Main idea or strategy to evaluate some limit $\lim_{s\to0}\zeta^{(k)}(s)$, where $\zeta(s)$ is the Riemann zeta function

Question

Seems that it is possible to get using Wolfram Alpha online calculator the limit as $s$ tends to zero of the derivative of the Riemann Zeta function $$\lim_{s\to0}\zeta^{(k)}(s).\tag{1}$$

For example write lim RiemannZeta'''(s), as s-->0 or lim RiemannZeta''''(s), as s-->0

Question. Imagine that one need to justify a limit of previous kind, say us $$\lim_{s\to0}\zeta^{(iv)}(s),\tag{2}$$ what is the way to get such closed-form in terms of constants? Isn't required the full expression, only is required how to start with the evaluation of this kind of limit, in this example say us $(2)$. Many thanks.

I am asking about a sketch to get the limit, for example what formula I need and what calculations will be need. Thus I don't require all details, since I think that it is tedious, and if you known it from the literature please refer it answering this as a reference request.

Answer 1

"Formulas for Higher Derivatives of the Zeta Function," by Tom M. Apostol, Math. Comp., Vol. 44, Num 169, January 1985, pp.223-232. According to the abstract, a closed form formula is given for $\zeta^{(k)}(0)$ is given, along with along with numerical values to 15D for $k=0\dots 18$

I found this by Googling "Higher derivatives of the zeta function," and then looking at the paper on JSTOR.

Answer 2

We may start from finding an integral representation for $\zeta(s)$ in the region $\text{Re}(s)>-1$.
For any $s$ such that $\text{Re}(s)>0$ we have $$ \zeta(s)=\left(1-\frac{2}{2^s}\right)^{-1}\eta(s)= \frac{2^s}{\Gamma(s)(2^s-2)}\int_{0}^{+\infty}\frac{x^{s-1}}{e^x+1}\,dx\tag{1}$$ and by integration by parts $$ \zeta(s) = \frac{2^s}{\Gamma(s+1)(2^s-2)}\int_{0}^{+\infty}\frac{x^s e^x}{(e^x+1)^2}\,dx \tag{2} $$ provides an analytic continuation to the wanted region. For instance, by evaluating $(2)$ at $s=0$ we have $$ \zeta(0) = -\int_{0}^{+\infty}\frac{e^x}{(e^x+1)^2}\,dx =-\frac{1}{2} \tag{3}$$ and if we need to find $\zeta^{(k)}(0)$ we just have to apply $\frac{d^k}{ds^k}$ to the RHS of $(2)$, then evaluate at $s=0$.
Of course we need the chain rule and the fact that $$ \frac{d^j}{ds^j}\int_{0}^{+\infty}\frac{x^s e^x}{(e^x+1)^2}\,dx = \int_{0}^{+\infty}\frac{x^s\left(\log x\right)^j e^x}{(e^x+1)^2}\,dx,$$ plus $\frac{d}{dx}\Gamma(x+1) = \Gamma(x+1)\psi(x+1)$.

$$ \zeta'(0)=-\log\sqrt{2\pi},\qquad \zeta''(0)=\frac{\gamma^2}{2}-\frac{\pi^2}{24}-\frac{\log^2(2\pi)}{2}+\gamma_1,\qquad \ldots $$ An alternative approach is to directly exploit the reflection formula for the $\zeta$ function, relating the derivatives at $s=0$ with the derivatives of $\eta(s)$ at $s=1$, which are given by conditionally convergent series of the form $\sum_{n\geq 1}\frac{(-1)^{n+1}\left(\log n\right)^k}{n}$.