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 :

```
Table[{eps,
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] }]
```

Same for the `Zeta`

function as a check :

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

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? – Nasser – 2014-02-25T03:20:23.097LMFDB is a database of information about $L$-functions and related structures: http://www.lmfdb.org/.

– A l'Maeaux – 2014-02-25T03:30:05.263`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).1This...and this.. might be of help. You can calculate the steps manually to gain better understanding. – Sejwal – 2014-02-25T03:47:45.080