Here is the definition of $\xi(s,\chi)$:


$\xi(s,\chi)= \left(\frac{s(s-1)}{2} \right)^{1\_{\chi=1}} (q/\pi)^{\frac{s+a}{2}} \Gamma \left( \frac{s+a}{2} \right) L(s,\chi)$


Here is the definition of the **Hadamard product applied to $\xi(s,\chi)$:**


Since it is an entire function of order one, there exists constants $A\_\chi, B\_\chi \in \mathbb{C}$ such that
$\xi(s,\chi)=e^{A\_\chi+B\_\chi s} \prod\_{p} \left(1-\frac{s}{p} \right) e^{s/\rho} $


with the product running over every non-trivial zero of $L(s,\chi)$.


**Why does this imply there are infinitely many non-trivial zeros for $L(s,\chi)?$** 


FYI: I know there non trivial zeros are in one to one correspondence with the zeros of $\xi(s,\chi)$



infinitely many non-trivial zeros for $L(s,\chi)$ using Hadamard product for $\xi(s,\chi)$
The existence of the Hadamard product by itself doesn't show the function has any zeroes (it might just be an exponential!) -- you need a little more.


The idea is to study the logarithmic derivative, which by your two expressions above satisfies ($\delta\_\chi = 1$ for the principal character):


$$ -\frac{\xi'}{\xi}(s;\chi) = \delta\_\chi\left(\frac{1}{s}-\frac{1}{1-s}\right)-\frac{1}{2}\log\frac{q}{\pi} -\frac{1}{2}\frac{\Gamma'}{\Gamma}\left(\frac{s+a}{2}\right) -\frac{L'}{L}(s;\chi) $$
$$ = -B\_\chi - \sum\_\rho \left(\frac{1}{s-\rho}+\frac{1}{\rho}\right)\,.$$


Now plug in $s=2+iT$ and examine the terms as $T\to\infty$. In the first row, all terms are bounded except for the logarithmic derivative of the Gamma function, which grows like $\log T$. On the second row, individual terms are bounded in $T$ so there must be infinitely many.


