On the asymptotic expansion of the q-dilogarithm

Abstract

In this work, we study some asymptotic expansion of the q-dilogarithm at $q=1$ and $q=0$ by using the Mellin transform and an adequate decomposition allowed by the Lerch functional equation.

1 Introduction

Euler’s dilogarithm is defined by [1]

$$ \mathit{Li}_{2}(z)=\sum_{n=1}^{\infty}\frac{z^{n}}{n^{2}}, \quad |z|< 1. $$

(1.1)

In [2], Kirillov defines the following q-analog of the dilogarithm $\mathit{Li}_{2}(z)$:

$$ \mathit{Li}_{2}(z;q)=\sum_{n=1}^{\infty}\frac{z^{n}}{n(1-q^{n})}, \quad |z|< 1, 0 < q < 1, $$

(1.2)

and he observes the following remarkable formula ([2], Section 2.5, Lemma 8):

$$ \sum_{n=0}^{\infty}\frac{z^{n}}{(q,q)_{n}}=\exp \bigl(\mathit{Li}_{2}(z,q)\bigr), \quad |z|< 1, |q|< 1, $$

(1.3)

where

$$ (q,q)_{0}=1, \qquad (q,q)_{n}=\prod _{k=0}^{n-1}\bigl(1-q^{k}\bigr), \quad n=1, 2,\ldots . $$

(1.4)

It seems a precise formulation of (1.3) going back to Ramanujan (see [3], Chapter 27, Entry 6) is given an asymptotic series for $\mathit{Li}_{2}(z; q)$ and Hardy and Littlewood [4] proved that for $|q|=1$, the identity holds inside the radius of convergence of either series.

Let $\omega=e^{zx+2\mathrm{i}\theta}$ with $\operatorname{Re} (z)>1$, $x>0$, and $0<\theta<1$. The main result of this work is the following complete asymptotic expansion of the q-dilogarithm function $\mathit{Li}_{2}(\omega;e^{-x})$ at ${x\rightarrow0}$:

$$\begin{aligned} \mathit{Li}_{2}\bigl(\omega,e^{-x}\bigr) \sim& \mathit{Ci}_{2}( \theta) \frac{1}{x}+\biggl(\frac{1}{2}-z\biggr)\mathit{Ci}_{1}( \theta)+\sum_{n=1} ^{\infty}\frac{(-1)^{n+1}}{(n+1)(n+1)!} \\ &{}\times B_{n+1} (z)B_{n+1}\bigl(1,e^{2\mathrm{i}\pi\theta} \bigr)x^{n} \quad \text{as } x\rightarrow0 \end{aligned}$$

(1.5)

and

$$ \mathit{Li}_{2}\bigl(\omega,e^{-x}\bigr) \sim\frac{4\gamma}{\pi}B_{2}( \theta)\frac{\mathrm{i}}{x}+4\sum_{n=1}^{\infty} \mathrm{i}^{n}\frac{\psi^{(n-1)}(z)B_{n+1} (\theta)}{(n+1)!}\biggl(\frac{2\pi}{x} \biggr)^{n},\quad x\rightarrow\infty. $$

(1.6)

In Section 2.5, Corollary 10 of [2], Kirillov and Ueno and Nishizawa derived the asymptotic expansion (1.5) by using the Euler-Maclaurin summation formula; see also [5], an integral representation of Barnes type for the q-dilogarithm. Second, we use the Lerch functional equation to decompose the integrand and to apply the Cauchy theorem.

2 q-Dilogarithm

The polylogarithm is defined in the unit disk by the absolutely convergent series [1]

$$ \mathit{Li}_{s}(z)=\sum_{n=1}^{\infty}\frac{z^{n}}{n^{s}},\quad |z|< 1. $$

(2.1)

Several functional identities satisfied by the polylogarithm are available in the literature (see [6]). For $n=2, \ldots$ , the function $\mathit{Li}_{n}(z)$ can also be represented as

$$ \mathit{Li}_{n}(z)= \int_{0}^{z}\frac{\mathit{Li}_{n-1}(t)}{t}\, dt, \quad n\in \mathbb {N}, \qquad \mathit{Li}_{1}(z)=-\log(1-z)= \int_{0}^{z}\frac{dt}{1-t}, $$

(2.2)

which is valid for all z in the cut plane $\mathbb{C}\setminus [1,\infty)$.

The notation $F(\theta,s)$ is used for the polylogarithm $\mathit{Li}_{s}(e^{2\mathrm{i}n\pi \theta})$ with θ real, called the periodic zeta function (see [7], Section 25.13) and is given by the Dirichlet series

$$ F(\theta,s )=\sum_{n=1}^{\infty}\frac{e^{2\mathrm{i}n\pi \theta}}{n^{s}}, \quad \theta\in\mathbb{R}, $$

(2.3)

it converges for $\operatorname{Re} s>1$ if $\theta\in\mathbb{Z}$, and for $\operatorname{Re} s>0$ if $\theta\in\mathbb{R}/\mathbb{Z}$. This function may be expressed in terms of the Clausen functions $\mathit{Ci}_{s}(\theta)$ and $\mathit{Si}_{s}(\theta)$, and vice versa (see [1], Section 27.8):

$$ \mathit{Li}_{s}\bigl(e^{\pm i \theta}\bigr) = \mathit{Ci}_{s}( \theta) \pm \mathrm{i} \mathit{Si}_{s}(\theta) . $$

(2.4)

In [8], Koornwinder defines the q-analog of the logarithm function

$$ -\log(1-z)=\sum_{n=1}^{\infty}\frac{z^{n}}{n},\quad |z|< 1, $$

as follows:

$$ \log_{q}(z)=\sum_{n=1}^{\infty}\frac{z^{n}}{1-q^{n}},\quad |z|< 1, 0< q< 1. $$

(2.5)

Recall that the q-analog of the ordinary integral (called Jackson’s integral) is defined by

$$ \int_{0}^{z}f(t) \, d_{q}t=(1-q)z\sum _{n=0}^{\infty}f\bigl(zq^{n} \bigr)q^{n}. $$

(2.6)

One can recover the ordinary Riemann integral as the limit of the Jackson integral for $q \uparrow1$.

Lemma 2.1

The function $\log_{q}(z)$ has the following q-integral representation:

$$ (1-q)\log_{q}(z)= \int_{0}^{z}\frac{1}{1-t}\, d_{q}t, \quad |z|< 1. $$

(2.7)

Moreover, it can be extended to any analytic function on $\mathbb{C}-\{ q^{-n}, n\in\mathbb{N}_{0}\}$.

Proof

Assume that $|z|<1$, then from (2.5) we have

$$\begin{aligned} (1-q)\log_{q}(z) &=(1-q)\sum_{n=1}^{\infty}\sum_{m=0}^{\infty}z^{n} q^{nm} \\ &=(1-q)z\sum_{m=0}^{\infty}q^{m}\sum _{n=0}^{\infty}z^{n} q^{nm} \\ &=(1-q)z\sum_{m=0}^{\infty}\frac{q^{m}}{1-zq^{m}}. \end{aligned}$$

The inversion of the order of summation is permitted, since the double series converges absolutely when $|z|<1$.

Let K be a compact subset of $\mathbb{C}-\{q^{-n}, n\in\mathbb {N}_{0}\}$. There exists $N\in\mathbb{N}$ such that, for all $z\in K$, $|q^{N}z|< q$. Then for $n\geq N$ we have

$$ \biggl\vert \frac{q^{m}}{1-zq^{m}}\biggr\vert \leq \frac{q^{m}}{1-q}. $$

(2.8)

Hence, the series $\sum_{m=N}^{\infty}\frac{q^{m}}{1-zq^{m}}$ converges uniformly in K. □

The q-dilogarithm (1.2) is related to Koornwinder’s q-logarithm (2.5) by

$$ \mathit{Li}_{2}(z,q)= \int_{0}^{z}\frac{\log_{q}(t)}{t}\, dt. $$

(2.9)

It follows that, for $n \geq2$, we can also define

$$ \mathit{Li}_{n}(z,q)= \int_{0}^{z}\frac{\mathit{Li}_{n-1}(t,q)}{t} \, dt. $$

(2.10)

This integral formula proves by induction that $\mathit{Li}_{n}(z,q)$ has an analytic continuation on $\mathbb {C}-[1, \infty)$. Moreover, for $|z|<1$, we have

$$ \mathit{Li}_{n}(z,q)=\sum_{k=1}^{\infty}\frac{z^{k}}{k^{n}(1-q^{k})}. $$

This converges absolutely for $|z| < 1$ and defines a germ of a holomorphic function in the neighborhood of the origin. Note that

$$\begin{aligned}& \lim_{q\uparrow1}(1-q)\mathit{Li}_{2}\bigl((1-q)z,q \bigr)=\mathit{Li}_{2}(z), \\& \lim_{q\downarrow0}(1-q)\mathit{Li}_{2}(z,q)=-\operatorname{Log}(1-z),\quad |z|< 1. \end{aligned}$$

Let $\omega=e^{-zx+2\mathrm{i}\pi\theta}$, $\theta\in\mathbb{R}$, and $\operatorname{Re} z>0$, we define

$$\begin{aligned}& \mathit{Ci}_{2}(\omega,q)=\sum_{n=1}^{\infty}\frac{e^{-zx}\cos(2\pi n\theta)}{n(1-q^{n})}, \end{aligned}$$

(2.11)

$$\begin{aligned}& \mathit{Si}_{2}(\omega,q)= \sum_{n=1}^{\infty}\frac{e^{-zx}\sin(2\pi n \theta)}{n(1-q^{n})}. \end{aligned}$$

(2.12)

Note that these functions can be considered as q-analogs of the Clausen functions (2.4) and are related to the q-dilogarithm by

$$ \mathit{Li}_{2}(\omega,q)=\mathit{Ci}_{2}(\omega,q)+\mathrm{i} \mathit{Si}_{2}(\omega,q). $$

(2.13)

Now, we will use the Mellin transform method to obtain the integral representation

$$ \mathit{Li}_{2}\bigl(\omega,e^{-x}\bigr)=\frac{1}{2\mathrm{i}\pi} \int_{c-\mathrm{i}\infty }^{c+\mathrm{i}\infty} \zeta(s,z)F(\theta,s) \Gamma(s)x^{-s} \, ds, \quad c>1, $$

(2.14)

where

$$\omega=e^{-zx+2\mathrm{i}\pi\theta},\qquad x>0, \qquad \operatorname{Re} z>1,\qquad 0< \theta< 1. $$

Recall that the Mellin transform for a locally integrable function $f(x)$ on $(0,\infty)$ is defined by

$$ M(f,s)= \int_{0}^{\infty}f(x)x^{s-1}\, dx, $$

(2.15)

which converges absolutely and defines an analytic function in the strip

$$a< \operatorname{Re} s< b, $$

where a and b are real constants (with $a< b$) such that, for $\varepsilon>0$,

$$ f(x)= \textstyle\begin{cases} \mathcal{O}(x^{-a-\varepsilon}) & \text{as } x\rightarrow 0^{+}, \\ \mathcal{O}(x^{-b-\varepsilon}) & \text{as } x\rightarrow +\infty. \end{cases} $$

(2.16)

The inversion formula reads

$$ f(x)=\frac{1}{2\mathrm{i}\pi} \int_{c-\mathrm{i}\infty}^{c+\mathrm{i}\infty} M(f,s)x^{-s} ds, $$

(2.17)

where c satisfies $a< c< b$. Equation (2.17) is valid at all points $x\geq0$ where $f(x)$ is continuous.

We first compute the Mellin transform $M(\psi_{n}(x),s)$, where

$$ \psi_{n}(x)=\frac{e^{-nzx}}{n(1-e^{-nx})}, \quad x>0, \operatorname{Re} z>1, n\in\mathbb{N}. $$

(2.18)

Since

$$\begin{aligned}& \psi_{n}(x)\sim\frac{1}{nx}, \quad x\rightarrow0^{+}, \end{aligned}$$

(2.19)

$$\begin{aligned}& \psi_{n}(x)\sim\frac{1}{n}e^{-nx(z-1)}, \quad x\rightarrow+ \infty. \end{aligned}$$

(2.20)

We concluded that $M(\psi_{n}(x),s)$ is defined in the half-plane $\operatorname{Re} s>0$. That is, the constants a and b satisfy $a=1$ and $b=+\infty$, which values can be used for all $n\geq1$ and $\operatorname{Re} z>1$. The Mellin transform of $\psi_{n}(x)$ can be obtained from the following integral representation of the Hurwitz zeta function $\zeta (s, z)$:

$$ \zeta(s,z)=\frac{1}{\Gamma(s)} \int_{0}^{\infty}\frac {e^{-zx}}{1-e^{-x}}x^{s-1} \, dx \quad \bigl(\operatorname{Re} s>0, \bigl\vert \arg(1-z)\bigr\vert < \pi; \operatorname{Re} s>1, z=1\bigr). $$

(2.21)

Note that $\zeta(s, z)$ is expressed also by the series

$$ \zeta(s,z)=\sum_{k=1}^{\infty}\frac{1}{(z+k)^{s}}, \quad \operatorname{Re} s>1, z\neq-1, -2, \ldots . $$

(2.22)

For the other values of z, $\zeta(s,z)$ is defined by analytic continuation. It has a meromorphic continuation in the s-plane, its only singularity in $\mathbb{C}$ being a simple pole at $s=1$,

$$ \zeta(s,z)=\frac{1}{s-1}-\psi(z)+ \mathcal{O}(s-1). $$

(2.23)

Applying the Mellin inversion theorem to the integral (2.21), we then find

$$ \psi_{n}(x)=\frac{1}{2\mathrm{i}\pi} \int_{c-\mathrm{i}\infty}^{c+\mathrm{i}\infty} \zeta(s,z)\Gamma(s) (nx)^{-s}\, ds. $$

(2.24)

We use the Stirling formula, which shows that, for finite σ,

$$ \Gamma(\sigma+\mathrm{i} t)=\mathcal{O}\bigl(|t|^{\sigma-1}e^{-\frac {1}{2}\pi|t|} \bigr)\quad \bigl(\vert t\vert \rightarrow+\infty\bigr) $$

(2.25)

and the well-known behavior of $\zeta(s,z)$ (see [9])

$$ \zeta(s,z)=\mathcal{O}\bigl(|t|^{\tau(\sigma)}\log|t|\bigr), $$

(2.26)

where

$$ \tau(\sigma)= \textstyle\begin{cases} \frac{1}{2}-\sigma, &\sigma\leq0, \\ \frac{1}{2},& 0\leq\sigma\leq\frac{1}{2}, \\ 1-\sigma,& \frac{1}{2}\leq\sigma\leq1, \\ 0,& \sigma\geq1. \end{cases} $$

Then we obtain the following majorization of the modulus of the integrand in (2.24):

$$ \mathcal{O}\bigl(|t|^{\tau(\sigma)+\sigma-1}\log|t|\bigr). $$

(2.27)

Consequently, the integral (2.24) converges absolutely in the whole vertical strip of the half-plane $\operatorname{Re} s>0$. Then we replace x by nx, where n is a positive integer, and sum over n, and we then obtain

$$ \mathit{Li}_{2}\bigl(\omega,e^{-x}\bigr) =\frac{1}{2\mathrm{i}\pi} \int_{c-\mathrm{i}\infty}^{c+\mathrm{i}\infty} \zeta(s,z)F(\theta,s+1) \Gamma(s)x^{-s}\, ds, \quad c>1, $$

(2.28)

where

$$\omega=e^{-zx+2\mathrm{i}\pi\theta},\qquad x>0,\qquad \operatorname{Re} z>1, \qquad 0< \theta< 1. $$

3 Asymptotic at $q=1$

The integral (2.28) will be used to derive asymptotic expansions of the q-dilogarithm. The contour of integration is moved at first to the left to obtain an asymptotic expansion at $q=1$ and then to the right to get an asymptotic expansion at $q=0$.

Let us consider the function

$$ g(s)=\zeta(s,z)F(\theta,s+1)\Gamma(s). $$

(3.1)

The periodic function zeta function $F(\theta,s)$ has an extension to an entire function in the s-plane (see [10]). Hence, the function $g(s)$ has a meromorphic continuation in the s-plane, its only singularity in $\mathbb{C}$ coincides with the pole of $\Gamma (s)$ and $\zeta(s,z)$ being a simple pole at $s=1, 0,-1,-2,\ldots$ .

Now we compute the residues of the poles. The special values at $s=-1, -2 \ldots$ of the periodic zeta function are reduced to the Apostol-Bernoulli polynomials (see [10]),

$$ F(\theta,-n )=-\frac{B_{n+1}(1,e^{2\mathrm{i}\pi\theta})}{n+1}. $$

(3.2)

We need also the following asymptotic expansions of $\Gamma(s)$ and $\zeta(s)$ at $s=0$:

$$\begin{aligned}& \Gamma(s)=\frac{1}{s}-\gamma+ \mathcal{O}\bigl(s^{2} \bigr), \end{aligned}$$

(3.3)

$$\begin{aligned}& \zeta(s)=\frac{1}{2}-z+s\log\frac{\Gamma(z)}{2\pi} + \mathcal{O} \bigl(s^{2}\bigr). \end{aligned}$$

(3.4)

Hence,

$$\begin{aligned}& \lim_{s\rightarrow1}(s-1) g(s)=\mathit{Li}_{2}\bigl(e^{2\mathrm{i}\pi\theta} \bigr), \\& \lim_{s\rightarrow-n}(s+n) g(s)=\frac{(-1)^{n}}{(n+1)(n+1)!}B_{n+1} (z)B_{n+1}\bigl(1,e^{2\mathrm{i}\pi\theta}\bigr). \end{aligned}$$

Here $B_{n}(z)$ is the Bernoulli polynomial (see [1]).

Let N be an integer and d real number such that $-N-1< d<-N$. We consider the integral taken round the rectangular contour with vertices at $d \pm \mathrm{i}A$ and $c \pm \mathrm{i}A$, so that the side in $\operatorname{Re}(s) < 0$ parallel to the imaginary axis passes midway between the poles $s=1, 0 -1, -2, \ldots, -N$. The contribution from the upper and lower sides $s = \sigma\pm\mathrm{i}A$ vanishes as $|A|\rightarrow+\infty$, since the modulus of the integrand is controlled by

$$ \mathcal{O}\bigl(|A|^{\tau(\sigma)+\sigma-1/2}\log |A| e^{-\frac{1}{2}\pi|A|}\bigr). $$

(3.5)

This follows from Stirling’s formula (2.25), the behavior $\zeta (s,z)$ being given by (2.26), and the following estimation:

$$\bigl\vert F(\theta,s+1)\bigr\vert \leq\zeta(\sigma+1)=\mathcal{O}(1),\quad \vert A\vert \rightarrow+\infty. $$

Displacement of the contour (2.28) to the left then yields

$$\begin{aligned} \mathit{Li}_{2}\bigl(\omega,e^{-x}\bigr) =& \mathit{Ci}_{2}(\theta) \frac{1}{x}+\biggl(\frac{1}{2}-z\biggr)\mathit{Ci}_{1}(\theta) \\ &{}+\sum _{n=1} ^{N}\frac{(-1)^{n+1}}{(n+1)(n+1)!}B_{n+1} (z)B_{n+1}\bigl(1,e^{2\mathrm{i}\pi\theta}\bigr)x^{n} + R_{N}(x), \end{aligned}$$

(3.6)

where the remainder integral $R_{N}(z)$ is given by

$$ R_{N}(x)=\frac{1}{2\mathrm{i}\pi} \int_{d-\mathrm{i}\infty}^{d+\mathrm{i}\infty} \zeta(s,z)F(\theta,s+1) \Gamma(s)x^{-s} \, ds, \quad x>0, \operatorname{Re} z>1.$$

(3.7)

From (3.5), we find

$$\bigl\vert R_{N}(x)\bigr\vert =\mathcal{O}\biggl( \frac{1}{x^{N+1}}\biggr). $$

4 Asymptotic at $q=0$

Recall that the periodic zeta function satisfies the functional equation (see [10])

$$\begin{aligned}& F(\theta,s)=\frac{\Gamma(1-s)}{(2\pi )^{1-s}} \bigl\{ e^{\frac{\pi\mathrm{i}(1-s)}{2}}\zeta(1-s, \theta)+ e^{\frac{\pi\mathrm{i}(s-1)}{2}}\zeta(1-s,1-\theta) \bigr\} \\& \quad (\operatorname{Re} s>1, 0< \theta< 1), \end{aligned}$$

(4.1)

first given by Lerch, whose proof follows the lines of the first Riemann proof of the functional equation for $\zeta(x)$.

It is well known that the asymptotic expansion near infinity via the Mellin transform is obtained by displacement of the contour of integration in the Mellin inversion formulas (2.16) to the right-hand side (see [11]). However, the integrand (2.28) has no poles in the half-plane $\operatorname{Re} s>1$. The periodic zeta function $F(\theta,s)$ has an analytic continuation to the whole s-space for $0<\theta<1$. Moreover, the poles of $\Gamma(1-s)$ in equation (4.1) at $s=-1, -2 \ldots$ are canceled by the zeros of the function

$$e^{\frac{\pi\mathrm{i}(1-s)}{2}}\zeta(1-s,\theta)+ e^{\frac{\pi\mathrm{i}(s-1)}{2}}\zeta(1-s,1-\theta). $$

On the other hand from (4.1) we easily obtain

$$ \Gamma (s)\bigl\{ F(\theta,s+1)+F(1-\theta,s+1)\bigr\} =-\frac{(2\pi)^{s+1}}{2s \sin\frac{\pi s}{2}}\bigl\{ \zeta(-s,\theta)+ \zeta(-s,1-\theta)\bigr\} , $$

(4.2)

where we are able to simplify (4.2) by the well-known reflection formulas

$$ \frac{\pi}{\sin\pi s}=\Gamma(s)\Gamma(1-s),\qquad \frac{\sin\pi s}{\pi}= \frac{2}{\pi}\sin\frac{\pi s}{2} \sin \frac{\pi(1- s)}{2}. $$

Proceeding similar to above we also obtain

$$ \Gamma(s)\bigl\{ F(\theta,s)-F(1-\theta,s)\bigr\} = \frac{(2\pi)^{s+1}}{2s \cos\frac{\pi(s)}{2}}\bigl\{ \zeta(-s,\theta)- \zeta(-s,1-\theta)\bigr\} . $$

(4.3)

Moreover, the integral representation (2.28) is valid for all $0<\theta<1$. So we can replace θ by $1-\theta$ in its integrand. Using the above decomposition (4.2) and (4.3), we then obtain

$$ \mathit{Ci}_{2}\bigl(\omega,e^{-x}\bigr)=-\frac{1}{2\mathrm{i}\pi} \int _{c-\mathrm{i} \infty}^{c+\mathrm{i}\infty}\frac{(2\pi)^{s+1}\zeta(s,z)}{2s\sin \frac{\pi s}{2}}\bigl\{ \zeta(-s, \theta)+ \zeta(-s,1-\theta)\bigr\} \frac{ds}{x^{s}} $$

(4.4)

and

$$ \mathit{Si}_{2}\bigl(\omega,e^{-x}\bigr)= \frac{1}{2\mathrm{i}\pi} \int_{c-\mathrm{i} \infty}^{c+\mathrm{i}\infty}\frac{(2\pi)^{s+1}\zeta(s,z)}{2s\cos \frac{\pi s}{2}}\bigl\{ \zeta(-s, \theta)- \zeta(-s,1-\theta)\bigr\} \frac{ds}{x^{s}}, $$

(4.5)

where $\omega=e^{-zx+2\mathrm{i}\pi\theta}$, $0< x$, $0<\theta<1$, $0<\operatorname{Re} z$, and $1< c<2$.

Note that the special values $\zeta(n,z)$ ($n\in\mathbb{N}_{0}$) are expressed in terms of the polygamma function $\psi(z)$,

$$ \zeta(n+1,z)=\frac{(-1)^{n+1}}{n!}\psi^{(n)}(z) , \quad z \neq 0, -1, -2, \ldots , $$

(4.6)

and $\zeta(-n,z)$ ($n\in\mathbb{N}$) is reduced to the Bernoulli polynomial

$$ \zeta(-n,z)=-\frac{B_{n+1}(z)}{n+1}. $$

(4.7)

Applying the identities for the Bernoulli polynomial

$$ B_{n}(1-\theta)=(-1)^{n}B_{n}(\theta), $$

we obtain

$$\begin{aligned}& \zeta(-n,\theta)+\zeta(-n,1-\theta)=\bigl((-1)^{n+1}-1\bigr) \frac {B_{n+1}(\theta)}{n+1}, \end{aligned}$$

(4.8)

$$\begin{aligned}& \zeta(-n,\theta)-\zeta(-n,1-\theta)=\bigl((-1)^{n}-1\bigr) \frac {B_{n+1}(\theta)}{n+1}. \end{aligned}$$

(4.9)

The integrand in (4.4) has a meromorphic continuation in the s-plane, its only singularity in the half-plane $\operatorname{Re} s>0$ coincides with the pole of $1/\sin{\frac{\pi s}{2}}$ being a simple pole at $s=2, 4,\ldots$ . Then by the Cauchy integral, we can shift the contour in (4.4) to the right, picking up the residues at $s=2, \ldots, 2N$ , with the result

$$ \mathit{Ci}_{2}\bigl(\omega,e^{-x}\bigr)=4\sum _{n=1}^{N}(-1)^{n}\frac{\psi^{(2n-1)}(z)B_{2n+1} (\theta)}{(2n+1)!} \biggl(\frac{2\pi}{x}\biggr)^{2n}+Q_{N}(x), $$

(4.10)

where

$$ Q_{N}(x)=-\frac{1}{2\mathrm{i}\pi} \int_{c+2N-\mathrm{i} \infty}^{c+2N+\mathrm{i}\infty}\frac{(2\pi)^{s+1}\zeta(s,z)}{2s\sin \frac{\pi s}{2}}\bigl\{ \zeta(-s, \theta)+ \zeta(-s,1-\theta)\bigr\} \frac{ds}{x^{s}}. $$

(4.11)

Using the following estimations in a vertical strip $s=\sigma+\mathrm{i}t$, $\sigma\neq0, \pm 1, \pm2,\ldots$ ,

$$ \frac{1}{\sin\frac{\pi s}{2}}= \mathcal{O}\bigl(|t|^{-1}e^{-\frac{\pi}{2}|t|} \bigr), $$

(4.12)

we obtain

$$ \bigl\vert Q_{N}(x)\bigr\vert =\mathcal{O}\biggl( \frac{1}{x^{2N+1}}\biggr). $$

(4.13)

Similarly,

$$ \mathit{Si}_{2}\bigl(\omega,e^{-x}\bigr)=\frac{4\gamma}{\pi}B_{2}( \theta)\frac {1}{x}+4\sum_{n=1}^{N}(-1)^{n} \frac{\psi^{(2n)}(z)B_{2n+2} (\theta)}{(2n+2)!}\biggl(\frac{2\pi}{x}\biggr)^{2n+1}+\mathcal{O} \biggl(\frac{1}{x^{2N+2}}\biggr). $$

(4.14)

Proposition 4.1

Let $\omega=e^{-zx+2\mathrm{i}\pi\theta}$, $x>0$, $\operatorname{Re} z>1$ and $0<\theta<1$. Then

$$ \mathit{Li}_{2}\bigl(\omega,e^{-x}\bigr) \sim\frac{4\gamma}{\pi}B_{2}( \theta)\frac{\mathrm{i}}{x}+4\sum_{n=1}^{\infty} \mathrm{i}^{n}\frac{\psi^{(n-1)}(z)B_{n+1} (\theta)}{(n+1)!}\biggl(\frac{2\pi}{x} \biggr)^{n}, \quad x\rightarrow\infty. $$

(4.15)