 Solution

The permitted zero-count bound $N(T)\ll T\log (T+2)$ implies, by partial summation,Indeed, the zeros with $|\rho |\le 1$ contribute a bounded amount since $\xi (0)\ne 0$; the rest contribute $N(T)/T+{\int }_1^{T}N(u)u^{-2}du$. For every zero in the sum, the preceding zero-free region of the Riemann zeta function givesThe supplied truncated explicit formula for the second Chebyshev function therefore impliesChoose $T=\exp (a\sqrt{\log x})$, with any fixed positive $a$ and then, for example, $a=\sqrt{c}$. Both terms have exponential decay in $\sqrt{\log x}$, and the logarithmic factors can be absorbed by slightly decreasing its coefficient. Hence, for some absolute $b>0$,If the truncated formula is taken only at admissible heights and non-prime-power endpoints, choose such heights of the same size and first apply it at half-integers. Monotonicity and the $O(\log x)$ jump of $\psi$ at an integer transfer this bound to all $x$; these convention choices do not change the conclusion.

To pass from prime powers to primes, let $\vartheta (x)={\sum }_{p\le x}\log p$. The higher prime-power contribution satisfiesThus $\vartheta (x)\sim x$. Apply partial summation again:The integral is $O(x/{\log }^{2}x)$: use $\vartheta (t)\ll t$ and split at $\sqrt{x}$. It is therefore smaller than the first term, which is asymptotic to $x/\log x$. This proves the Prime number theorem in the requested form