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 $a\neq b$?

  • $\begingroup$ Are you looking for a closed form for the coefficients? $\endgroup$ Feb 22, 2015 at 20:47
  • $\begingroup$ 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 $0 < |s| < \delta$. $\endgroup$ Feb 22, 2015 at 23:35
  • $\begingroup$ @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 $(1)$. $\endgroup$ Feb 23, 2015 at 9:20
  • $\begingroup$ @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 ? $\endgroup$
    – Lucian
    Feb 18, 2017 at 16:26
  • $\begingroup$ @Lucian Even if I've some difficulty to grasp your definition, I would say keep on working on your intuition. $\endgroup$ Feb 19, 2017 at 21:51

2 Answers 2


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 $0 < |s| < 1$

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

  • 1
    $\begingroup$ 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 $\gamma_1(a,b)$. $\endgroup$ Feb 24, 2015 at 10:47

We may follow Euler's lead.

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 $s \neq 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.