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Generally there is no Euler product for Dirichlet L-functions $L(\chi,s)$ in the critical strip.(cf Is the Euler product formula always divergent for 0<Re(s)<1?)

But I would like to know if there is an Euler product for the following product (with $\chi$ a primitive characater):

$$L(\chi,s) L(\overline{\chi},1-s)= \sum\limits_{n=1}^{\infty} \sum\limits_{k=1}^{\infty} \chi(n) \frac{1}{n^{1-s}} \; \overline{ \chi(k)} \frac{1}{k^{s}}$$

in the specific case where $s=\frac{1}{2}+it$

as in this case (product is on prime numbers):

$$S_P(\chi)=\prod\limits_{p=1}^{P} (1- \chi(p) p^{-s})^{-1} (1- \overline{ \chi(p)} p^{s-1})^{-1}= \prod\limits_{p=1}^{P} (1- ( \chi(p) p^{-\frac{1}{2}-it} +\overline{ \chi(p)} p^{-\frac{1}{2}+it})+ p^{-1})^{-1} $$

So the term $( \chi(p) p^{-\frac{1}{2}-it} +\overline{ \chi(p)} p^{-\frac{1}{2}+it})$ is real and if I take the log of the product it seems I have a sum with divergent terms going to $-\infty$ meaning the product converges.

$ln(S_P(\chi))= - \sum\limits_{p=1}^{P}- ( \chi(p) p^{-\frac{1}{2}-it} +\overline{ \chi(p)} p^{-\frac{1}{2}+it})- \sum\limits_{p=1}^{P} p^{-1}+ ...$

So, is this correct and in this specific case an Euler product exits ?

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    $\begingroup$ It is sloppy to say "there is no Euler product" when you really mean to say "there is no proof of convergence of the Euler product." It'd be like saying the Riemann hypothesis is not true when you mean to say it is not proved yet to be true. For nontrivial $\chi$ it is plausible that the Euler product for $L(\chi,s)$ does converge when ${\rm Re}(s) > 1/2$, but there is no proof of that. On the critical line is a surprise: if $\prod_{p} 1/(1 - \chi(p)/\sqrt{p})$ converges to a nonzero value and $\chi$ is quadratic, that value must be not $L(\chi,1/2)$, but $\sqrt{2}L(\chi,1/2)$. $\endgroup$
    – KConrad
    May 15, 2015 at 16:55
  • $\begingroup$ Dear Professor Conrad, Is there a reason why the factor $\sqrt{2}$ occurs? @KConrad $\endgroup$ May 15, 2015 at 17:01
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    $\begingroup$ @DianbinBao, yes, see math.uconn.edu/~kconrad/articles/eulerprod.pdf. Computationally it is a "second moment" effect. $\endgroup$
    – KConrad
    May 15, 2015 at 17:08
  • $\begingroup$ Dear Professor Conrad, you mean convergence of $\sum\limits_{p=1}^{P}- ( \chi(p) p^{-\frac{1}{2}-it} +\overline{ \chi(p)} p^{-\frac{1}{2}+it})$ is not proved and therefore proof that the proposed Euler product converges is pending also ? @KConrad $\endgroup$
    – Bertrand
    May 15, 2015 at 17:27
  • $\begingroup$ Certainly convergence is not yet proved, but more to the point I think it is completely unrealistic to expect convergence to be proved in our lifetime. Convergence of a partial Euler product for $L(\chi,s)$ at any point $s_0$ where $1/2 < {\rm Re}(s_0) < 1$ would imply $L(\chi,s)$ is nonvanishing for ${\rm Re}(s) > {\rm Re}(s_0)$, which goes beyond what anyone has ever shown (i.e., a vertical zero-free strip of positive width to the left of ${\rm Re}(s) = 1$). That already looks very hard, so proving something similar at a point on the critical line seems even more remote. [Continued...] $\endgroup$
    – KConrad
    May 15, 2015 at 22:23

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