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

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.

• 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. – Vincent Granville Jan 3 at 20:38
• 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 $k$. – Steven Clark Jan 4 at 2:46
• 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 $\sum _{m=1}^{\infty } \frac{(-1)^{m+1} \log ^2(m)}{m^{0.8}}=-0.0668616$. – Steven Clark Jan 4 at 3:08
• @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. – Vincent Granville Jan 4 at 21:16
• I'm not sure what's going on with your calculations. Mathematica 12.2.0.0 gives me the result $\sum\limits_{m=1}^{100000} \frac{(-1)^{m+1} \log ^2(m)}{m^{0.8}}=-0.073489$. – Steven Clark Jan 4 at 23:16

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.