About the coefficients of Taylor series for the complex Riemann Zeta function $\zeta(s)$

Question

The following real-valued functions are closely related to the zeros of $\zeta(s)$ in the critical strip $\frac{1}{2}<\Re(s) < 1$.

$$\phi_1(\sigma, t) = \sum_{n=1}^\infty (-1)^{n+1}\frac{\cos(t\log n)}{n^\sigma},\\ \phi_2(\sigma, t) = \sum_{n=1}^\infty (-1)^{n+1}\frac{\sin(t\log n)}{n^\sigma}. $$

We use the standard notation $s=\sigma + it$ for a complex number $s$, in the context of the Riemann Hypothesis; $\sigma+it$ is a non-trivial root of $\zeta(s)$ if and only if $\phi_1(\sigma, t)=\phi_2(\sigma, t)=0$, see here. The Taylor series for $\phi_1,\phi_2$, assuming $\sigma$ is fixed and $t$ is the variable, are

$$\phi_1(\sigma, t)=\sum_{k=0}^\infty (-1)^k\frac{\mu_1(k,\sigma)}{2k!}t^{2k}, \mbox { } \mbox { } \mbox {with } \mu_1(k,\sigma)=\sum_{m=1}^\infty (-1)^{m+1} \frac{(\log m)^{2k}}{m^\sigma},\\ \phi_2(\sigma, t)=\sum_{k=0}^\infty (-1)^k\frac{\mu_2(k,\sigma)}{(2k+1)!}t^{2k+1}, \mbox { } \mbox { } \mbox {with } \mu_2(k,\sigma)=\sum_{m=1}^\infty (-1)^{m+1} \frac{(\log m)^{2k+1}}{m^\sigma}.\\ $$

My question

Are $\mu_1(k,\sigma)$ and $\mu_2(k,\sigma)$ known functions? The series that define them (albeit alternating) converge very, very slowly if $k$ is large and $\sigma$ barely above $\frac{1}{2}$. Can you find tabulated values somewhere? If I use Mathematica, it returns a complex number with the imaginary part extremely close to zero; it looks like Mathematica "knows" it comes from a family of special function defined on the complex plane, but it does not tell me which one. See here or picture below.

More difficult question

Can you get bounds for the error term if you only use the first $N$ terms in the Taylor expansion? Also, can this lead to integral representations for the functions $\phi_1, \phi_2$ if you use the Euler-Maclaurin summation formula applied to the Taylor series, or by other means? For that purpose, you might want to replace $(-1)^k$ by $\cos k\pi$, and $(2k)!$ by $\Gamma(2k+1)$, in the Taylor series expansion.

Answer 1

I believe there are closed form representations of the form:

$$\mu_1(k,\sigma)=2^{-\sigma}\,\sum\limits_{j=0}^{2 k}b_{k,j}\,\zeta^{(j)}(\sigma)$$

$$\mu_2(k,\sigma)=2^{-\sigma}\,\sum_{j=0}^{2 k+1}c_{k,j}\,\zeta^{(j)}(\sigma)$$

where the sums are over $\zeta(\sigma)$ and it's derivatives so it doesn't look very promising with respect to finding an efficient way to calculate these for large values of $k$.

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Mathematica gives the following closed form representations of $\mu_1(k,\sigma)=\sum\limits_{m=1}^\infty\frac{(-1)^{m+1}\log^{2 k}(m)}{m^{\sigma}}$ for the first few values of $k$ and the result $\mu_1(1,0.8)=-0.0668616$.

$\begin{array}{cc} \text{k} & \text{$\mu_1(k,\sigma $)} \\ 0 & 2^{-\sigma } \left(2^{\sigma }-2\right) \zeta (\sigma ) \\ 1 & 2^{-\sigma } \left(\log (16) \zeta '(\sigma )+\left(2^{\sigma }-2\right) \zeta ''(\sigma )-2 \log ^2(2) \zeta (\sigma )\right) \\ 2 & 2^{-\sigma } \left(4 \log (2) \left(\log (2) \log (4) \zeta '(\sigma )-\log (8) \zeta ''(\sigma )+2 \zeta ^{(3)}(\sigma )\right)+\left(2^{\sigma }-2\right) \zeta ^{(4)}(\sigma )-2 \log ^4(2) \zeta (\sigma )\right) \\ 3 & 2^{-\sigma } \left(12 \log ^5(2) \zeta '(\sigma )-30 \log ^4(2) \zeta ''(\sigma )+\left(2^{\sigma }-2\right) \zeta ^{(6)}(\sigma )+\log (4) \left(6 \zeta ^{(5)}(\sigma )+5 \log (2) \left(\log (16) \zeta ^{(3)}(\sigma )-3 \zeta ^{(4)}(\sigma )\right)\right)-2 \log ^6(2) \zeta (\sigma )\right) \\ 4 & 2^{-\sigma } \left(2 \log (4) \left(4 \log ^6(2) \zeta '(\sigma )-7 \log (2) \left(\log ^2(2) \left(\log (2) \log (4) \zeta ''(\sigma )+5 \zeta ^{(4)}(\sigma )-2 \log (4) \zeta ^{(3)}(\sigma )\right)+2 \zeta ^{(6)}(\sigma )-\log (16) \zeta ^{(5)}(\sigma )\right)+4 \zeta ^{(7)}(\sigma )\right)+\left(2^{\sigma }-2\right) \zeta ^{(8)}(\sigma )-2 \log ^8(2) \zeta (\sigma )\right) \\ 5 & 2^{-\sigma } \left(20 \log ^9(2) \zeta '(\sigma )-90 \log ^8(2) \zeta ''(\sigma )+\left(2^{\sigma }-2\right) \zeta ^{(10)}(\sigma )+6 \log ^2(2) \left(\log (4) \left(20 \zeta ^{(7)}(\sigma )+\log (2) \left(\log (2) \left(42 \zeta ^{(5)}(\sigma )+5 \log (2) \left(\log (16) \zeta ^{(3)}(\sigma )-7 \zeta ^{(4)}(\sigma )\right)\right)-35 \zeta ^{(6)}(\sigma )\right)\right)-15 \zeta ^{(8)}(\sigma )\right)+20 \log (2) \zeta ^{(9)}(\sigma )-2 \log ^{10}(2) \zeta (\sigma )\right) \\ \end{array}$

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Mathematica gives the following closed form representations of $\mu_2(k,\sigma)=\sum\limits_{m=1}^\infty\frac{(-1)^{m+1}\log^{2 k+1}(m)}{m^{\sigma}}$ for the first few values of $k$.

$\begin{array}{cc} \text{k} & \text{$\mu_2(k,\sigma $)} \\ 0 & 2^{-\sigma } \left(-\left(2^{\sigma }-2\right) \zeta '(\sigma )-\log (4) \zeta (\sigma )\right) \\ 1 & -2^{-\sigma } \left(-6 \log ^2(2) \zeta '(\sigma )+\log (64) \zeta ''(\sigma )+\left(2^{\sigma }-2\right) \zeta ^{(3)}(\sigma )+2 \log ^3(2) \zeta (\sigma )\right) \\ 2 & 2^{-\sigma } \left(10 \log (2) \left(\log ^3(2) \zeta '(\sigma )-2 \log ^2(2) \zeta ''(\sigma )-\zeta ^{(4)}(\sigma )+\log (4) \zeta ^{(3)}(\sigma )\right)-\left(2^{\sigma }-2\right) \zeta ^{(5)}(\sigma )-2 \log ^5(2) \zeta (\sigma )\right) \\ 3 & 2^{-\sigma } \left(14 \log ^6(2) \zeta '(\sigma )-14 \log (2) \left(3 \log ^4(2) \zeta ''(\sigma )+\zeta ^{(6)}(\sigma )+5 \log ^2(2) \left(\zeta ^{(4)}(\sigma )-\log (2) \zeta ^{(3)}(\sigma )\right)-\log (8) \zeta ^{(5)}(\sigma )\right)-\left(2^{\sigma }-2\right) \zeta ^{(7)}(\sigma )-2 \log ^7(2) \zeta (\sigma )\right) \\ 4 & 2^{-\sigma } \left(6 \log (2) \left(3 \log ^7(2) \zeta '(\sigma )+\log (4) \left(-6 \log ^5(2) \zeta ''(\sigma )+6 \zeta ^{(7)}(\sigma )+7 \log (2) \left(-2 \zeta ^{(6)}(\sigma )+\log ^2(2) \left(\log (4) \zeta ^{(3)}(\sigma )-3 \zeta ^{(4)}(\sigma )\right)+\log (8) \zeta ^{(5)}(\sigma )\right)\right)-3 \zeta ^{(8)}(\sigma )\right)-\left(2^{\sigma }-2\right) \zeta ^{(9)}(\sigma )-2 \log ^9(2) \zeta (\sigma )\right) \\ 5 & 2^{-\sigma } \left(22 \log ^{10}(2) \zeta '(\sigma )-22 \log (2) \left(5 \log ^8(2) \zeta ''(\sigma )+\zeta ^{(10)}(\sigma )+\log (2) \left(\log (8) \left(5 \zeta ^{(8)}(\sigma )-5 \log ^5(2) \zeta ^{(3)}(\sigma )+\log (4) \left(-5 \zeta ^{(7)}(\sigma )+5 \log ^3(2) \zeta ^{(4)}(\sigma )+7 \log (2) \left(\zeta ^{(6)}(\sigma )-\log (2) \zeta ^{(5)}(\sigma )\right)\right)\right)-5 \zeta ^{(9)}(\sigma )\right)\right)-\left(2^{\sigma }-2\right) \zeta ^{(11)}(\sigma )-2 \log ^{11}(2) \zeta (\sigma )\right) \\ \end{array}$

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I believe the results above can be stated more concisely as follows:

$$\mu_1(k,\sigma)=\frac{\partial^{2 k}\,\eta(\sigma)}{\partial\sigma^{2 k}}$$

$$\mu_2(k,\sigma)=-\frac{\partial^{2 k+1}\,\eta(\sigma)}{\partial\sigma^{2 k+1}}$$

where $\eta(\sigma)=\left(1-2^{1-\sigma}\right)\ \zeta(\sigma)$ is the Dirichlet eta function.