A limit involving Riemann zeta function $\zeta(s)$

Question

I am reading Apostol's "Introduction to Analytic Number Theory". In Chapter 3, he defines Riemann Zeta Function, $\zeta(s)$, by the equation \begin{equation} \zeta(s) = \sum_{n = 1}^{\infty} \frac{1}{n^s} \quad \text{if } s > 1, \end{equation} and by the equation \begin{equation} \zeta(s) = \lim_{x \to \infty} \left( \sum_{n \leq x} \frac{1}{n^s} - \frac{x^{1 - s}}{1 -s} \right) \quad \text{if } 0 < s < 1. \end{equation} Then he derives the following asymptotic formulas. \begin{equation} \label{harmonic_partial_sums} \sum_{n \leq x} \frac{1}{n} = \log x + C + O\left(\frac{1}{x}\right), \end{equation} \begin{equation} \label{1/ns partial sums} \sum_{n \leq x} \frac{1}{n^s} = \zeta(s) + \frac{x^{1 - s}}{1 - s} + O(x^{-s}), \end{equation} where $C$ is the Euler-Mascheroni constant. Looking at these formulas, one might suspect (naively) that \begin{equation} \lim_{s \to 1} \left( \zeta(s) + \frac{x^{1 - s}}{1 -s} \right) = \log x + C. \end{equation} I tried to prove this by going back to the Abel summation formula but came across the integral \begin{equation} \int_{x}^{\infty} \frac{\{t\}}{t^{s+1}} \, dt \end{equation} whose limit does not exist as $s$ approaches $1$. I would highly appreciate if I could know whether the limit exists and if it exists what's its value.

Answer 1

The Riemann zeta function has a simple pole at $s=1$, with residue $1$. If you take $x\gt0$ a real number, as I think you are, then: $$\begin{align}\lim_{s\to1}(\zeta(s)+x^{1-s}(1-s)^{-1})&=\lim_{s\to1}\left[\frac{x^{1-s}-1}{1-s}+\underset{\longrightarrow\gamma}{\underbrace{(\zeta(s)-(s-1)^{-1})}}\right]\\&\overset{L.H}=\gamma+\lim_{s\to1}\frac{(-\ln x)x^{1-s}}{-1}\\&=\ln(x)+\gamma\end{align}$$

As you expected, where the constant $C$ is the $\gamma$ the Euler-Mascheroni constant, or the first Stieltjes constant.

To see where these constants are coming from, I will reference for you an answer someone gave me a few months ago, here. No complex analysis is involved, really - you should only understand that "analytic continuation is unique" to understand why Conrad's zeta is the same as your zeta.

$$\gamma=\lim_{n\to\infty}\left[-\ln n+\sum_{m=1}^n\frac{1}{m}\right]$$