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Derivation of the Laurent Expansion of the Riemann Zeta Function 📂Functions

Derivation of the Laurent Expansion of the Riemann Zeta Function

Theorem

The Laurent expansion of the Riemann zeta function $\zeta$ is as follows:

$$ \zeta (s) = {{ 1 } \over { s-1 }} + \sum_{n=0}^{\infty} \gamma_{n} {{ (1-s)^{n} } \over { n! }} \qquad , s > 1 $$

Here, $\gamma_{n}$ is the nth Stieltjes constant, defined as follows:

$$ \gamma_{n} := \lim_{m \to \infty} \sum_{k=1}^{m} \left( {{ \left( \log k \right)^{n} } \over { k }} - {{ \left( \log m \right)^{n} } \over { n+1 }} \right) $$

Description

In particular, the Stieltjes constants are $\gamma_{0} = \gamma$ for $n=0$, which is the Euler-Mascheroni constant.

According to this series expansion, the residue of $\zeta (1-s)$ is $1$.

Derivation1

$$ \zeta_{m} (s) := \sum_{k=1}^{m} \left( k^{-s} - {{ m^{1-s} } \over { 1-s }} \right) $$

If we define $\zeta_{m}(s)$ as above, since $\displaystyle \zeta (s) = \sum_{k \in \mathbb{N}} k^{-s}$, if $s>1$ then $m \to \infty$ when $\zeta_{m}(s) \to \zeta (s)$. By reversing the base and exponent on the right side and performing a Taylor expansion about the exponential function $\displaystyle { { e ^ x } }=\sum _{ n=0 }^{ \infty }{ \frac { { x } ^{ n } }{ n! } }$, we obtain

$$ \begin{align*} & \zeta_{m} (s) + 0 \\ =& \sum_{k=1}^{m} \left( k^{-s} - {{ m^{1-s} } \over { 1-s }} \right) + \left( {{ 1 } \over { 1-s }} - {{ 1 } \over { 1-s }} \right) \\ =& {{ 1 } \over { s - 1 }} + \sum_{k=1}^{m} \left[ k^{-s} - {{ m^{1-s} - 1} \over { 1 - s }} \right] \\ =& {{ 1 } \over { s - 1 }} + \sum_{k=1}^{m} \left[ {{ 1 } \over { k }} k^{1-s} - {{ e^{(1-s) \log m} - 1} \over { 1 - s }} \right] \\ =& {{ 1 } \over { s - 1 }} + \sum_{k=1}^{m} \left[ {{ 1 } \over { k }} e^{(1-s) \log k} - {{ 1 + \sum_{n = 0}^{\infty} \left[ (1-s) \log m \right]^{n+1} - 1} \over { 1 - s }} \right] \\ =& {{ 1 } \over { s - 1 }} + \sum_{n = 0}^{\infty} \left[ \sum_{k=1}^{m} \left( {{ 1 } \over { k }} {{ \left[ (1-s) \log k \right]^{n} } \over { n! }} - {{ 1 } \over { 1-s }} {{ \left[ (1-s) \log m \right]^{n+1} } \over { (n+1)! }} \right) \right] \\ =& {{ 1 } \over { s - 1 }} + \sum_{n = 0}^{\infty} \left[ \sum_{k=1}^{m} \left( {{ 1 } \over { k }} {{ (1-s)^{n} \left[ \log k \right]^{n} } \over { n! }} - {{ (1-s)^{n} \left[ \log m \right]^{n+1} } \over { (n+1)! }} \right) \right] \\ =& {{ 1 } \over { s - 1 }} + \sum_{n = 0}^{\infty} {{ (1-s)^{n} } \over { n! }} \left[ \sum_{k=1}^{m} \left( {{ \left[ \log k \right]^{n} } \over { k }} - {{ \left[ \log m \right]^{n+1} } \over { n+1 }} \right) \right] \end{align*} $$

which falls out neatly. Now, by taking $m \to \infty$ if $s>1$, we get the following:

$$ {{ 1 } \over { s - 1 }} + \sum_{n = 0}^{\infty} \gamma_{n} {{ (1-s)^{n} } \over { n! }} = \lim_{m \to \infty} \zeta_{m} (s) = \zeta (s) $$