

Example 3.3.

Example 3.4.

Example 3.5.

Theorem 3.6 ([1]).

3.1. A connection to Eisenstein series

Conjecture 3.7.

4. Poles of ζf​(s)\zeta_{f}(s)

Theorem 4.1.

Proof.

Corollary 4.2.

Proof.

Example 4.3.

Proposition 4.4.

Proof.

Remark 4.5.

Theorem 4.6.

Proof.

Conjecture 4.7.

5. Derivative at s=0s=0

Proposition 5.1.

Lemma 5.2.

Proof.

Lemma 5.3.

Proof.

Lemma 5.4.

Proof.

Example 5.5.

Theorem 5.6.

6. Number of representations for Lie algebra G2G_{2}

Theorem 6.1.

7. Appendix: evaluation of a definite integral

Lemma 7.1.

Proof.

References

4. Poles of $\zeta_{f}(s)$

5. Derivative at $s=0$

6. Number of representations for Lie algebra $G_{2}$