How to compute $\zeta'(0)$?

Question

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.

Answer 1

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 $$ \bbox[5px,border:2px solid #C0A000]{\zeta'(0)=-\frac12\log(2\pi)}\tag3 $$

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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 $$ \bbox[5px,border:2px solid #C0A000]{\zeta'(0)=-\frac12\log(2\pi)}\tag7 $$

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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

Answer 2

I don't think using the reflection-like formula for the Riemann function is the way to go. I'm not entirely sure how much you know, but this is the most straightforward way to derive the identity for $\zeta'(0)$.

First, start off with the infinite product for $\sin x$ $$\sin x=x\prod\limits_{n\geq1}\left(1-\frac {x^2}{\pi^2n^2}\right)\tag1$$

Let $x=\frac {\pi}2$ and simplify to see that$$\begin{align*}\frac {\pi}2 & =\prod\limits_{n\geq1}\left(\frac {4n^2}{4n^2-1}\right)\\ & =\frac 21\times\frac 23\times\frac 43\times\frac 45\cdots\tag2\end{align*}$$

We will be using $(2)$ later on. Now recall the expansion for the polylogarithmic function

$$-\mathrm{Li}_s(-1)=\sum\limits_{n\geq1}\frac {(-1)^{n-1}}{n^s}=\zeta(s)\left(1-2^{1-s}\right)$$Differentiate with respect to $s$ and rearrange the infinite sum to get$$\sum\limits_{n\geq1}(-1)^n\frac {\log n}{n^s}=\left(1-2^{1-s}\right)\zeta'(s)+2^{1-s}\zeta(s)\log2$$Set $s=0$ so the right-hand side becomes$$-\zeta'(0)-\log 2=\frac 12\log\left(\frac {\pi}2\right)\tag3$$after using what we derived in $(2)$. The equation now reduces down to$$\zeta'(0)\color{blue}{=-\frac 12\log 2\pi}$$

Answer 3

Using Euler-Maclaurin formula, one can deduce that

$$ \zeta(s)={1\over s-1}+\frac12-s\int_1^\infty{\overline B_1(x)\over x^{s+1}}\mathrm dx $$

remains valid as long as the latter integral converges. Differentiating on both side, we have

$$ \zeta'(0)=-1-\int_1^\infty{\overline B_1(x)\over x}\mathrm dx $$

To determine the value of the latter integral, consider applying Euler-Maclaurin formula to factorials:

$$ \log N!=\left(N+\frac12\right)\log N-N+1+\int_1^N{\overline B_1(x)\over x}\mathrm dx $$

Now, it follows from Stirling's approximation that

$$ 1+\int_1^N{\overline B_1(x)\over x}\mathrm dx=\frac12\log2\pi+\mathcal O\left(\frac1N\right) $$

As a result, we deduce

$$ \zeta'(0)=-\frac12\log2\pi $$