Using Mathematica to find series expansions for partial derivatives of the generalized Riemann zeta function

Question

I am trying to use Mathematica to find a suitable series expansion for the expression $$ \zeta ^{(1,0)}\left(-1,1-\frac{i}{2}\right) - \zeta^{(1,0)}\left(-1,1+\frac{i}{2}\right),$$ which Mathematica evaluates numerically as 0.484427... times the imaginary unit $i$, where the superscripts in the expression displayed above indicate that the generalized Riemann zeta function $\zeta(s,a)$ is being differentiated with respect to $s$, and then evaluated at the indicated arguments. It seems that some issues arise if we consider how this numerical evaluation is obtained through a direct application of the definition of the function $\zeta(s,a)$, which Mathematica defines so that $$\zeta(s, a) = \sum_{k=0}^{\infty} (k + a)^{-s},$$ omitting the cases whereby $k + a = 0$. It would seem that we should have that $$\zeta^{(1,0)}(t,b) = \sum _{k=0}^{\infty } -(b+k)^{-t} \log (b+k)$$ whenever the above series converges. However, the series $$\sum _{k=0}^{\infty } \left(\left(-1+\frac{i}{2}\right)-k\right) \log \left(\left(1-\frac{i}{2}\right)+k\right)$$ does not converge, but Mathematica evaluates $\zeta ^{(1,0)}\left(-1,1-\frac{i}{2}\right)$ numerically as $(-0.224051...) + (0.242213...) i$. So, since the above series does not converge, why does Mathematica evaluate $\zeta ^{(1,0)}\left(-1,1-\frac{i}{2}\right)$ as having a finite real part and a finite imaginary part? What does the expression $\zeta ^{(1,0)}\left(-1,1-\frac{i}{2}\right)$ even mean if the corresponding series does not converge? What would be a suitable way of writing this expression as an infinite sum derived from the series definition for the generalized Riemann zeta function?

Thank you.

Answer 1

As @somos says you need an alternative series expansion that converges for your point of interest. The one at http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/Zeta2/06/05/01/01/0001/ is more useful in this case. Summing only for $k\le n$:

myzeta[s_, a_, n_] := Sum[(-1)^j*Binomial[k,j]*(a+j)^(1-s)/((k+1)*(s-1)),
                          {k,0,n}, {j,0,k}]

Taking the first derivative of each term with respect to $s$:

myzeta1[s_, a_, n_] := -Sum[(-1)^j*Binomial[k,j]*(a+j)^(1-s)*(Log[a+j]+1/(s-1))/((k+1)*(s-1)),
                            {k,0,n}, {j,0,k}]

Your formula is thus, for example with $k\le10$:

With[{n = 10}, myzeta1[-1, 1-I/2, n] - myzeta1[-1, 1+I/2, n]] // N
(* 0. + 0.490021 I *)

Replacing // N with // ComplexExpand gives you a formula that you can interpret as a series expression:

With[{n = 10}, myzeta1[-1, 1-I/2, n] - myzeta1[-1, 1+I/2, n]] // ComplexExpand
(* I (1/4 - 483/44 ArcTan[1/22] + 44289/440 ArcTan[1/20]
      - (1631473 ArcTan[1/18])/3960 + (1045007 ArcTan[1/16])/1056
      - (11394539 ArcTan[1/14])/7392 + (16353181 ArcTan[1/12])/10080
      - (1303391 ArcTan[1/10])/1120 + (983509 ArcTan[1/8])/1760
      - (541091 ArcTan[1/6])/3168 + (221209 ArcTan[1/4])/7392
      - (83711 ArcTan[1/2])/36960 - (83711 Log[5/4])/55440
      + (221209 Log[17/4])/27720 - (541091 Log[37/4])/18480
      + (983509 Log[65/4])/13860 - (1303391 Log[101/4])/11088
      + (1257937 Log[145/4])/9240 - (876503 Log[197/4])/7920
      + 61471/990 Log[257/4] - 5051/220 Log[325/4]
      + 111/22 Log[401/4] - 1/2 Log[485/4]) *)