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 $s \neq 0$.

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