TITLE: How to compute $\zeta'(0)$?
QUESTION [8 upvotes]: I want to compute $\zeta'(0)$. I read that $\zeta'(0) = -\frac{\ln(2\pi)}{2}$. I try to use the fact that
$$
\zeta(1-s) = \frac{1}{\pi} (2 \pi)^{1-s} \cos(\frac{\pi s}{2}) \Gamma(s) \zeta(s)
$$
Take derivative from both sides:
\begin{align*}
-\zeta'(1-s) = (\frac{1}{\pi}(2 \pi)^{1-s} \Gamma(s)\zeta(s))'\cos(\frac{\pi s}{2}) + (\frac{1}{\pi}(2 \pi)^{1-s} \Gamma(s)\zeta(s))(-\sin(\frac{\pi s}{2}) \frac{\pi}{2})
\end{align*}
I try to let $s \to 1$ to get the answer, but $\zeta(s)\sin(\frac{\pi s}{2})$ diverges, and I don't know what will happen to the first term. I am not so familiar with the expansion of $\Gamma(s)$ or $\zeta(s)$. Sorry for asking this but I just don't know where to start.

REPLY [2 votes]: Euler-Maclaurin Sum Formula
In this answer, I use the Euler-Maclaurin Sum Formula to get, for $s\gt-1$,
$$
\hspace{-1cm}\sum_{k=1}^n\log(k)\,k^{-s}=\frac{\log(n)\,n^{1-s}}{1-s}-\frac{n^{1-s}}{(1-s)^2}-\zeta'(s)+\frac12\log(n)\,n^{-s}+O\!\left(\log(n)\,n^{-1-s}\right)\tag1
$$
Setting $s=0$ gives the logarithm of Stirling's Formula:
$$
\sum_{k=1}^n\log(k)=\log(n)\,n-n-\zeta'(0)+\frac12\log(n)+O\!\left(\log(n)\,n^{-1}\right)\tag2
$$
Matching the constant in Stirling's Formula gives
$$
\bbox[5px,border:2px solid #C0A000]{\zeta'(0)=-\frac12\log(2\pi)}\tag3
$$

Dirichlet $\boldsymbol{\eta}$
In this answer, it is shown that
$$
\eta'(0)=\frac12\log\left(\frac\pi2\right)\tag4
$$
Differentiating
$$
\eta(s)=\left(1-2^{1-s}\right)\zeta(s)\tag5
$$
gives
$$
\eta'(s)=\left(1-2^{1-s}\right)\zeta'(s)+\log(2)2^{1-s}\zeta(s)\tag6
$$
Plugging in $s=0$, and using $(4)$ and $\zeta(0)=-\frac12$, as shown in this answer, gives
$$
\bbox[5px,border:2px solid #C0A000]{\zeta'(0)=-\frac12\log(2\pi)}\tag7
$$

Integration by Parts with a Riemann-Stieltjes Integral
$$
\begin{align}
\sum_{k=1}^n\frac1{k^s}
&=\int_{1^-}^{n^+}\frac{d\lfloor x\rfloor}{x^s}\tag{8a}\\
&=\frac1{s-1}-\frac1{(s-1)\,n^{s-1}}-\int_{1^-}^{n^+}\frac{d\left(\{x\}-\frac12\right)}{x^s}\tag{8b}\\
&=\frac1{s-1}-\frac1{(s-1)\,n^{s-1}}+\frac1{2n^s}+\frac12-s\int_1^n\frac{\{x\}-\tfrac12}{x^{s+1}}\,\mathrm{d}x\tag{8c}
\end{align}
$$
For $s\gt-1$, the integral in $\text{(1c)}$ converges as $n\to\infty$.
As discussed in this answer, $\zeta(s)$ is the constant term in $\text{(8c)}$. Therefore, for $\mathrm{Re}(s)\gt-1$,
$$
\zeta(s)=\frac1{s-1}+\frac12-s\int_1^\infty\frac{\{x\}-\tfrac12}{x^{s+1}}\,\mathrm{d}x\tag9
$$
and therefore,
$$
\zeta'(s)=-\frac1{(s-1)^2}-\int_1^\infty\frac{\{x\}-\tfrac12}{x^{s+1}}+s\int_1^\infty\frac{\{x\}-\tfrac12}{x^{s+1}}\,\log(x)\,\mathrm{d}x\tag{10}
$$
Plugging in $s=0$ gives
$$
\begin{align}
\zeta'(0)
&=-1-\sum_{k=1}^\infty\int_0^1\frac{x-\frac12}{k+x}\,\mathrm{d}x\tag{11a}\\
&=-1-\sum_{k=1}^\infty\left(\color{#C00}{1}-\left(k+\frac12\right)\log\left(\frac{\color{#090}{k+1}}{\color{#00F}{k}}\right)\right)\tag{11b}\\
&=-1-\lim_{n\to\infty}\left(\color{#C00}{n}-\color{#090}{\sum_{k=2}^{n+1}\left(k-\frac12\right)\log(k)}+\color{#00F}{\sum_{k=1}^n\left(k+\frac12\right)\log(k)}\right)\tag{11c}\\
&=-1-\lim_{n\to\infty}\left(n-\left(n+\frac12\right)\log(n+1)+\sum_{k=1}^n\log(k)\right)\tag{11d}\\
&=\bbox[5px,border:2px solid #C0A000]{-\frac12\log(2\pi)}\tag{11e}
\end{align}
$$
Explanation:
$\text{(11a)}$: break the domain into unit intervals and sum
$\text{(11b)}$: integrate
$\text{(11c)}$: write the infinite sum as a limit
$\phantom{\text{(11c):}}$ substitute $k\mapsto k-1$ in the green sum
$\text{(11d)}$: pull out the $k=n+1$ term and combine the rest
$\text{(11e)}$: apply Stirling's Formula