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Write $F(s)=-{\zeta }^{\prime }(s)/\zeta (s)$ and ${\psi }_{\Gamma }={\Gamma }^{\prime }/\Gamma$. The global partial-fraction expansion of the zeta logarithmic derivative isHere $\gamma$ is the Euler--Mascheroni constant, and the sum includes multiplicities of the Nontrivial zeros of the Riemann zeta function. The paired terms are locally absolutely convergent away from their poles, by the Riemann–von Mangoldt formula. This identity follows by differentiating the genus-one Hadamard factorization of the Riemann xi function and using the Gamma function recurrence.

We first prove the classical zero-free region of the Riemann zeta function. The Euler-product nonvanishing of the Riemann zeta function and its functional equation place every nontrivial zero in the closed strip $0\le \beta \le 1$; no assertion about zeros on its boundary is needed yet. The real logarithmic derivative of Riemann xi and the digamma function asymptotic imply, for $1<\sigma \le 2$ and $|t|\ge 1$,Every summand being subtracted is nonnegative. If $\beta +it$ is a zero, retaining its summand gives $\Re F(\sigma +it)\le C\log (|t|+2)-1/(\sigma -\beta )$; at height $2t$ the upper bound $C\log (|t|+2)$ suffices. At real $\sigma$ the simple pole of $\zeta$ gives $F(\sigma )=1/(\sigma -1)+O(1)$.

For $\sigma >1$, the Dirichlet series $F(s)={\sum }_n\Lambda (n)n^{-s}$ has nonnegative coefficients. The identity $3+4\cos u+\cos 2u=2(1+\cos u{)}^2\ge 0$ therefore gives the three-four-one zero-free-region argumentWith $L=\log (|t|+2)$, a zero at $\beta +it$ would consequently satisfyChoose $\sigma =1+a/L$, with an absolute $a>0$ sufficiently small, and put $c=a/8$. If $\beta \ge 1-c/L$, this inequality would imply $4/(a+c)\le 3/a+C$. But $4/(a+a/8)-3/a=5/(9a)>C$ for small enough $a$, a contradiction.

For any fixed $t\ne 0$, the same positivity inequality as $\sigma \downarrow 1$ excludes a zero at $1+it$: its negative contribution has coefficient at least four, whereas the real pole contributes only three. Any zero at $1+2it$ only strengthens the contradiction. The pole at $s=1$ has a zero-free punctured neighbourhood. By compactness, the remaining bounded-height portion of the line has a zero-free neighbourhood as well. Shrink $c$ to combine these facts with the large-height bound. Hencewith the understanding that $s=1$ is a pole.

The Riemann–von Mangoldt explicit formula for the Second Chebyshev function, with half weight at a prime-power endpoint, isThe zero sum uses symmetric height truncation. To estimate it, take a half-integer $x$ and use the truncated explicit formula for the second Chebyshev functionThe zero-free region gives $|x^{\rho }|\le x\exp (-c\log x/\log (T+2))$ for these zeros. The Riemann–von Mangoldt formula and partial summation give ${\sum }_{|\Im \rho |\le T}|\rho {|}^{-1}\ll {\log }^2(T+2)$. Choose $T=\exp (\sqrt{\log x})$. Both the zero sum and the truncation error are then $O(x{e}^{-c^{\prime }\sqrt{\log x}})$ after reducing the positive constant to absorb logarithmic factors. Replacing an arbitrary $X$ by $\lfloor X\rfloor +1/2$ preserves the Von Mangoldt function sum and changes the main term by at most one. This proves the requested form of the Prime number theorem:

For $\sigma >1$, unique prime factorization and absolute convergence give the Euler productEvery factor is nonzero and the product converges to a nonzero limit. Equivalently, the absolutely convergent identityprovides a reciprocal. This proves the Euler-product nonvanishing of the Riemann zeta function.