The Prime Number Theorem

Filip Jaskovic

February 13, 2026

Abstract

This note presents a concise analytic proof of the prime number theorem in the Newman style, emphasizing the non-vanishing of $\zeta(s)$ on $\Re(s)=1$ and a short Tauberian argument.

The prime number theorem, that the number of primes $\le x$ is asymptotic to $x/\log x$ , was proved (independently) by Hadamard and de la Vallée Poussin in 1896. Their proof had two elements: showing that Riemann’s zeta function $\zeta(s)$ has no zeros with $\Re(s)=1$ , and deducing the prime number theorem from this. An ingenious short proof of the first assertion was found soon afterwards by the same authors and by Mertens and is reproduced here, but the deduction of the prime number theorem continued to involve difficult analysis. A proof that was elementary in a technical sense, avoiding the use of complex analysis, was found in 1949 by Selberg and Erdős, but this proof is intricate and much less clearly motivated than the analytic one. A few years later, D. J. Newman found a very simple version of the Tauberian argument needed for an analytic proof of the prime number theorem. We describe that proof, which has a beautifully simple structure and uses little beyond Cauchy’s theorem.

Recall that $f(x)\sim g(x)$ means $\lim_{x\to\infty} f(x)/g(x)=1$ , and that $\mathrm{O}(f)$ denotes a quantity bounded in absolute value by a fixed multiple of $f$ . We denote by $\pi(x)$ the number of primes $\le x$ .

\pi(x)\sim\dfrac{x}{\log x}

x\to\infty

We present the argument in a sequence of steps by studying the three functions

\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}, \quad \Phi(s)=\sum_p\frac{\log p}{p^s}, \quad \vartheta(x)=\sum_{p\le x}\log p \quad (s\in\mathbb{C},\;x\in\mathbb{R}),

where $p$ always denotes a prime. The series defining $\zeta(s)$ and $\Phi(s)$ are absolutely and locally uniformly convergent for $\Re(s)>1$ , so they define holomorphic functions on that half-plane.

\zeta(s)=\prod_p(1-p^{-s})^{-1}

for

\Re(s)>1

\zeta(s) =\sum_{r_2,r_3,\dots\ge0}\left(2^{r_2}3^{r_3}\cdots\right)^{-s} =\prod_p\left(\sum_{r\ge0}p^{-rs}\right) =\prod_p\frac{1}{1-p^{-s}} \qquad (\Re(s)>1).

\zeta(s)-\dfrac{1}{s-1}

extends holomorphically to

\Re(s)>0

\zeta(s)-\frac{1}{s-1} =\sum_{n=1}^{\infty}\frac{1}{n^s}-\int_1^{\infty}\frac{1}{x^s}\,dx =\sum_{n=1}^{\infty}\int_n^{n+1}\left(\frac{1}{n^s}-\frac{1}{x^s}\right)dx.

The series on the right converges absolutely for $\Re(s)>0$ because

\left|\int_n^{n+1}\left(\frac{1}{n^s}-\frac{1}{x^s}\right)dx\right| =\left|s\int_n^{n+1}\int_n^x\frac{du}{u^{s+1}}\,dx\right| \le\max_{n\le u\le n+1}\left|\frac{s}{u^{s+1}}\right| =\frac{|s|}{n^{\Re(s)+1}},

by the mean value theorem.

\vartheta(x)=\mathrm{O}(x)

2^{2n}=(1+1)^{2n} =\binom{2n}{0}+\cdots+\binom{2n}{2n} \ge\binom{2n}{n} \ge\prod_{n<p\le2n}p =e^{\vartheta(2n)-\vartheta(n)}.

Hence, since $\vartheta(x)$ changes by $\mathrm{O}(\log x)$ when $x$ changes by $\mathrm{O}(1)$ , we get $\vartheta(x)-\vartheta(x/2)\le Cx$ for any $C>\log 2$ and all $x\ge x_0(C)$ . Summing over $x,x/2,\ldots,x/2^r$ with $x/2^r\ge x_0>x/2^{r+1}$ gives $\vartheta(x)\le2Cx+\mathrm{O}(1)$ .

\zeta(s)\neq0

and

\Phi(s)-\dfrac{1}{s-1}

is holomorphic for

\Re(s)\ge1

-\frac{\zeta'(s)}{\zeta(s)} =\sum_p\frac{\log p}{p^s-1} =\Phi(s)+\sum_p\frac{\log p}{p^s(p^s-1)}.

The final sum converges for $\Re(s)>\tfrac12$ , so together with (II) this gives a meromorphic continuation of $\Phi(s)$ to $\Re(s)>\tfrac12$ , with poles only at $s=1$ and zeros of $\zeta(s)$ . If $\zeta(s)$ has a zero of order $\mu$ at $s=1+i\alpha$ ( $\alpha\in\mathbb{R}$ , $\alpha\ne0$ ) and a zero of order $\nu$ at $1+2i\alpha$ , then

\lim_{\epsilon\searrow0}\epsilon\Phi(1+\epsilon)=1, \quad \lim_{\epsilon\searrow0}\epsilon\Phi(1+\epsilon\pm i\alpha)=-\mu, \quad \lim_{\epsilon\searrow0}\epsilon\Phi(1+\epsilon\pm2i\alpha)=-\nu.

The inequality

\sum_{r=-2}^{2}\binom{4}{2+r}\Phi(1+\epsilon+ir\alpha) =\sum_p\frac{\log p}{p^{1+\epsilon}}\left(p^{i\alpha/2}+p^{-i\alpha/2}\right)^4 \ge0

implies $6-8\mu-2\nu\ge0$ , so $\mu=0$ , i.e. $\zeta(1+i\alpha)\ne0$ .

\displaystyle\int_1^{\infty}\frac{\vartheta(x)-x}{x^2}\,dx

converges.

\Phi(s) =\sum_p\frac{\log p}{p^s} =\int_1^{\infty}\frac{d\vartheta(x)}{x^s} =s\int_1^{\infty}\frac{\vartheta(x)}{x^{s+1}}\,dx =s\int_0^{\infty}e^{-st}\vartheta(e^t)\,dt.

Therefore (V) follows by applying the analytic theorem below to $f(t)=\vartheta(e^t)e^{-t}-1$ and $g(z)=\Phi(z+1)/(z+1)-1/z$ , using (III) and (IV).

Let

f(t)

(

t\ge0

) be bounded and locally integrable, and suppose

g(z)=\int_0^{\infty}f(t)e^{-zt}\,dt

for

\Re(z)>0

extends holomorphically to

\Re(z)\ge0

. Then

\int_0^{\infty}f(t)\,dt

exists and equals

g(0)

\vartheta(x)\sim x

Since $\vartheta$ is non-decreasing,

\int_x^{\lambda x}\frac{\vartheta(t)-t}{t^2}\,dt \ge\int_x^{\lambda x}\frac{\lambda x-t}{t^2}\,dt>0,

for such $x$ , contradicting (V). Similarly, if for some $\lambda<1$ one had $\vartheta(x)\le\lambda x$ for all sufficiently large $x$ , then

\int_x^{\lambda^{-1}x}\frac{\vartheta(t)-t}{t^2}\,dt<0,

again contradicting (V). Therefore $\vartheta(x)\sim x$ .

The prime number theorem follows from (VI):

\vartheta(x)=\sum_{p\le x}\log p,

and partial summation yields $\pi(x)\sim x/\log x$ .

g_T(z)=\int_0^T f(t)e^{-zt}\,dt.

This is entire. We must show $\lim_{T\to\infty}g_T(0)=g(0)$ . Let $R$ be large, and let $C$ be the boundary of

\{z\in\mathbb{C}:|z|\le R,\,\Re(z)\ge-\delta\},

where $\delta>0$ is small enough that $g$ is holomorphic on and inside $C$ . By Cauchy’s theorem,

g(0)-g_T(0) =\frac{1}{2\pi i}\int_C\left(g(z)-g_T(z)\right) \left(1+\frac{z^2}{R^2}\right)\frac{dz}{z}.

On the right semicircle ( $\Re(z)>0$ ), the integrand is bounded by $2B/R^2$ , where $B=\sup_{t\ge0}|f(t)|$ , so that contribution is $\ll B/R$ . For the left part ( $\Re(z)<0$ ), treat $g$ and $g_T$ separately. Since $g_T$ is entire, its contour may be replaced by the left semicircle $C'_- = \{z:|z|=R,\Re(z)<0\}$ , and the same $\ll B/R$ bound follows because

|g_T(z)| =\left|\int_0^T f(t)e^{-zt}\,dt\right| \le B\int_0^T e^{-\Re(z)t}\,dt \le\frac{Be^{-\Re(z)T}}{|\Re(z)|} \quad(\Re(z)<0).

Finally, the remaining integral over $C_-$ tends to $0$ as $T\to\infty$ because its integrand is $g(z)(1+z^2/R^2)/z$ times $e^{zT}$ , and $e^{zT}\to0$ uniformly on compact subsets of $\Re(z)<0$ . Hence

\limsup_{T\to\infty}|g(0)-g_T(0)|\le\frac{2B}{R}.

Since $R$ is arbitrary, this proves the theorem.

Historical remarks

The “Riemann” zeta function $\zeta(s)$ was first introduced and studied by Euler, and the product representation in (I) is his. The connection with the prime number theorem was found by Riemann, who made a deep study of the analytic properties of $\zeta(s)$ . For the present proof, however, the nearly trivial continuation property (II) is sufficient. The ingenious estimate in (III) is essentially due to Chebyshev, who proved refined bounds showing that $\vartheta(x)/x$ (and hence also $\pi(x)/(x/\log x)$ ) stays between roughly $0.92$ and $1.11$ for all sufficiently large $x$ . This remained the best result until the full prime number theorem was proved in 1896 by de la Vallée Poussin and Hadamard. The short proof of non-vanishing on $\Re(s)=1$ in (IV) is in essence Hadamard’s, later refined by de la Vallée Poussin and Mertens. The analytic theorem and its use in (V) and (VI) are due to D. J. Newman. Apart from minor simplifications, this exposition follows Newman and Korevaar.

For broader historical perspective, see Bateman and Diamond.

References

[B] P. Bateman and H. Diamond, A hundred years of prime numbers, Amer. Math. Monthly 103 (1996), 729-741.
[K] J. Korevaar, On Newman’s quick way to the prime number theorem, Math. Intelligencer 4(3) (1982), 108-115.
[N] D. J. Newman, Simple analytic proof of the prime number theorem, Amer. Math. Monthly 87 (1980), 693-696.
[T] E. C. Titchmarsh, The Theory of the Riemann Zeta Function, Oxford, 1951.