# derivatives of Artin L-functions

This is a vague question: I'm sorry for that.

Let's start with $\chi$ a (primitive odd) Dirichlet character modulo $n$ and look at the corresponding L-function $$L(s, \chi)=\sum \frac{\chi(n)}{n^s}.$$ A classical formula due to Hurwitz computes the logarithmic derivative of $L(s, \chi)$ at $s=0$ in terms of the Gamma function: $$\frac{L'(0, \chi)}{L(0,\chi)}=-\log n + \frac{\sum \log \Gamma(1-\frac{u}{n})\chi(u)}{\sum \chi(u) \frac{u}{n}}$$ (By the way, does anybody know of some good references for this formula?)

That's the abelian case. Now imagine $L/\mathbb{Q}$ is a Galois field extension with non-abelian Galois group $G$, and let $\chi$ be a character of $G$. Now one has the Artin L-function.

Question: Is there any similar formula know for $\frac{L'(0, \chi)}{L(0, \chi)}$?

My naive guess would be the following: by Brauer's theorem one knows that $\chi$ can be written as a sum with rational coefficients of characters induced by cyclic subgroups of $G$, so $\log L(s, \chi)$ should be a sum with rational coefficients of some Dirichlet L-functions and for each one there is a such a formula. However, I was unable to find something like that in the literature. Is there some mistake here? If rational coefficients create some trouble, one can also write $\chi$ as a combination with integer coefficients of characters induced from elementary subgroups. Does it help?

Another question: is there any relation between Jacobi sums and these derivatives?

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In many (most?) cases L(chi, 0) will be zero, so how are you defining its logarithmic derivative? I do not see how your reduction to linear combinations of Dirichlet L functions will work - surely you get rather linear combinations of logs of Hecke L functions of Groessencharacters of number fields? –  David Loeffler Jan 22 '13 at 19:20
I agree with David that you have to put some assumption on the Artin representation $\rho$ in order for the log derivative at 0 to be well-defined. You need $L(\rho,0)\neq 0$, and a condition for that is worked out in my answer to this question mathoverflow.net/questions/67747/… Assuming my computations are correct, you need that complex conjugation acts as $-1$ on the representation space. –  François Brunault Jan 23 '13 at 8:37
See H. Yoshida, On absolute CM periods II, American J. of Math. 120 (1998) and the references therein, esp. to Shintani's work on the polygamma functions. –  Damian Rössler Jan 23 '13 at 21:53
May I suggest that you use $n$ for the summation index and something else for the conductor? –  Julien Puydt Aug 12 '13 at 7:38
I wish I could upvote @JulienPuydt's comment multiple times. –  Steven Landsburg Sep 11 '13 at 0:54