# Asymptotic expansion of $\zeta(s \mid a,b)= \sum_{n=1}^{\infty} \frac{1}{(n+a)^{s}(n+b)}$

I'm interested in an asymptotic expansion of the following Riemann zeta-type function \begin{align} \displaystyle \zeta(s \mid a,b) := \sum_{n=1}^{\infty} \frac{1}{(n+a)^{s}(n+b)}, \quad \Re a >-1, \, \Re b >-1, \, s>0, \tag1 \end{align} as $s \to 0^+$.

The case $a=b$ in $(1)$ leads to the Riemann Hurwitz zeta function with the Laurent expansion near $0$:
\begin{align} \sum_{n=1}^{\infty} \frac{1}{(n+a)^{s+1}} = \frac{1}{s}+\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!} \gamma_{k}(a+1)s^{k}, \quad s>0, \tag2 \end{align} where $\displaystyle \gamma_{k}(a+1)$ are the generalized Stieltjes constants with $\displaystyle \gamma_{0}(a+1)=-\Gamma'(a+1)/\Gamma(a+1)$.

What is an asymptotic expansion, as $s \to 0^+$, of $\displaystyle \zeta(s \mid a,b)$ when $a\neq b$?

• Are you looking for a closed form for the coefficients? Feb 22 '15 at 20:47
• Such a series exists, at least. There are coefficients $c_k(a,b)$ such that, for some $\delta > 0$, $$\zeta(s | a,b) = \frac{1}{s} + \sum_{k=0}^{\infty} c_k(a,b) s^k$$ for all $0 < s < \delta$. The series here can be used to analytically continue $\zeta(s|a,b)$ to the annulus $0 < |s| < \delta$. Feb 22 '15 at 23:35
• @AntonioVargas Yes, I have the same conjecture, I've defined some Laurent-Stieltjes type constants by $$\zeta(s \mid a,b) = \frac{1}{s} + \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!}\gamma_k(a,b) s^k \tag1$$ near $0$. Then we may extend $ζ(⋅∣a,b)$ to a meromorphic function on $\mathbb{C}$, as is the classic Hurwitz zeta function. The point is to prove $(1)$. Maybe the Euler–Maclaurin formula could be an interesting tool to prove $(1)$. Feb 23 '15 at 9:20
• @OlivierOloa: I've noticed your fondness for various generalizations of $\zeta$ functions, based on infinite series. In your opinion, would a generalization of the same functions, but based on infinite products rather than infinite series, hold any merit ? Feb 18 '17 at 16:26
• @Lucian Even if I've some difficulty to grasp your definition, I would say keep on working on your intuition. Feb 19 '17 at 21:51

For $s > 0$ we have

$$\sum_{n=1}^{\infty} \frac{1}{(n+a)^s(n+b)} - \sum_{n=1}^{\infty} \frac{1}{(n+a)^{s+1}} = \sum_{n=1}^{\infty} \frac{a-b}{(n+a)^{s+1}(n+b)} =: g(s).$$

The series on the right-hand side converges and is analytic on $\operatorname{Re} s > -1$, so the difference on the left-hand side can be analytically continued to this region. Consequently, the analytic continuation $\zeta(s \mid a,b)$ of your sum satisfies

$$\zeta(s \mid a,b) = \zeta(s+1,a+1) + g(s)$$

for $\operatorname{Re} s > -1$, $s \neq 0$ and thus has a simple pole at $s=0$ with residue $1$.

For $|s| < 1$ we have

$$g(s) = (a-b)\sum_{k=0}^\infty \left( \sum_{n=1}^\infty \frac{[\log(n+a)]^k}{(n+a)(n+b)} \right) \frac{(-s)^k}{k!},$$

so for $0 < |s| < 1$

$$\zeta(s \mid a,b) = \frac{1}{s} + \sum_{k=0}^{\infty} \left( \gamma_{k}(a+1) + (a-b)\sum_{n=1}^\infty \frac{[\log(n+a)]^k}{(n+a)(n+b)} \right) \frac{(-s)^k}{k!}.$$

• Thank you Antonio. +1 and accepted. There is a typo in your result, the factor $a-b$ should appear in your expansion. Observe that, using the notations in my previous comment, we get $$\gamma_0(a,b)=-\psi(b+1)$$ due to the classic evaluation $$\sum_{n=1}^\infty \frac{a-b}{(n+a)(n+b)} =\psi(a+1)-\psi(b+1).$$ I'm looking for an 'interesting' evaluation for $\gamma_1(a,b)$. Feb 24 '15 at 10:47

Euler was the first to define a constant of the form (1734) \begin{align} \gamma & = \lim_{N\to\infty}\left(1+\frac12+\frac13+\cdots+\frac1N-\log N\right)=0.577215\ldots. \tag1 \end{align}
Later Stieltjes found (1885) that the Laurent series expansion around $1$ of the Riemann zeta function, $$\zeta(1+s) = \frac{1}{s} + \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!}\gamma_k s^k, \quad s \neq 0,\tag2$$ is such that the scaled coefficients of the regular part of the expansion, now called the Stieltjes constants, are given by \begin{align} \gamma_k& = \lim_{N\to \infty}\left(\sum_{n=1}^N \frac{\log^k n}{n}-\frac{\log^{k+1} \!N}{k+1}\right). \end{align} \tag3
In the same vein, J.B. Wilton (1927) and B. Berndt (1972) established that the Laurent series expansion in the neighbourhood of $1$ of the Hurwitz zeta function \begin{align} \zeta(1+s,a) = \frac1s+\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!} \gamma_{k}(a)\:s^{k}, \quad \Re a>0, \,s\neq 0, \tag4 \end{align} is such that the scaled coefficients of the regular part of the expansion, called the generalized Stieltjes constants, are given by \begin{align} \gamma_k(a)& = \lim_{N\to \infty}\left(\sum_{n=0}^N \frac{\log^k (n+a)}{n+a}-\frac{\log^{k+1} (N+a)}{k+1}\right), \quad \Re a>0. \end{align} \tag5 Do we have a form resembling the original definition of Euler's constant for our coefficients?
Theorem. Let $a,b$ be complex numbers such that $\Re a >-1, \, \Re b >-1$. Consider the Riemann zeta type function initially defined as \begin{align} \zeta(s\mid a,b) := \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)}, \quad \Re s>0. \tag6 \end{align} Then the meromorphic extension of $\displaystyle \zeta(\cdot\mid a,b)$ admits the following Laurent series expansion around $0$, $$\zeta(s \mid a,b) = \frac{1}{s} + \sum_{k=0}^{+\infty} \frac{(-1)^{k}}{k!}\gamma_k(a,b) s^k, \quad s \neq 0,\tag7$$ and \begin{align} \gamma_k(a,b)& = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log^k (n+a)}{n+b}-\frac{\log^{k+1} \!N}{k+1}\right). \end{align} \tag8
To see this, let $a,b$ be complex numbers such that $\Re a >-1, \, \Re b >-1$.
We first assume $\Re s>0$. Observing that, for each $n \geq 1$, $$\left|\sum_{k=0}^{\infty}\frac{\log^k(n+a)}{n+b}\frac{(-1)^{k}}{k!}s^k\right| \leq \sum_{k=0}^{\infty}\left|\frac{\log^k(n+a)}{n+b}\right|\frac{|s|^k }{k!}<\infty$$ and that $$\sum_{n=1}^{\infty}\left|\sum_{k=0}^{\infty}\frac{\log^k(n+a)}{n+b}\frac{(-1)^{k}}{k!}s^k\right|=\sum_{n=1}^{\infty}\left|\frac1{(n+a)^s(n+b)}\right| = \sum_{n=1}^{\infty}\frac1{|n+a|^{\Re s}|n+b|}<\infty,$$ we obtain \begin{align} &\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!}\lim_{N\to+\infty}\left(\sum_{n=1}^N\frac{\log^k(n+a)}{n+b}-\frac{\log^{k+1} \!N}{k+1}\right) s^k \\\\ &= \lim_{N\to+\infty}\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!}\left(\sum_{n=1}^N\frac{\log^k(n+a)}{n+b}-\frac{\log^{k+1} \!N}{k+1}\right) s^k \\\\ &=\lim_{N\to+\infty}\sum_{k=0}^{\infty}\left(\sum_{n=1}^N\frac{(-1)^{k}}{k!}\frac{\log^k(n+a)}{n+b}s^k -\frac{(-1)^{k}}{k!}\frac{\log^{k+1} \!N}{k+1}s^k\right) \\\\ &=\lim_{N\to+\infty}\left(\sum_{n=1}^N\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!}\frac{\log^k(n+a)}{n+b}s^k -\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!}\frac{\log^{k+1} \!N}{k+1}s^k\right) \\\\ &=\lim_{N\to+\infty}\left(\sum_{n=1}^N\frac1{(n+a)^s(n+b)} +\frac1{N^s}-\frac1s\right) \\\\ &=\zeta(s \mid a,b)-\frac1{s}. \end{align} Then we extend the preceding identity by meromorphic continuation to all $s \neq 0$.