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It is well known that Euler-Mascheroni constant has an alternative definition in terms of zeta function:

$$ \gamma=\lim_{s\to1^+}f(s)\equiv\lim_{s\to1^+}\left[\zeta(s)-{1\over s-1}\right] $$

Using this definition, I would like to derive the Laurent series expansion for Riemann zeta function, but I am currently stuck because I am in a hard time trying to show the convergence of the following limit:

$$ \lim_{s\to1^+}{\mathrm{d}^n\over\mathrm{d}s^n}f(s) $$

My attempt was to show that the function $f(s)$ is analytic in $\mathbb{C}\setminus\{1\}$ and continuous in $\mathbb{C}$ (by setting $f(1)=\gamma$), and prove that it is entire using Morera's theorem. Using analytic continuation, I am able to express zeta functon in terms of a contour integral:

$$ \zeta(s)=\Gamma(1-s)g(s)\equiv{\Gamma(1-s)\over2\pi i}\oint_C{z^{s-1}\over e^{-z}-1}\mathrm{d}z $$

where the contour $C$ is shown below: Contour used for analytic continuation

Therefore, $f(s)=\Gamma(1-s)g(s)-{1\over s-1}$, which is already analytic every where except for $z=1$, so how can I show that it is continuous at $z=1$ so that I can prove $f(s)$ to be entire using Morera's theorem?

TravorLZH
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    $\sum_{n\ge 1} n^{-s}-\int_n^{n+1} x^{-s}dx$ converges for $\Re(s) > 0$ – reuns Apr 25 '20 at 20:19
  • @reuns Is that sufficient to show the limit of their derivatives at s=1 exist – TravorLZH Apr 25 '20 at 20:35
  • It is not hard to show it and all the derivatives converge and are analytic. For $\Re(s) > 1$ it is $(\sum_n n^{-s})-1/(s-1)$ and its derivatives, for $\Re(s)\in (0,1]$ it is the analytic continuation $\zeta(s)-1/(s-1)$ – reuns Apr 25 '20 at 22:12
  • @reuns I know that zeta(s) can be analytically continued to everywhere but s=1. However, is it enough to show that $\zeta(s)-1/(s-1)$ is entire given $$\gamma=\lim_{s\to1^+}\left[\zeta(s)-{1\over s-1}\right]$$. – TravorLZH Apr 25 '20 at 22:36
  • It is unclear what you are not sure about. That $f(s)=\zeta(s)-1/(s-1)$ is entire means that $f(s)=\sum_k \frac{f^{(k)}(1)}{k!}(s-1)^k$ where $f^{(k)}(1)=\lim_{s\to 1} f^{(k)}(s)$. This is the definition of analytic + Cauchy integral formula from which analytic everywhere implies that the Taylor series converges to the function everywhere. From the above formula the same reasoning holds for $\Re(s) >0$. – reuns Apr 25 '20 at 22:38
  • @reuns I know what "entire" means, but I just don't know how to show $f(s)$ is entire. – TravorLZH Apr 25 '20 at 22:51
  • Why do you want to show it ? Either the functional equation or analytic continuation to every strip following the same idea as above – reuns Apr 26 '20 at 00:00
  • @reuns I would like to show that the formula for Stieljes constants converges, and that is why I created this question – TravorLZH Apr 26 '20 at 12:59

1 Answers1

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After learning more knowledge in this area, I found that Riemann-Stieltjes integration does the job. For $\Re(s)>1$, we have

$$ \begin{aligned} \zeta(s) &=1+\int_1^\infty{\mathrm d\lfloor x\rfloor\over x^s} \\ &=1+\int_1^\infty{\mathrm dx\over x^s}-\int_1^\infty{\mathrm d\{x\}\over x^{s+1}} \\ &={s\over s-1}-s\int_1^\infty{\{x\}\over x^{s+1}}\mathrm dx \end{aligned} $$

Since the last integral converges for all $\Re(s)>0$:

$$ \begin{aligned} \left|\int_1^\infty{\{x\}\over x^{s+1}}\mathrm dx\right| &\le\int_1^\infty{\mathrm dx\over x^{\Re(s)+1}}={1\over\Re(s)}<\infty \end{aligned} $$

we conclude that

$$ \zeta(s)-{1\over s-1}=1-s\int_1^\infty{\{x\}\over x^{s+1}}\mathrm dx $$

converges and is continuous at $s=1$.

TravorLZH
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