# Question related to derived formula for zeta-zero counting function

This question assumes the definitions of the Mellin and Fourier transforms illusrated in (1) and (2) below and the corresponding relationship illustrated in (3) below.

(1) $$\quad\mathcal{M}_x[f(x)](s)=\int\limits_0^\infty f(x)\,x^{s-1}\,dx$$

(2) $$\quad\mathcal{F}_x[f(x)](t)=\int\limits_{-\infty}^\infty f(x)\,e^{-i\,t\,x}\,dx$$

(3) $$\quad\mathcal{F}_u\left[f\left(e^u\right)\right](s)=\mathcal{M}_x[f(x)](-i\,s)$$

The von Mangoldt explicit formula for the second Chebyshev function and its first-order derivative are as follows.

(4) $$\quad\psi_o(x)=x-\sum\limits_{k=1}^\infty\left(\frac{x^{\rho_k}}{\rho_k}+\frac{x^{\rho_{-k}}}{\rho_{-k}}\right)-\log(2\,\pi)-\sum\limits_{n=1}^\infty\frac{x^{-2\,n}}{-2\,n}\\$$ $$\qquad\qquad\quad=x-\sum\limits_{k=1}^\infty\left(\frac{x^{\rho_k}}{\rho_k}+\frac{x^{\rho_{-k}}}{\rho_{-k}}\right)-\log(2\,\pi)-\frac{1}{2}\log\left(1-\frac{1}{x^2}\right)$$

(5) $$\quad\psi_o'(x)=1-\sum\limits_{k=1}^\infty\left(x^{\rho_k-1}+x^{\rho_{-k}-1}\right)-\sum\limits_{n=1}^\infty x^{-2\,n-1}\\$$ $$\qquad\qquad\quad=1-\sum\limits_{k=1}^\infty\left(x^{\rho_k-1}+x^{\rho_{-k}-1}\right)+\frac{1}{x-x^3}$$

I've been investigating derivation of a zeta-zero counting function based on the $$f(x)$$ function defined in (6) below, the Mellin transform defined in (7) below, and the subsequent integral illustrated in (8) below all of which assume the Riemann Hypothesis (as does this entire question). Note $$f(x)$$ defined in (6) below is related to $$\psi_o'(x)$$ defined in (5) above.

(6) $$\quad f(x)=\sum\limits_{k=1}^\infty\left(x^{\rho_k-1}+x^{\rho_{-k}-1}\right)=\sum\limits_{k=1}^\infty\left(x^{-1/2+i\,\Im(\rho_k)}+x^{-1/2-i\,\Im(\rho_k)}\right)$$

(7) $$\quad F(t)=\frac{1}{2\,\pi}\mathcal{M}_x\left[x^{1/2}f(x)\right](-i\,t)=\sum\limits_{k=1}^\infty\left(\delta\left(t-\Im\left(\rho_k\right)\right)+\delta\left(t+\Im\left(\rho_k\right)\right)\right)$$

(8) $$\quad N(T)=\int\limits_0^T F(t)\,dt=\sum\limits_{k=1}^\infty \left(\theta\left(T-\Im\left(\rho_k\right)\right)+\theta\left(T+\Im\left(\rho_k\right)\right)\right)$$

The Mellin transform illustrated in (7) above follows from the Mellin transform illustrated in (9) below which can also be understood in terms of the equivalent Fourier transform illustrated in (10) below.

(9) $$\,\,\,\mathcal{M}_x\left[x^{1/2}\left(x^{-1/2+i\,\Im\left(\rho_k\right)}+x^{-1/2-i\,\Im\left(\rho_k\right)}\right)\right](-i\,t)=2\,\pi\left(\delta\left(t-\Im\left(\rho_k\right)\right)+\delta\left(t+\Im\left(\rho_k\right)\right)\right)$$

(10) $$\quad\mathcal{F}_u\left[\left(e^u\right)^{1/2}\left(\left(e^u\right)^{-1/2+i\,\Im\left(\rho_k\right)}+\left(e^u\right)^{-1/2-i\,\Im\left(\rho_k\right)}\right)\right](t)=\\$$ $$\qquad\quad\mathcal{F}_u\left[e^{u/2}\left(2\,e^{-u/2}\,\cos\left(\Im\left(\rho_k\right)\log \left(e^u\right) \right)\right)\right](t)=2\,\pi\left(\delta \left(t-\Im\left(\rho_k\right)\right)+\delta\left(t+\Im\left(\rho_k\right)\right)\right)$$

The Mellin transform defined in (7) above can be approximated using finite integration limits as illustrated in (10) below, and the result can then be integrated as illustrated in (11) below to approximate the $$N(T)$$ function defined in (8) above.

(11) $$\quad\overset{\text{~}}{F}(t)=\frac{1}{2\,\pi}\int\limits_{1/X}^X x^{1/2}\,f(x)\,x^{-i\,t-1}\,dt=\frac{i}{2\,\pi}\sum\limits_{k=1}^K\left(\frac{X^{-i\,\left(t-\Im\left(\rho_k\right)\right)}-X^{i\,\left(t-\Im\left(\rho_k\right)\right)}}{t-\Im\left(\rho_k\right)}+\frac{X^{-i\,\left(t+\Im\left(\rho_k\right)\right)}-X^{i\,\left(t+\Im\left(\rho_k\right)\right)}}{t+\Im\left(\rho_k\right)}\right)\\$$ $$\qquad\qquad\qquad\qquad\qquad=\sum\limits_{k=1}^K\left(\frac{\sin\left(\log(X)\,\left(t-\Im\left(\rho_k\right)\right)\right)}{\pi\,\left(t-\Im\left(\rho_k\right)\right)}+\frac{\sin\left(\log(X)\,\left(t+\Im\left(\rho _k\right)\right)\right)}{\pi\,\left(t+\Im\left(\rho _k\right)\right)}\right),\quad X\to\infty\land K\to\infty$$

(12) $$\quad \overset{\text{~}}{N}(T)=\int\limits_0^T\overset{\text{~}}{F}(t)\,dt=\sum\limits_{k=1}^K\left(\frac{\text{Si}\left(\log(X)\left(T-\Im\left(\rho_k\right)\right)\right)}{\pi}+\frac{\text{Si}\left(\log(X)\left(T+\Im\left(\rho_k\right)\right)\right)}{\pi}\right),\quad X\to\infty\land K\to\infty$$

The following plot illustrates $$\overset{\text{~}}{F}(t)$$ defined in formula (11) above evaluated at $$X=10,000$$ and $$K=10$$. The red discrete portion of the plot illustrates $$\frac{1}{2}\left(\overset{\text{~}}{F}\left(\Im\left(\rho_k\right)-\epsilon\right)+\overset{\text{~}}{F}\left(\Im\left(\rho_k\right)+\epsilon\right)\right)$$ where $$\epsilon=0.01$$.

Figure (1): Illustration of $$\overset{\text{~}}{F}(t)$$ defined in Formula (11)

The following plot illustrates $$\overset{\text{~}}{N}(T)$$ defined in formula (12) above evaluated at $$X=10,000$$ and $$K=10$$. The red discrete portion of the plot illustrates $$\frac{1}{2}\left(\overset{\text{~}}{N}\left(\Im\left(\rho_k\right)-\epsilon\right)+\overset{\text{~}}{N}\left(\Im\left(\rho_k\right)+\epsilon\right)\right)$$ where $$\epsilon=0.01$$.

Figure (2): Illustration of $$\overset{\text{~}}{N}(T)$$ defined in Formula (12)

Question (1): Can a zeta-zero counting function be derived from a non-zeta-zero representation of the first-order derivative $$\psi'(x)$$ of the second Chebyshev function?

I've been attempting to derive a zeta-zero counting function based on (6), (7), and (8) above using a non-zeta-zero representation of the first-order derivative $$\psi'(x)$$ of the second Chebyshev function and have not yet been successful, but the approach I've been investigating is outlined below.

Note the $$f(x)$$ function defined in (6) above obeys the functional equation illustrated in (13) below which follows from (14) below. Note when the right-side of (14) below is simplified, the two terms on the right-side are in reverse-order of the two terms on the left-side.

(13) $$\quad f(x)=\frac{1}{x}f(\frac{1}{x})$$

(14) $$\quad\left(x^{-1/2+i\,\Im(\rho_k)}+x^{-1/2-i\,\Im(\rho_k)}\right)=\frac{1}{x}\left(\left(\frac{1}{x}\right)^{-1/2+i\,\Im(\rho_k)}+\left(\frac{1}{x}\right)^{-1/2-i\,\Im(\rho_k)}\right)$$

Assuming the definition of $$g(x)$$ defined in (15) below, $$f(x)$$ is approximated in the intervals for $$x>1$$ and $$0 as illustrated in (16) and (17) below. Note (16) below follows from (5) above (since $$\psi'(x)$$ is approximated by $$\psi_o'(x)$$ for $$x>1$$), and (17) below follows from (13) above.

(15) $$\quad g(x)=1-\psi'(x)-\sum\limits_{n=1}^\infty x^{-2\,n-1}=1-\psi'(x)+\frac{1}{x-x^3}$$

(16) $$\quad\overset{\text{~}}{f}(x)=g(x)\,,\qquad 1

(17) $$\quad\overset{\text{~}}{f}(x)=\frac{1}{x}g(\frac{1}{x})\,,\quad 0

Based on the approximations defined in (16) and (17) above, I've been thinking perhaps $$F(t)$$ defined in (7) above can be approximated as illustrated in (18) below (using a non-zeta-zero representation of $$\psi'(x)$$ in $$g(x)$$ defined in (15) above), and the result can then be integrated as illustrated in (19) below to approximate $$N(T)$$ defined in (8) above.

(18) $$\quad\overset{\text{~}}{F}(t)=\frac{1}{2\,\pi}\left(\int\limits_1^\infty x^{1/2}\,g(x)\,x^{-i\,t-1}\,dx+\int\limits_0^1 x^{1/2}\left(\frac{1}{x}\,g(\frac{1}{x})\right) x^{-i\,t-1}\,dx\right)\\$$ $$\qquad\qquad\quad=\frac{1}{2\,\pi}\left(\int\limits_1^\infty x^{1/2}\,g(x)\,x^{-i\,t-1}\,dx+\int\limits_0^1 x^{-1/2}\,g(x^{-1})\,x^{-i\,t-1}\,dx\right)$$

(19) $$\quad\overset{\text{~}}{N}(T)=\int\limits_0^T\overset{\text{~}}{F}(t)\,dt$$

The approach outlined above is an oversimplification since there are convergence issues that need to be addressed which are described in the two answers posted by reuns below. After reading the two answers posted by reuns below, I realized the approach I've been investigating is related to Weil's explicit formula, and the two answers posted below are probably a more correct way to approach the derivation of a zeta-zero counting function from a non-zeta-zero representation of $$\psi'(x)$$.

About the explicit formulas and the possible mistakes you'll find here and there. The main point is to look at $$\psi_{1/2}(x)= \sum_{n \le x} \Lambda(n) n^{-1/2}$$ For $$\sigma > 1/2$$ and under the RH for $$\sigma > 0$$ $$\psi_{1/2}(x)- 2 x^{1/2} 1_{x > 1} = 1_{x > 1}\frac{1}{2i\pi} \int_{\sigma-i\infty}^{\sigma+i\infty}( -\frac{\zeta'(s+1/2)}{\zeta(s+1/2)} -\frac{1}{s-1/2}) \frac{x^s}{s}ds$$ $$= 1_{x > 1} \sum Res(( -\frac{\zeta'(s+1/2)}{\zeta(s+1/2)} -\frac{1}{s-1/2}) \frac{x^s}{s})\\= -1_{x > 1} (\sum_t \frac{x^{it}}{it}+\sum_{k=1}^\infty \frac{x^{-2k-1/2}}{-2k-1/2}-\frac{\zeta'(1/2)}{\zeta(1/2)}-\frac{1}{-1/2})$$

As you know those things don't converge absolutely and to show they do and the formulas are valid you need a lot of complicated estimates. Now the density of zeros implies the series over the zeros is a tempered distribution and for things like $$\phi(s) = \Gamma(s)=\mathcal{M}[ e^{-x}](s)$$ (meromorphic and Schwartz on vertical lines) it is much easier to look at the regularized version

$$(\psi_{1/2}(x)- 2 x^{1/2} 1_{x > 1}) \ast_{\mathcal{M}} \mathcal{M}^{-1}[\phi(s)](x) = \frac{1}{2i\pi} \int_{\sigma-i\infty}^{\sigma+i\infty} \phi(s)( -\frac{\zeta'(s+1/2)}{\zeta(s+1/2)} -\frac{1}{s-1/2}) \frac{x^s}{s}ds$$ $$= \sum Res(\phi(s)( -\frac{\zeta'(s+1/2)}{\zeta(s+1/2)} -\frac{1}{s-1/2}) \frac{x^s}{s})$$ because this time the convergence is absolute and the residue theorem applies without problems. And it implies the formula above are true in the sense of distributions, from which the Weil explicit formula follows :

Letting $$x = e^u$$ and extending by odd symmetry we have the tempered distribution. $$F(u) = -(\psi_{1/2}(e^u)- 2 e^{u/2} 1_{u > 0})+(\psi_{1/2}(e^{-u})- 2 e^{-u/2} 1_{u > 0})$$ Its distributional derivative is even $$F'(u) = 4 \delta(u)+e^{|u|/2}-\sum_n \Lambda(n) n^{-1/2}(\delta(u-\ln n)+\delta(u+\ln n))$$ Differentiating the explicit formula, first for $$u > 0$$, then for $$u <0$$, then adding the missing $$\delta(u)$$ at $$0$$ we obtain $$F'(u) = 4 \delta(u)+ 1_{u > 0} (\sum_\rho e^{itu}+\sum_{k=1}^\infty e^{-(2k+1/2)u})+1_{u < 0} (\sum_\rho e^{-itu}+\sum_{k=1}^\infty e^{(2k+1/2)u})$$ $$=4 \delta(u)+ \sum_t e^{it u} + \sum_{k=1}^\infty e^{-(2k+1/2)|u|}=4 \delta(u)+ \sum_t e^{it u} +pv.(\frac{e^{-|u|/2}}{e^{2|u|}-1})$$

Overall it means for any Schwartz function $$\varphi$$ then $$\lim_{N \to \infty}\int_{-N}^N \varphi(u) e^{|u|/2}du-\sum_{n=1}^{e^N} \Lambda(n) n^{-1/2}(\varphi(\ln n)+\varphi(-\ln n)) \\= \lim_{N \to \infty}\int_{-N}^N \varphi(u) (4 \delta(u)+ \sum_t e^{it u} +pv.(\frac{e^{-|u|/2}}{e^{2|u|}-1})) du\\ = \sum_t \hat{\varphi}(t) + 4 \varphi(0) + \lim_{c \to 0}\int_{-\infty}^{-c}+\int_c^\infty\varphi(u) \frac{e^{-|u|/2}}{e^{2|u|}-1} du$$ which is the Weil explicit formula.

• Your formula is supposed to be useful for what ? Knowing the zeros you can find the zeros, this is tautologic. If $$h(u) = \sum_l \frac{1}{g(w_l)}e^{i u w_l}$$ then $$\sum_{ |w_l| < T} 1 = \frac{1}{2\pi}\int_{-\infty}^\infty \mathcal{F}[h](w) 1_{|w| < T} g(w)dw= \int_{-\infty}^\infty h(u) \overline{\mathcal{F}^{-1}[1_{|w| < T} \overline{g(w)}](u)} du$$

This is how you obtained your formula. If the series converges only in the sense of distributions you need to replace $$1_{|w| < T}$$ by a $$C^\infty_c$$ approximation of it (being careful with the $$w_l$$ close to $$T$$)

• The explicit formula says that under the RH, the Fourier transform of $$\sum_n \frac{\Lambda(n)}{n^{1/2}} (\delta(u-\ln n)+\delta(u+\ln n))=e^{-u/2}\psi'(e^u)+e^{u/2}\psi'(e^{-u})$$ is $$f(w)+2\pi\sum_{t = \Im(\rho)} \delta(w- t)$$ where $$f$$ is a simple term involving the pole at $$s=1$$ and a sum over the trivial zeros.

Let $$\phi \in C^\infty_c([-1,1]),\int_{-\infty}^\infty \phi(w)dw = 1$$ and $$\psi_{T,m} = 1_{[-T,T]} \ast m\phi(m.)$$ a $$C^\infty_c$$ approximation of $$1_{[-T,T]}$$ (can you replace $$C^\infty_c$$ by $$C^1_c$$ ?).

If there are no zeros on $$1/2+i[T-1/m,T+1/m]$$ then

$$N(T) = \sum_{|\Im(\rho)| < T} 1 = \int_{-\infty}^\infty \psi_{T,m}(w) (\sum_{t = \Im(\rho)} \delta(w- t))dw$$ $$= \frac{1}{2\pi}\int_{-\infty}^\infty \mathcal{F}^{-1}[\psi_{T,m}](u) (\sum_n \frac{\Lambda(n)}{n^{1/2}} (\delta(u-\ln n)+\delta(u+\ln n)))du - \int_{-\infty}^\infty \psi_{T,m}(w)f(w)dw$$

Approximating $$e^{-u/2}\psi'(e^u)+e^{u/2}\psi'(e^{-u})$$ with the approximate explicit formula for $$\psi$$ you get your formula again.

• As I said my ultimate objective is to derive a formula for a zeta-zero counting function from a non-zeta-zero representation of $\psi(x)$ or $\psi'(x)$. The purpose of the formula derived from the zeta-zero representation of $\psi'(x)$ is to validate and explore the approach I have in mind and gain insight as to whether this approach can be translated to a non-zeta-zero representation. – Steven Clark Feb 17 at 5:09
• Shouldn't $e^{-u/2}\psi'(e^u)+e^{u/2}\psi'(e^{-u})$ be $e^{u/2}\psi'(e^u)+e^{-u/2}\psi'(e^{-u})$ since $\mathcal{F}_u\left[\frac{\Lambda(n)}{\sqrt{n}}(\delta(u-\log(n))+\delta(u+\log(n)))\right](w)=\\$ $\mathcal{F}_u\left[\Lambda(n)\left(e^{u/2}\delta\left(e^u-n\right)+e^{-\frac{u}{2}}\delta\left(e^{-u}-n\right)\right)\right](w)=\\$ $\mathcal{M}_x\left[\Lambda(n)\left(x^{1/2}\delta(x-n)+x^{-1/2}\delta\left(x^{-1}-n\right)\right)\right](-i\,w)=\\$ $\Lambda(n)\left(n^{-\frac{1}{2}+i\,w}+n^{-\frac{1}{2}-i\, w}\right)$? – Steven Clark Mar 1 at 4:58