Laurent expansion of $\zeta(s)$ I am studying the proof of the Prime Number Theorem and I want to show that the function $\frac{\zeta'(s)}{\zeta(s)}$ has a simple pole at $s=1$.
I think that if I can find the Laurent series expansion of $\zeta(s)$, I could then find the same for $\frac{\zeta'(s)}{\zeta(s)}$ and then conclude that it has a simple pole at $s=1$.(Correct me if I am wrong.)
But, how do I find the Laurent expansion ? I know that $\zeta(s)$ has a simple pole at $s=1$ but how can I use this to find the complete expansion ?
Also, do I even need to find the complete expansion to show that $\frac{\zeta'(s)}{\zeta(s)}$ has a simple pole at $s=1$ ? Is there any other way ?
Please help. Any help/hint shall be highly appreciated.
 A: Finding Laurent expansion for $\zeta(s)$ is equivalent to finding a power series representation for
$$
F(s)=\zeta(s)-{1\over s-1}
$$
at $s=1$. This means that we need to develop strategies allowing us to deduce a formula for $F^{(k)}(s)$. This can be done by plugging the Dirichlet series representation of $\zeta(s)$ into Euler-Maclaurin formula:
$$
\zeta(s)=\sum_{n=1}^\infty{1\over n^s}={1\over s-1}+\frac12-s\int_1^\infty{\overline B_1(x)\over x^{s+1}}\mathrm dx
$$
Consequently we have for all $k>1$ that
$$
F^{(k)}(1)={\mathrm d^k\over\mathrm ds^k}\left[-s\int_1^\infty{\overline B_1(x)\over x^{s+1}}\mathrm dx\right]_{s=1}
$$
After simplifications, we can observe that
$$
\begin{aligned}
{\partial^k\over\partial s^k}[-sx^{-s-1}]
&=(-s)(-\log x)^kx^{-s-1}-k(-\log x)^{k-1}x^{-s-1} \\
&=(-1)^kx^{-s-1}[(-s)\log^kx+k\log^{k-1}x] \\
\end{aligned}
$$
As a result, we have
$$
\begin{aligned}
F^{(k)}(1)
&=(-1)^k\int_1^\infty{\overline B_1(x)[k\log^{k-1}x-\log^kx]\over x^2}\mathrm dx \\
&=(-1)^k\int_1^\infty\overline B_1(x)\mathrm d\left(\log^kx\over x\right)
\end{aligned}
$$

One can verify that this quantity converges

To eliminate $\overline B_1(x)$ in the above integral, we apply Euler-Maclaurin formula to $\log^kx/x$ (for $k>1$):
\begin{aligned}
\sum_{n=1}^N{\log^kn\over n}
&={\log^{k+1}N\over k+1}+{\log^kN\over2N}+\int_1^N\overline B_1(x)\mathrm d\left(\log^kx\over x\right) \\
&={\log^{k+1}N\over k+1}+\gamma_k+o(1)
\end{aligned}
where $\gamma_k$ is the Stieltjes constants:
$$
\gamma_k=\int_1^\infty\overline B_1(x)\mathrm d\left(\log^kx\over x\right)
$$
As a result, we can plug Stieltjes constants back into $F(s)$ to get
$$
F(s)=F(1)+\sum_{k=1}^\infty{(-1)^k\gamma_k\over k!}(s-1)^k
$$
Now it remains to determine $F(1)$, and using Euler-Maclaurin again on harmonic series allows us to determine $F(1)=\gamma$. Consequently, the Laurent expansion of $\zeta(s)$ at $s=1$ is as follows
$$
\zeta(s)={1\over s-1}+\sum_{k=0}^\infty{(-1)^k\gamma_k\over k!}(s-1)^k
$$
where
$$
\gamma_k=\lim_{N\to\infty}\left\{\sum_{n=1}^N{\log^kn\over n}-{\log^{k+1}N\over k+1}\right\}
$$

$\gamma_0=\gamma$ makes the above expression valid for $k\ge0$.

A: If all one cares about is knowing that there is a simple pole at $s=1$ (and perhaps what its residue is), this can actually be done quite quickly using some standard complex-analytic results. For a reference, see these notes by Terry Tao, and in particular Exercise 14. By partial summation (or Euler-Maclaurin; they're essentially the same thing), we have
$$
\zeta(s) = \frac{1}{s-1} + s\int_{1}^{\infty} \frac{\left\{ x\right\}}{x^{s+1}} dx,
$$
where $\left\{ x\right\}$ denotes the fractional part of $s$. The integral converges absolutely for $\Re s > 0$, and so has no poles in this region. Thus $\zeta(s)$ has a simple pole at $s=1$ with residue $1$.
For a meromorphic function $f$, the only poles of $f'/f$ are simple poles occuring at the poles and zeros of $f$. Thus $\zeta'(s)/\zeta(s)$ has a simple pole at $s=1$. In fact, the residue is the negative order of pole, so the residue at $s=1$ of $\zeta'(s)/\zeta(s)$ is $-1$.
A: This answer was posted to another question which dealt only with the constant term of the Laurent expansion, but it actually answers this question, so I will repost it here.

A Simple Derivation of the Laurent Series for Zeta
$$
\begin{align}
&\frac1{s-1}+\sum_{k=1}^mk^{-s}-\frac{m^{1-s}-1}{1-s}\tag1\\
&=\frac1{s-1}+\sum_{k=1}^m\frac1ke^{(1-s)\log(k)}-\frac{e^{(1-s)\log(m)}-1}{1-s}\tag2\\
&=\frac1{s-1}+\sum_{n=0}^\infty\left[\sum_{k=1}^m\frac1k\frac{(1-s)^n\log(k)^n}{n!}-\frac{(1-s)^n\log(m)^{n+1}}{(n+1)!}\right]\tag3\\
&=\frac1{s-1}+\sum_{n=0}^\infty\frac{(1-s)^n}{n!}\left[\sum_{k=1}^m\frac{\log(k)^n}k-\frac{\log(m)^{n+1}}{n+1}\right]\tag4
\end{align}
$$
Explanation:
$(2)$: convert powers to exponentials
$(3)$: expand exponentials about $s=1$
$(4)$: pull out a common factor
Taking the limit as $m\to\infty$, for $s\gt1$,
$$
\bbox[5px,border:2px solid #C0A000]{\zeta(s)=\frac1{s-1}+\sum_{n=0}^\infty\frac{(1-s)^n}{n!}\,\gamma_n}\tag5
$$
where
$$
\bbox[5px,border:2px solid #C0A000]{\gamma_n=\lim_{m\to\infty}\left[\sum_{k=1}^m\frac{\log(k)^n}k-\frac{\log(m)^{n+1}}{n+1}\right]}\tag6
$$
$\gamma_n$ is the $n^\text{th}$ Stieltjes constant; $\gamma_0$ is the Euler-Mascheroni constant.
