The input


gives 0 even when s0 is a root. For example, from LMFDB, I found s0 = 0.5 + 1.51608375316006 I is an approximate root of DirichletL(19,10,s). (In LMFDB this character is actually indexed 18, though.)

For the Riemann zeta function, we can get around this by using ZetaZero[1] to represent s0. What can be done for other $L$-functions?

  • $\begingroup$ what is LMFDB?....from help it says about Residue Laurent expansion of expr. What is the Laurent expansion of 1/DirichletL[19,10,s]? does it have a Laurent expansion? $\endgroup$ – Nasser Feb 25 '14 at 3:20
  • $\begingroup$ LMFDB is a database of information about $L$-functions and related structures: lmfdb.org. DirichletL[19,10,s] is a specific $L$-function, $L(\chi,s)$, where the modulus of the Dirichlet character $\chi$ is $19$. It has a simple zero at s0, so it's reciprocal should have a pole there (and therefore a Laurent expansion). $\endgroup$ – A l'Maeaux Feb 25 '14 at 3:30
  • 1
    $\begingroup$ This...and this.. might be of help. You can calculate the steps manually to gain better understanding. $\endgroup$ – Sejwal Feb 25 '14 at 3:47

You can use Cauchy's theorem.

Define the approximate zero of your function :

zero = FindRoot[DirichletL[19, 10, s], {s, 0.5 + I}][[1, 2]]
(* 0.5 + 1.51608 I *) 

Series will not consider this a pole of 1/DirichletL[19, 10, s] and I think this is why you get a zero residue.

However, integrating on a small square around that pole one finds :

       NIntegrate[1/DirichletL[19, 10, s], 
         {s, zero + eps (1+I), zero + eps (-1+I), zero + eps (-1-I), 
             zero + eps (1-I), zero + eps (1+1 I)}]/(2 Pi I)}, 
 {eps, 10^Range[0., -5, -1] }]

enter image description here

Same for the Zeta function as a check :

Residue[1/Zeta[s], {s, ZetaZero[1]}] // N
(* 1.2451 - 0.198218 I *)

         {s, ZetaZero[1] + eps (1+I), ZetaZero[1] + eps (-1+I), 
             ZetaZero[1] + eps (-1-I), ZetaZero[1] + eps (1-I), 
             ZetaZero[1] + eps (1+I)}]/(2 Pi I)}, 
{eps, 10^Range[0, -5, -1] }]

enter image description here

  • $\begingroup$ Good method for a numerical solution. But it is possible to get the residue as a symbolic expression (as a function of s0) ? $\endgroup$ – Vaclav Kotesovec Feb 17 '17 at 21:46
  • $\begingroup$ For example for ZetaZero[1] we have the residue 1/Derivative[1][Zeta][ZetaZero[1]]. But in general ? $\endgroup$ – Vaclav Kotesovec Feb 17 '17 at 21:53
  • $\begingroup$ @VaclavKotesovec I don't know; unlike for the Zeta function, there is no symbolic expression of the zeros for DirichletL. $\endgroup$ – b.gates.you.know.what Feb 18 '17 at 9:07

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