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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?

Thanks for your help.

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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
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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
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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
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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

1 Answer 1

François is absolutely right. In my question I was assuming that the logarithmic derivative exists. So assume the condition he worked out is satisfied. What can be said about the values of this derivatives in terms of Gamma values or Jacobi sums?

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@magnilor : Please edit your question instead of posting this as an "answer". –  Chandan Singh Dalawat Jan 25 '13 at 9:16

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