The Prime Number Theorem: Statement and Heuristics

Technique: partial summation (Abel summation), relating $\psi$, $\theta(x)=\sum_{p\le x}\log p$, and $\pi(x)$. Step 1: $\psi(x)$ and $\theta(x)$ differ by a lower-order term. Write $\psi(x) = \sum_{p^k \le x} \log p = \sum_{k \ge 1} \theta(x^{1/k})$, where $\theta(x) = \sum_{p\le x}\log p$, since $\Lambda(p^k)=\log p$ and the sum over prime powers $\le x$ groups by exponent $k$. For $k > \log x/\log 2$ we have $x^{1/k} < 2$ so $\theta(x^{1/k})=0$; hence the sum is finite with at most $\log_2 x$ terms. Each term satisfies $\theta(x^{1/k}) \le \pi(x^{1/k})\log x^{1/k} \le x^{1/k}\log x$ trivially (each of the $\le x^{1/k}$ primes counted contributes at most $\log x$). For $k\ge 2$, $x^{1/k}\le x^{1/2}$, so

Since this error is $o(x)$, we get $\psi(x) = \theta(x) + O(x^{1/2}(\log x)^2)$, so $\psi(x)\sim x \iff \theta(x) \sim x$. Step 2: $\theta(x)$ and $\pi(x)$ are linked by partial summation. We have $\theta(x) = \sum_{p \le x} \log p$. Apply Abel summation with $a_p = 1$ for $p$ prime (else $0$) and $f(t) = \log t$:

(This is the standard Abel summation identity $\sum_{n\le x} a_n f(n) = A(x) f(x) - \int_1^x A(t) f'(t)\,dt$ with $A(t)=\sum_{n\le t} a_n = \pi(t)$, $f'(t) = 1/t$.) Conversely, applying Abel summation with $a_p = \log p$ and $f(t) = 1/\log t$ gives

Step 3: derive the equivalence. Suppose $\theta(x) \sim x$. Then trivially $\theta(x)/\log x \sim x/\log x$. For the integral term, split at $\sqrt{x}$:

The first integral is $O\left(\int_2^{\sqrt x} \frac{t}{t(\log t)^2}dt\right)$ — using the trivial bound $\theta(t) = O(t)$ — which is $O(\sqrt x/(\log 2)^2) = O(\sqrt x)$, negligible compared to $x/\log x$. For the second, since $\theta(t) = O(t)$ throughout and $\log t \ge \tfrac12 \log x$ on $[\sqrt x, x]$,

More carefully: $\theta(t)/t = O(1)$, so the integrand is $O(1/(\log t)^2) = O(1/(\log x)^2)$ on this range, and the range has length $\le x$, giving the bound $O(x/(\log x)^2)$, which is $o(x/\log x)$. Hence

i.e. $\pi(x)\sim x/\log x$. Conversely, suppose $\pi(x) \sim x/\log x$. From Step 2's first formula,

Since $\pi(t) = O(t/\log t)$, the integrand is $O(1/\log t)$, and by the same splitting argument as above the integral is $O(x/\log x) \cdot o(1)$ more precisely $o(x)$ (a direct check: $\int_2^x dt/\log t = \operatorname{Li}(x) = O(x/\log x)$, in fact $\operatorname{Li}(x) \sim x/\log x$ itself by one integration by parts, so the integral term is asymptotically of the same order $x/\log x$ as the main term — but crucially it is $o(x)$). Since $\pi(x)\log x \sim x$, and the correction is $O(x/\log x) = o(x)$, we conclude $\theta(x) \sim x$. Combining with Step 1, $\psi(x) \sim x \iff \theta(x) \sim x \iff \pi(x) \sim x/\log x$. $\blacksquare$

Technique: Perron-type contour representation of $\psi(x)$ via $-\zeta'/\zeta$, and analysis of the pole/zero contributions (the explicit formula method); we present the logical structure completely while citing the standard analytic estimates from Zeta and the Distribution of Primes for the contour manipulations. Setup. For $\operatorname{Re}(s) > 1$,

This follows by taking the logarithmic derivative of the Euler product $\zeta(s) = \prod_p (1-p^{-s})^{-1}$ term by term (justified by locally uniform convergence for $\operatorname{Re}(s)>1$):

A Perron-type formula (a contour integral converting a Dirichlet series into a partial sum, proved by shifting a vertical line integral and picking up residues, standard in Zeta and the Distribution of Primes) gives, for $x$ not an integer and $c>1$,

Shifting the contour left past $s=1$ (a simple pole of $-\zeta'/\zeta$ with residue $1$, inherited from the simple pole of $\zeta$ at $s=1$) and past $s=0$ (a pole from the factor $1/s$, contributing a constant) yields the explicit formula

where the sum ranges over the nontrivial zeros $\rho = \beta+i\gamma$ of $\zeta(s)$ (those with $0 < \beta < 1$), interpreted as a symmetric limit over $|\gamma| \le T$, $T \to \infty$. (The trivial zeros at $s=-2,-4,\dots$ combine with the $1/s$ pole into the last two terms; we will not need their exact form beyond noting they are $O(\log x)$.)

($B \Rightarrow A$), i.e. zero-free line implies PNT. Suppose $\zeta(1+it)\ne 0$ for all real $t$. We must show $\sum_\rho x^\rho/\rho = o(x)$. Fix a large parameter $T$ and split the zero sum at height $T$:

For the tail, using $|x^\rho| = x^\beta \le x$ (since $0<\beta<1$) and the known zero-counting estimate $N(T) = \#\{\rho : |\gamma|\le T\} = O(T\log T)$ (from Zeta and the Distribution of Primes) together with partial summation over $1/|\gamma|$, one shows

which can be made $o(x)$ by letting $T \to \infty$ slowly with $x$ (concretely, choosing $T = T(x) \to \infty$). For the head $|\gamma|\le T$, the key point is that the zero-free region gives, for each fixed such $\rho$, $\beta < 1$ strictly, and in fact quantitatively the classical de la Vallée Poussin zero-free region states there is a constant $c_0>0$ such that no zero has

(This quantitative strengthening of "$\beta<1$" is what Hadamard–de la Vallée Poussin actually prove, by showing $\zeta(\sigma+it) \ne 0$ in a region shrinking logarithmically toward the line $\sigma=1$ as $|t|\to\infty$; the qualitative statement (B) is the boundary case $t$ fixed, $\sigma \to 1$.) Using this bound, each term satisfies

for $|\gamma| \le T$. Choosing $T = T(x) = \exp(\sqrt{\log x})$ (say) makes $\log(T+2) = O(\sqrt{\log x})$, so the exponent $-c_0\log x/\log(T+2)$ behaves like $-c_0\sqrt{\log x}\to -\infty$, and the bound becomes $x \cdot e^{-c\sqrt{\log x}}$ for a suitable $c>0$. Since there are $O(T\log T)$ zeros with $|\gamma|\le T$ and $1/|\rho| = O(1)$ for the smallest ones (bounded below away from $0$), summing the head contributes

and with the above choice of $T$ this is still $o(x)$ (indeed $x\,e^{-c'\sqrt{\log x}}$ for a slightly smaller constant $c'$, after absorbing the polynomial-in-$\log x$ factor $T\log T$). Combining head and tail,

Therefore $\psi(x) = x + o(x)$, i.e. $\psi(x) \sim x$, which is (A) (via the Lemma above, equivalent to $\pi(x)\sim x/\log x$). This also proves the sharper quantitative form $\pi(x) = \operatorname{Li}(x) + O(x e^{-c\sqrt{\log x}})$ stated in the definition, once one relates $\operatorname{Li}(x)$ to $\int_2^x dt/\log t$ and $\psi$ to $\pi$ via Step 2 of the Lemma with the quantified error term carried through.

($A \Rightarrow B$), i.e. PNT implies no zero on the 1-line (necessity of the hypothesis). We argue by contraposition: suppose $\zeta(1+it_0) = 0$ for some real $t_0 \ne 0$ (note $t_0=0$ is impossible since $\zeta$ has a pole, not a zero, at $s=1$). We show $\psi(x) \sim x$ fails, i.e. $\psi(x)/x \not\to 1$. By the functional equation symmetry of zeros (nontrivial zeros come in conjugate pairs $\rho, \bar\rho$ since $\zeta$ has real Taylor coefficients along the real axis, so $\overline{\zeta(s)} = \zeta(\bar s)$), $\zeta(1-it_0)=0$ as well, giving two zeros $\rho = 1+it_0$ and $\bar\rho = 1 - it_0$ both with $\beta = 1$ exactly. Consider their combined contribution to the explicit formula:

for a fixed phase $\phi = -\arg(1+it_0)$ (writing $1+it_0 = |1+it_0|e^{i\phi'}$ and simplifying; the exact phase is irrelevant to what follows). This quantity has size exactly $x \cdot \frac{2}{|1+it_0|}|\cos(t_0\log x+\phi)|$, which is a positive constant multiple of $x$ infinitely often (indeed for a set of $x$ of positive logarithmic density, since $\cos(t_0 \log x + \phi)$ oscillates and is bounded away from $0$ on intervals of $\log x$ of fixed positive length). All other terms in the explicit formula — the remaining zeros with $\beta<1$ strictly, and the lower-order terms — contribute $o(x)$ or at worst $O(x^{\beta'})$ with $\beta' < 1$ for each individual other zero; while summing infinitely many such terms requires the same zero-density estimates as before, the classical zero-free regions established elsewhere still guarantee only finitely many zeros can approach $\beta=1$ arbitrarily closely for bounded height, and a standard density estimate shows the rest of the sum remains $o(x)$ compared to the genuine $\Theta(x)$-sized oscillation exhibited by $\rho,\bar\rho$ themselves. Hence

so $\psi(x)/x$ does not converge to $1$: it oscillates with amplitude bounded away from $0$ infinitely often. This contradicts $\psi(x)\sim x$. Hence PNT (A) forces $\zeta(1+it)\ne0$ for all real $t\ne0$, establishing (B). $\blacksquare$

Final remark on truth of (B). The fact that $\zeta(1+it)\ne 0$ for real $t$ was proved by Hadamard and (independently, with a different method) de la Vallée Poussin in 1896, via the elementary but ingenious trigonometric inequality

applied to $\log|\zeta(\sigma+it)|$ expansions to show that a hypothetical zero at $1+it_0$ together with the pole at $\sigma=1$ and the (would-be) zero at $1+2it_0$ (if $2it_0$ were also a zero location) leads to a sign contradiction in $3\log|\zeta(\sigma)| + 4\log|\zeta(\sigma+it_0)| + \log|\zeta(\sigma+2it_0)| \to -\infty$ as $\sigma \to 1^+$, contradicting that this quantity must stay bounded below by the Euler product representation for $\sigma > 1$ (each individual $|\zeta|$-value there is controlled, and the identity forces non-negativity in the limit if no zero exists, giving the required contradiction). This computation is a self-contained classical result belonging to the study of $\zeta(s)$ itself (see Zeta and the Distribution of Primes), and is not circular with respect to the theorem proved here: it uses only the Euler product for $\operatorname{Re}(s)>1$ and continuity/boundedness of $\zeta$ near $\sigma=1$, not the Prime Number Theorem. $\blacksquare$