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 $$|\rho| (on purpose I do not use the more standard $$(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.

• At least I'd agree that it is not trivial to prove what you want... – paul garrett Aug 30 at 22:51
• Is it clear that $\sum_{|\rho|<T}1/\rho$ has a limit as $T\to\infty$? I only know how to prove this in the self-dual case. – Aurel Aug 31 at 8:56
• @Aurel: I think this is standard, but otherwise include it in the conjectural statement. – Henri Cohen Aug 31 at 9:30
• If I'm not mistaken, your equality should be invariant under the map $s\mapsto 1-\bar{s}$. Does it help? – Sylvain JULIEN Aug 31 at 11:16
• @Sylvain: yes, this proves that $b$ is purely imaginary, but nothing more as far as I can see. – Henri Cohen Aug 31 at 12:40

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 $$R\to \infty$$.

• Another derivation appears in Corollary 10.18 of Montgomery-Vaughan; see also the discussion at the end of Section 10.3, where the proof is attributed to Vorhauer. Related to this is this question and the long discussion in the comments: mathoverflow.net/questions/343248/b-chi-l1-chi-l1-chi-dotsc – Peter Humphries Aug 31 at 22:51
• @PeterHumphries: I recall that other MO question and the discussion, but I don't quite see that Corollary 10.18 is the same as this question. Here one seems to say that $B(\chi)$ is simply $-\sum_\rho 1/\rho$ with the zeros counted as $|\rho|\le T$ (not just that the real parts match, which is obvious). It's this part that I found surprising and don't recall seeing before. – Lucia Sep 1 at 0:01
• I believe that a slightly different argument due to Ihara-Murty-Shimura ("On the log derivatives of Dirichlet L-functions at $s=1$") implies $b=0$ for (finite order) Hecke L-functions. It's Theorem 2 of the paper (subsecs 2.3, 2.4, pp. 4-5), and a key part of the proof is Weil's explicit formula. – Alufat Sep 4 at 17:33
• @Alufat: Thanks for that reference! – Lucia Sep 4 at 17:39
• @Alufat: so my own proof may be correct after all (but too long to post). Thanks for the reference. – Henri Cohen Sep 4 at 19:24