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 ?