TITLE: Geometric Mean of $L(1,\chi)$ for quadratic Dirichlet characters
QUESTION [14 upvotes]: Let $S = \{D_1, D_2, D_3, \ldots \}$ be the set of all prime discriminants 
(or positive prime discriminants) of quadratic number fields. For such a 
discriminant let $\chi_j(n) = (\frac{D_j}n)$ be the associated Dirichlet 
character and 
$$ L(1,\chi_j) = \sum_{n \ge 1} \frac1n \Big(\frac{D_j}{n}\Big) $$
the value of Dirichlet's L-series at $s = 1$. If we assume that
about half prime discriminants $D$ have $(D/p) = +1$ and the other half
satisfy $(D/p) = -1$, and if we interchange the limits, then the
geometric mean of the values of $L(1,\chi_j)$ is given by 
\begin{align*}
 \lim_{k \to \infty} \bigg( \prod_{j=1}^k L(1,\chi_j) \bigg)^{1/k} 
   & = \lim_k \prod_p \Big(\frac{p}{p-1}\Big)^{\frac{k}{2k}}
        \cdot \Big(\frac{p}{p+1}\Big)^{\frac{k}{2k}} \\
   &= \prod_p \Big(\frac{p^2}{p^2-1}\Big)^{1/2} = \sqrt{\zeta(2)}  
     = \frac{\pi}{\sqrt{6}}. 
\end{align*}
Has this result due to Scholz been studied anywhere?
Scholz also believed that if $S$ denotes the set of all fundamental
discriminants, then the corresponding limit is equal to
$$ \prod \Big( \frac{p^2}{p^2-1} \Big)^{\frac{p}{2p+2}}. $$

REPLY [15 votes]: The distribution of values of $L(1,\chi_d)$ as $d$ varies over fundamental discriminants has been extensively studied.  For example, see this paper of Granville and Soundararajan which gives uniform such results (and discusses other references and history).  The main result there shows that $L(1,\chi_d)$ is distributed like a random Euler product $L(1,X)= \prod_p (1-X(p)/p)^{-1}$ where $X(p)$ for primes $p$ denote independent random variables with $X(p)=1$ with probability $p/(2(p+1))$, $-1$ with probability $p/(2(p+1))$ and $0$ with probability $1/(p+1)$.   From this the last assertion you make about fundamental discriminants follows: you want to compute (for large $D$)
$$ 
\exp\Big( \frac{1}{|\{|d|\le D\}|} \sum_{|d|\le D} \log L(1,\chi_d) \Big)
\sim \exp\Big({\Bbb E} (\log L(1,X)) \Big) = \prod_p \Big(\frac{p^2}{p^2-1}\Big)^{\frac{p}{2p+2}}.
$$
Small modifications to the same techniques would allow you to study the family of prime discriminants that you mentioned -- the only difference is in adjusting the probabilistic model to reflect the fact that very few prime discriminants will be divisible a given prime $p$ (as opposed to all fundamental discriminants where this proportion is $1/(p+1)$).