5
$\begingroup$
  • Let $$W(z)=\sum_{n=1}^\infty \frac{\Lambda(n)}{n^{1/2}} e^{-nz}, \qquad\Re(z) > 0$$ For $\frac{y}{2\pi}=\frac{a}{q} \in \mathbb{Q}$, as $x \to 0^+$ we have $$W(x-iy) -{\scriptstyle \underset{(n,q) > 1}{\sum} \frac{\Lambda(n)}{n^{1/2}} e^{iny}e^{-nx} }= \sum_{\chi \bmod q} \frac{\widehat{\chi}(a)}{\phi(q)}\sum_{n=1}^\infty \frac{\Lambda(n)}{n^{1/2}} \chi(n) e^{-nx}\\= \frac{\sqrt{\pi}\frac{\mu(q)}{\phi(q)}}{x^{1/2}}+O(x^{1/2-\sigma_q-\epsilon})$$ where $\widehat{\chi}(a) = \sum_{n=1}^q \chi(n) e^{-2i \pi n a/q}$ and $\Re(s) > \sigma_q$ is the largest zero-free half-plane of the $L(s,\chi),\chi \bmod q$ (for simplicity assume two $L(s,\chi),\chi \bmod q$ don't have any non-trivial zero in common). The main term $\frac{\sqrt{\pi}\mu(q)/\phi(q)}{x^{1/2}}$ comes from the pole of $\zeta(s)$ and the mean-value of $\Lambda(n)e^{2i \pi n a/q}$. So $W(x-iy),x \to 0^+$ encodes the generalized Riemann hypothesis for the $\chi \bmod q$.

  • From an inverse Mellin transfom and the residue theorem we obtain the explicit formula $$W(z)=\frac{1}{2i\pi} \int_{2-i\infty}^{2+i\infty} \Gamma(s) \frac{-\zeta'}{\zeta}(s+1/2) z^{-s}ds=\sum Res[\Gamma(s) \frac{-\zeta'}{\zeta}(s+1/2)z^{-s}]$$ $$= \frac{\sqrt{\pi}}{z^{1/2}}-\sum_t \Gamma(it)z^{-it}+\sum_{k=0}^\infty z^{k/2} (b_k+c_k \log(z))\tag{2}$$

    where $it +1/2=\rho$ the sum being over the non-trivial zeros of $\zeta(s)$. The Stirling approximation and the density of zeros show it converges for $\Re(z) > 0$.

  • Note $x-iy \approx y e^{-i(\pi/2-x/y)}$ as $x \to 0^+$. From $(2)$ we find $$W(x-iy) = W(y e^{-i(\pi/2-x/y)})+O(x^{1/2-\epsilon})\\=\sum_{t > 0} \Gamma(it) (y e^{-i(\pi/2-x/y)})^{-it}+\alpha+O(x^{1/2-\epsilon}) $$ $$= \sqrt{-2i\pi}\sum_{t >0} e^{i t (\log (t)-1)} t^{-1/2} y^{-it} e^{-t x/y}+\alpha+O(x^{1/2-\epsilon}) \tag{3}$$

    the last step comes from the Stirling approximation. We keep only the sum over $t > 0$ because the sum over $t < 0$ is analytic around $x=0$ but not around $ x= \pi$

  • Taking the Mellin transform we obtain the generalized Dirichlet series $$ \sqrt{-2i\pi} \sum_{t > 0} e^{it (\log (t)-1)} y^{s-it}t^{-1/2-s} \simeq \frac{1}{\Gamma(s)}\int_0^\infty W(x-iy) x^{s-1}dx=\sum_{n=1}^\infty \frac{\Lambda(n)}{n^{1/2}}e^{iny} n^{-s} $$ where $\simeq$ means the difference is analytic for $\Re(s) > -1/2$ (except a pole at $s=0$). Under $y \mapsto y+2\pi$, the RHS is invariant, thus the LHS is almost invariant.

Question : How to explain that $\sum_{t > 0} e^{it (\log (t)-1)} y^{s-it}t^{-1/2-s}$ is almost invariant under $y \mapsto y+2\pi$, what does it mean in term of the distribution of the zeros ?

There is a plot showing $T \mapsto \sqrt{-2i\pi}\sum_{t \le T y} e^{it (\log (t)-1)} y^{-it}t^{-1/2}$ for $y = 2\pi,4\pi$ and $6\pi$ (the imaginary part is bounded and $400$ times smaller) enter image description here

Note a similar discussion holds when $\Lambda(n)$ is replaced by the coefficients of the logarithmic derivative of $L(s,\chi_5)+L(s,\overline{\chi_5})$, a Dirichlet series whose twists by Dirichlet characters are meromorphic and whose functional equation tell us some bounds on the logarithmic derivative needed for the explicit formula

$\endgroup$
  • $\begingroup$ You refer to the largest zero-free region of Dirichlet $L$-functions mod $q$ as ${\rm Re}(s) > \sigma_q$. That is a uniform right half-plane. Do you know some argument for $\sigma_q < 1$? $\endgroup$ – KConrad Oct 5 '18 at 3:39
  • $\begingroup$ @KConrad it is very possible $\sigma_q=1$. At first I was searching for a different proof of the PNT on the power series side. I don't know what is the estimate for $W(x)$ equivalent to $\sum_{n \le x} \Lambda(n) = x+O(\frac{x}{\log ^k x} )$ $\endgroup$ – reuns Oct 5 '18 at 3:56
  • $\begingroup$ I have not followed all the details in your post (for lack of time), but is the last display not the explanation you are looking for? I mean, the RHS is invariant under $y\mapsto y+2\pi$, hence the LHS is almost invariant. $\endgroup$ – GH from MO Oct 5 '18 at 14:11
  • $\begingroup$ @GHfromMO Yes that's why the LHS should be almost invariant too. I updated a few details. $\endgroup$ – reuns Oct 6 '18 at 13:25
  • $\begingroup$ @reuns: If you proved an identity, LHS=RHS, and the RHS is almost invariant, then the LHS is almost invariant, too. What other explanation are you looking for? To me, the words "explanation" and "proof" are synonyms. $\endgroup$ – GH from MO Oct 6 '18 at 22:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.