Higher derivatives of Zeta function for $s>1$?

Question

I'm searching for integral expressions for $\zeta^{(n)} (s)$ for $n \geq 1$ and general $s$.

I have found two useful papers 1 and 2 from which I was able to obtain the following two cases which seem to work great numerically:

$$1) \qquad \Re (s)<0$$

Denote $$s_1=1-s$$

Then we have:

$$(-1)^n \zeta^{(n)} (s)= \\ = \frac{1}{(2 \pi i)^{s_1} } \int_0^{\infty } \frac{t^{s_1-1} \left((-1)^{s_1} \left(\log t-\log (2 \pi )+\frac{\pi i}{2}\right)^n+\left(\log t-\log (2 \pi )-\frac{\pi i}{2}\right)^n\right)}{e^t-1} \, dt$$

It follows from a more general formula (2) in paper 1.*

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$$2) \qquad 0 < \Re (s)<1$$

Then we have:

$$\zeta^{(n)} (s)= \\ = \frac{1}{2 \pi i} \int_0^{\infty } \left(\frac{1}{2 (u+1)}+\log (u)-\psi (u+1)\right) \left(e^{i \pi s} \left(\log \left(\frac{1}{u}\right)+i \pi \right)^n- \\ -e^{-i \pi s} \left(\log \left(\frac{1}{u}\right)-i \pi \right)^n\right) \frac{du}{u^s} $$

Which directly follows from formula (1.7) in paper 2.

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My question is: can we find a similar explicit expression for $\zeta^{(n)} (s)$ with $\Re (s)>1$?

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$^*$ For anyone who doesn't have access to paper 1, here's the original formula, valid for all complex $s$:

$$(-1)^n \zeta^{(n)} (1-s) = \sum_{k=0}^n \binom{n}{k} \left(e^{s z} z^{n-k}+e^{s z^*} (z^*)^{n-k} \right) \left( \Gamma(s) \zeta(s) \right)^{(k)}$$

Where $z=-\log (2 \pi) - \frac{\pi i}{2}$.

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A side question: what other ways exist to numerically evaluate these derivatives? How does Mathematica do it, in case you know?

Answer 1

I finally found a suitable integral expression valid for all $s \neq 1$:

$$\zeta (s)={\frac {1}{s-1}}+{\frac {1}{2}}+2\!\int _{0}^{\infty }\!\!\!{\frac {\sin(s\arctan t)}{\left(1+t^{2}\right)^{\frac {s}{2}}\left(e^{2\pi t}-1\right)}}\,\mathrm {d} t$$

Modifying it a little, I was able to obtain the following expression for the derivative, which seems to work better numerically than all the previous ones.

For $ n \geq 1$ and $s \neq 1$:

$$\zeta^{(n)}(s)=\frac{(-1)^n n!}{(s-1)^{n+1}}- \\ -i \int_0^{\infty } \frac{dt}{e^{2 \pi t}-1} \left(\frac{(1+i t)^s}{\left(1+t^2\right)^s} \log ^n\left(\frac{1+i t}{1+t^2}\right)-\frac{(1-i t)^s }{\left(1+t^2\right)^s}\log ^n\left(\frac{1-i t}{1+t^2}\right) \right)$$

Numerical check:

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To answer some confusion in the comments, I didn't know about such a nice integral when asking the question. Usual integral representations of Zeta contain Gamma function as well, and so the derivatives have very messy expressions. Not to mention, the integrals are not suitable for numerical evaluation for some parameters, while this one works like a charm.