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I am working through a problem set in an analytic number theory course, and the following problem was included:

Show that if $\chi$ is a non-principal Dirichlet character $\pmod{m}$, and where $L(s, \chi)$ is an $L$-function, then $$L(0, \chi) = \frac{-1}{m} \sum_{c=1}^m \chi(c) c,$$ and $$L'(0, \chi) = L(0, \chi) \log m + \sum_{c=1}^m \chi(c) \log \Gamma (\frac{c}{m}).$$

WARNING: The professor writing these problems has an unfortunate habit of teXing problems up incorrectly! Hence part of the "fun" for students taking the course is to figure out if the statement of the problem itself is correct, and if not, to figure out how to modify the statement to make it workable.

I am wondering if anyone visiting would be able to tell whether the problem as stated is right (and if so, suggest a strategy for proving it); if the statement is false, I am curious to know if anyone could either suggest how to modify the statement to be workable, or even point me in the direction of a text that contains a correct statement.

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up vote 3 down vote accepted

This problem is missing some stuff. To get that kind of equality you need $\chi(-1)=-1$, since if $\chi(-1)=1$, then $L(0,\chi)=0$. I believe the question should be something like this:

The question: If $\chi(-1)=-1$, prove that $$L(0,\chi)=\frac{-1}{G(\chi)\sqrt{q}} \sum_{a=1}^q \chi(a)a,$$ where $G(\chi)$ is the Gauss sum.

Hint: So how can you do this? Start by getting an expression for $L(1,\chi)$ by switching sums in the definition of $L(s,\chi)$ with the identity $$\chi(n)=\frac{1}{G\left(\overline{\chi}\right)}\sum_{a=1}^{q}\overline{\chi}(a)e\left(\frac{an}{q}\right),$$ where $e(x)=e^{2\pi i x}$. Be careful about justifying convergence. It takes some work, and you have to argue why a certain sum with the logarithm of the sin function will cancel out. (This requires $\chi(-1)=-1$, and using the symmetry) Then apply the functional equation for $L(s,\chi)$ to get $L(0,\chi)$ from $L(1,\chi)$. (The $\kappa$ in the $\sin(x)$ factor is what causes $L(0,\chi)=0$ when $\chi(-1)=1$)

I hope that helps.

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Even if $\chi(-1) = 1$, we still have: $$L(0,\chi) = -\frac{1}{m}\sum_{c=1}^m c\chi(c)$$ What's important is the assumption $\chi$ is not the principal character. – Eric Jul 11 '13 at 11:33

(I only prove the formula for $L(0,\chi)$, I'd like some hints for proving $L'(0,\chi)$ without the functional equation)

suppose $F(s) = \sum_{n=1}^\infty a_n n^{-s}$ converges for $Re(s) > 0$. with $A(x) = \sum_{n \le x} a_n$ :

$$F(s) = s \int_1^\infty A(x) x^{-s-1} dx = s \int_1^\infty \left(\overline{A}+(A(x)-\overline{A})\right) x^{-s-1} dx = \overline{A} + s\int_1^\infty (A(x)-\overline{A}) x^{-s-1} dx$$

suppose now that $A(x)$ is $q$ periodic, with $\overline{A} = \frac{1}q\int_0^q A(x) dx$ the mean-value of $A(x)$, $\int_1^\infty (A(x)-\overline{A}) x^{-s-1} dx$ converges for $Re(s) > -1$ and $$F(0)= \overline{A} =\frac{1}q\int_0^q A(x) dx = \frac1q\sum_{n=1}^q (q-n) \ a_n = -\frac1q\sum_{n=1}^q n \ a_n$$ since $\sum_{n=1}^q a_n = 0$. (in the same way we get $F'(0) = \int_1^\infty \frac{A(x)-\overline{A}}{x} dx$)

  • if $\chi$ is a non-principal character modulo $q$, choosing $a_n = \chi(n), F(s) = L(s,\chi)$ yields $A(x) = \sum_{n \le x} \chi(n)$ which is $q$ periodic because $\sum_{n=1}^q \chi(n) = 0$ hence $$L(0,\chi) = -\frac{1}{q}\sum_{n=1}^q n \chi(n)$$

  • if $\chi$ is the principal character modulo $q$ ( i.e. $\chi(n) = 1$ if $gcd(n,q) = 1$, $\chi(n) = 0$ otherwise) we get $L(s,\chi) = \sum_{n=1}^\infty \chi(n) n^{-s} = \zeta(s) \prod_{p | q} (1-p^{-s})$ and choosing $a_n = (-1)^{n+1}$ yields $F(s) = \eta(s) = (1-2^{1-s})\zeta(s)$, $A(x)$ is $2$ periodic $$\eta(0) = \overline{A}= -\frac{1}{2}\sum_{n=1}^2 n(-1)^{n+1} = 1/2$$ hence $$\zeta(0) = -1/2$$ and if $q \ge 2$ : $$L(0,\chi) = \lim_{s \to 0} \zeta(s) \prod_{p | q} (1-p^{-s}) = \zeta(0) \lim_{s \to 0} \prod_{p | q} (1-p^{-s}) = 0$$.

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