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:
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?
Best Answer
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 $s=1$.