I have already asked this question here in a different form, but really need an answer.
Let $L(s)$ be a "standard" $L$-function, say with Euler product, functional equation, etc...
(Selberg class if you like), of order 1, and let $\Lambda(s)$ be the completed $L$-function
with gamma factors. We thus have $\Lambda(k-s)=\omega\Lambda^\*(s)$, where $\Lambda^\*$ is
the "dual" Lambda function (example: if $L(s)$ corresponds to a Dirichlet character $\chi$,
$\Lambda^\*$ corresponds to its conjugate), and $\omega$ root number of modulus 1.


Assume for instance that there are no poles. Since $\Lambda$ has order $1$ it has a Hadamard
product $$\Lambda(s)=ae^{bs}\prod\_{\rho}(1-s/\rho)\;,$$
where the product is over the zeros of $\Lambda$ and understood as the limit as $T\to\infty$
of the product for $|\rho|<T$ (on purpose I do not use the more standard $(1-s/\rho)e^{s/\rho}$).


My question is this: do we always have $b=0$ ? This is trivial if $\Lambda^\*=\Lambda$
(self-dual), otherwise the only thing I can prove is that $b$ is purely imaginary.
I have experimented numerically with some non self-dual $L$ functions attached to Dirichlet
characters, and it seems to be true.


Remarks: 1) I may have a proof using the "explicit formula" of Weil, but I am not sure of its
validity, and it seems too complicated. 2) I have a vague memory of Harold Stark mentioning this
result 50 years ago.



Hadamard factorization of L-functions
I believe you are correct and $b$ is zero, although I find it inexplicable why this is not better known (certainly I didn't know it before). Let's stick to a primitive Dirichlet character $\mod q$, but what follows should be applicable in general. If we take logarithmic derivatives, then
$$ 
\frac{\Lambda^{\prime}}{\Lambda}(s) = b + \sum\_{\rho} \frac{1}{s-\rho}, 
$$
with the understanding that the zeros $\rho=\beta+i\gamma$ are counted with $|\gamma|\le T$, and then $T\to \infty$. Let's evaluate the above at $s=R$ for a large real number $R$, and focus just on the imaginary parts.


Now
$$ 
\text{Im} \Big( \frac{\Lambda^{\prime}}{\Lambda}(R)\Big)
$$
tends exponentially to $0$ as $R\to \infty$. So let's look at the imaginary part on the right hand side, which is
$$ 
\text{Im} (b) + \lim\_{T\to \infty} \sum\_{|\gamma|\le T} \frac{\gamma}{(R-\beta)^2 + \gamma^2}.
$$
Note that
$$ 
\sum\_{|\gamma|\le T} \frac{\gamma}{(R-\beta)^2+\gamma^2} 
= \sum\_{|\gamma|\le T}\Big( \frac{\gamma}{R^2+\gamma^2} + O\Big( \frac{R|\gamma|}{(R^2+\gamma^2)^2}\Big)\Big). \tag{1}
$$
To handle the error term, split into the terms $|\gamma|\le R$ and $|\gamma|>R$, obtaining that the error term is
$$ 
\ll \sum\_{|\gamma|\le R} \frac{1}{R^2} + \sum\_{R<|\gamma|} \frac{R}{|\gamma|^3} \ll \frac{\log qR}{R}, 
$$
upon recalling that there are $\ll \log q(|t|+1)$ zeros in an interval of length $1$ (we will recall this more precisely next).


Now the main term in (1) can be handled by partial summation. For $t>0$, put $N^+(t)$ to be the number of zeros of $\Lambda$ with imaginary part between $0$ and $t$, and $N^{-}(t)$ to be the number of zeros with imaginary part between $-t$ and $0$. Then both $N^+$ and $N^-$ satisfy by the argument principle the well known asymptotic formula (for $t\ge 1$)
$$ 
N^+(t), N^{-} (t) = \frac{t}{2\pi} \log \frac{qt}{2\pi e} +O(\log (q(t+1))).
$$
Thus for all $t>0$
$$ 
|N^+(t) - N^-(t)| = O(\log (q(2+t))). 
$$
Now by partial summation
\begin{align\*}
\sum\_{|\gamma|\le T} \frac{\gamma}{R^2+\gamma^2} &= \int\_0^{T} \frac{t}{R^2+t^2} dN^+(t) - \int\_0^T \frac{t}{R^2+t^2} dN^-(t) \\
&= \frac{T}{R^2+T^2} (N^+(T)-N^-(T)) - \int\_0^T (N^+(t)-N^-(t)) \Big( \frac{t}{R^2+t^2}\Big)^{\prime} dt \\ 
&= O\Big(\frac{T\log qT}{R^2+T^2} \Big) + O\Big(\int\_0^T (\log q(t+2)) \Big(\frac{1} {R^2+t^2} + \frac{2t^2}{(R^2+t^2)^2} \Big)dt \Big)\\
&= O\Big( \frac{\log qR}{R}\Big),
\end{align\*}
upon letting $T\to \infty$.


We conclude that the quantity in (1) is $O((\log qR)/R$, and so tends to $0$ as $R\to \infty$.


