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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.

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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.

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  • $\begingroup$ An interesting thing is that if the Taylor expansion leads to an integral representation of $\phi_1$ and $\phi_2$ via the Euler-Maclaurin formula, the variable $t$ and the parameter $\sigma$ will be separated in that integral. $\endgroup$ – Vincent Granville Jan 3 at 20:38
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    $\begingroup$ I believe there are closed form representations of the form $\mu_1(k,\sigma)=2^{-\sigma}\,\sum_{j=0}^{2 k}b_{k,j}\,\zeta^{(j)}(\sigma)$ and $\mu_2(k,\sigma)=2^{-\sigma}\,\sum_{j=0}^{2 k+1}c_{k,j}\,\zeta^{(j)}(\sigma)$ 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 $k$. $\endgroup$ – Steven Clark Jan 4 at 2:46
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    $\begingroup$ For your example where $k=1$, $\sum _{m=1}^{\infty } (-1)^{m+1} m^{-\sigma } \log ^2(m)=-2^{-\sigma } \left(-4 \log (2) \zeta '(\sigma )-\left(2^{\sigma }-2\right) \zeta ''(\sigma )+2 \log ^2(2) \zeta (\sigma )\right)$ and for $\sigma=0.8$ this gives $\sum _{m=1}^{\infty } \frac{(-1)^{m+1} \log ^2(m)}{m^{0.8}}=-0.0668616$. $\endgroup$ – Steven Clark Jan 4 at 3:08
  • $\begingroup$ @Steven: You are welcome to post your comments as an answer to my question. Still I don't know why Mathematica treats it as a complex function. $\endgroup$ – Vincent Granville Jan 4 at 21:16
  • $\begingroup$ I'm not sure what's going on with your calculations. Mathematica 12.2.0.0 gives me the result $\sum\limits_{m=1}^{100000} \frac{(-1)^{m+1} \log ^2(m)}{m^{0.8}}=-0.073489$. $\endgroup$ – Steven Clark Jan 4 at 23:16
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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 $k$.


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


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


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.

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