Can the prime number theorem be obtained from the explicit formula,


$\psi(x)=x-\sum\_{\zeta(\rho)=0}\frac{x^\rho}{\rho}+O(1)$?


Here, $\psi(x)=\sum\_{k=1}^\infty\sum\_{p^k<x}\log p$



Prime number theorem via the explicit formula
Expanding on my comments above, most textbooks will show how the Prime Number Theorem follows from $\psi(x)\sim x$. This does not require the full strength of the Explicit Formula for $\psi(x)$, and most textbooks will prove the PNT before the Explicit Formula. One needs more than just the real part of each $\rho$ is $<1$; one needs to understand the contribution of all of them. The fact that the infinite series is not absolutely convergent complicates matters.


However, one can consider instead
$$\psi\_1(x)=\int\_0^x\psi(u)\, du=\sum\_{n\le x}(x-n)\Lambda(n)$$
This function also has an Explicit Formula
$$\psi\_1(x)=\frac{x^2}{2}-\sum\_\rho \frac{x^{\rho+1}}{\rho(\rho+1)}-x\frac{\zeta^\prime}{\zeta}(0)+\frac{\zeta^\prime}{\zeta}(-1)-\frac{1}{2} x \log \left(1-\frac{1}{x^2}\right)-\coth ^{-1}(x)$$
and now the sum over $\rho$ *is* absolutely convergent. This is Theorem 28 in Ingham's "The Distribution of Prime Numbers". Ingham comments on p.74



> 
> A generalised form of Theorem 28 was the basis of de la Vallee
> Poussin's proof of the Prime Number Theorem...
> 
> 
> 




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(Updated to address the questions of Steven Clark below)


The explicit formula for $\psi\_1(x)$ is not obtained by integrating the explicit formula for $\psi(x)$ term by term; again that's not allowed without absolute convergence. Instead the approach is via an inverse Mellin transform
$$
I(m)=-\frac{1}{2\pi i}\int\_{C(m)}\frac{x^{s+1}}{s(s+1)}\frac{\zeta^\prime}{\zeta}(s)\, ds,
$$
where $C(m)$ is a large rectangle avoiding the zeros of $\zeta(s)$ The Residue Theorem, standard estimates and letting $m\to \infty$ give the explicit formula. (see Ingham for details.)


