proof of the prime number theorem




For x>0x \gt 0, let π(x)\pi(x) denote the number of primes pxp \leq x (in these notes, pp is always used to denote prime numbers). The celebrated Prime Number Theorem is the asymptotic statement

π(x)xlog(x)\pi(x) \sim \frac{x}{\log(x)}

where we recall that f(x)g(x)f(x) \sim g(x) means limxf(x)g(x)=1\underset{x \to \infty}{\lim} \frac{f(x)}{g(x)} = 1.

This was first proven in 1896 independently by Hadamard and de la Vallée-Poussin, first by proving a result on the zeroes of the Riemann zeta function ζ(s)\zeta(s) (that none occur on the line Re(s)=1Re(s) = 1), which can be proven without too much difficulty. They then deduced the prime number theorem from this result, by a fairly intricate analysis.

Over the decades this analysis was whittled down more and more until it reached a fairly definitive form, in the form of a succinct “Tauberian theorem” due to D.J. Newman (1980). A streamlined account was written up in the American Mathematical Monthly by Don Zagier (1997), who managed to produce a proof virtually from scratch in a little less than three pages (it requires just a bit of background in complex analysis, not much more than Cauchy’s theorem).

As in his famous one-sentence proof that primes of the form 4n+14n+1 are sums of two squares, Zagier seemed to be aiming explicitly for maximal compression while still keeping the argument (arguably) self-contained. It’s an impressive performance, alright, but its terseness will probably force most readers to follow up with pen and paper on the side to verify his arguments in detail. In this article we will mostly follow Zagier’s line, but with some slight expansions and glosses so that pen and paper is (hopefully) unnecessary.

Chebyshev’s estimate

In this section we prove a result essentially due to Chebyshev which puts us already in the neighborhood of the prime number theorem. Introduce a function

θ(x)= pxlog(p).\theta(x) = \sum_{p \leq x} \log(p).

By the following result, our job is done once we establish the relation θ(x)x\theta(x) \sim x.


The asymptotic statement θ(x)x\theta(x) \sim x implies the prime number theorem.


In one direction, we have an obvious inequality

θ(x)= pxlog(p) pxlog(x)=π(x)log(x).\theta(x) = \sum_{p \leq x} \log(p) \leq \sum_{p \leq x} \log(x) = \pi(x)\log(x).

Assuming that θ(x)x\theta(x) \sim x, this implies that given ϵ>0\epsilon \gt 0, the inequality

(1)1ϵ<π(x)log(x)x 1 - \epsilon \lt \frac{\pi(x) \log(x)}{x}

holds for all sufficiently large xx.

In the other direction, for any ϵ>0\epsilon \gt 0 we have

θ(x) x 1ϵ<pxlog(p) x 1ϵ<pxlog(x 1ϵ) = (1ϵ)log(x)(π(x)π(x 1ϵ)) (1ϵ)log(x)(π(x)x 1ϵ)\array{ \theta(x) & \geq & \sum_{x^{1-\epsilon} \lt p \leq x} \log(p) \\ & \geq & \sum_{x^{1-\epsilon} \lt p \leq x} \log(x^{1-\epsilon}) \\ & = & (1 - \epsilon)\log(x) \cdot (\pi(x) - \pi(x^{1 - \epsilon})) \\ & \geq & (1 - \epsilon)\log(x) \cdot (\pi(x) - x^{1 - \epsilon}) }

From the last line and (1) (which implies π(x)log(x)x\frac{\pi(x)\log(x)}{x} is bounded below by a positive constant), we obtain

θ(x)x (1ϵ)π(x)log(x)xx ϵ (12ϵ)π(x)log(x)x\array{ \frac{\theta(x)}{x} & \geq & (1-\epsilon)\frac{\pi(x)\log(x)}{x} - x^{-\epsilon} \\ & \geq & (1 - 2\epsilon)\frac{\pi(x)\log(x)}{x} }

for sufficiently large xx. Thus for any small ϵ>0\epsilon \gt 0 (say less than 1/31/3) and the assumption θ(x)x\theta(x) \sim x we have

1ϵπ(x)log(x)x(12ϵ) 1θ(x)x(13ϵ) 11 - \epsilon \leq \frac{\pi(x)\log(x)}{x} \leq (1 - 2\epsilon)^{-1}\frac{\theta(x)}{x} \leq (1 - 3\epsilon)^{-1}

or simply

1ϵπ(x)log(x)x(13ϵ) 11 - \epsilon \leq \frac{\pi(x)\log(x)}{x} \leq (1 - 3\epsilon)^{-1}

for all sufficiently large xx.

Proving the asymptotic result θ(x)x\theta(x) \sim x will involve complex analysis applied to the Riemann zeta function. An easier preliminary result gives information on the order of θ(x)\theta(x); it will later be used as a technical hypothesis in the Tauberian theorem with which we conclude. The order result is essentially due to Chebyshev, who actually established sharper estimates (which we won’t need). In fact, he also showed that if π(x)log(x)x\frac{\pi(x)\log(x)}{x} has a limit, then that limit must be 11.


θ(x)=O(x)\theta(x) = \mathrm{O}(x).


First let us observe that

2 2n(2nn) n<p<2np=exp(θ(2n)θ(n))2^{2n} \geq \binom{2n}{n} \geq \prod_{n \lt p \lt 2n} p = \exp(\theta(2n) - \theta(n))

so that

2nlog(2)θ(2n)θ(n).2n\log(2) \geq \theta(2n) - \theta(n).

If nx<n+1n \leq x \lt n+1, then θ(x)=θ(n)\theta(x) = \theta(n) and θ(2x)θ(2n)+log(2x)\theta(2x) \leq \theta(2n) + \log(2x), so that

2xlog(2)2nlog(2)θ(2n)θ(n)θ(2x)log(2x)θ(x)2x\log(2) \geq 2n\log(2) \geq \theta(2n) - \theta(n) \geq \theta(2x) - \log(2x) - \theta(x)

or simply

θ(x)θ(x/2)xlog(2)+log(x).\theta(x) - \theta(x/2) \leq x\log(2) + \log(x).

Taking a telescopic sum, we have

θ(x)= k=0 n1θ(x/2 k)θ(x/2 k+1) k=0 n1(x/2 k)log(2)+ k=0 n1log(x/2 k)\theta(x) = \sum_{k = 0}^{n-1} \theta(x/2^k) - \theta(x/2^{k+1}) \leq \sum_{k=0}^{n-1} (x/2^k)\log(2) + \sum_{k=0}^{n-1} \log(x/2^k)

where nn is chosen so that 1x/2 n<21 \leq x/2^n \lt 2 (so n=floor(log 2(x))n = floor(\log_2(x))). The first term on the right is bounded above by log(2)x\log(2) x. The second term is bounded above by log 2(x)log(x)=Klog(x) 2\log_2(x) \cdot \log(x) = K\log(x)^2 (where K=log 2(e)K = \log_2(e)), which considered asymptotically is small compared to xx. Thus, for any C>log(2)C \gt \log(2), we have

θ(x)Cx\theta(x) \leq C x

for all sufficiently large xx, which completes the proof.

Analytic continuation of the zeta function to Re(s)>0Re(s) \gt 0


(2)ζ(s)= n=1 1n s= p(11p s) 1 \zeta(s) = \sum_{n=1}^\infty \frac1{n^s} = \prod_p \left(1 - \frac1{p^s}\right)^{-1}

define the usual Riemann zeta function on the region Re(s)>1Re(s) \gt 1. First we establish a meromorphic continuation of ζ(s)\zeta(s) to the region Re(s)>0Re(s) \gt 0, or in fact a holomorphic continuation of ζ(s)1s1\zeta(s) - \frac1{s-1} to this region.


The function Ψ\Psi defined by the expression

Ψ(s)= n=1 n n+11n s1x sdx\Psi(s) = \sum_{n=1}^\infty \int_n^{n+1} \frac1{n^s} - \frac1{x^s}\; d x

converges absolutely for Re(s)>0Re(s) \gt 0, giving a holomorphic function in that region, and

Ψ(s)=ζ(s)1s1\Psi(s) = \zeta(s) - \frac1{s-1}

over the region Re(s)>1Re(s) \gt 1.


Let us bound the summands in the definition of Ψ(s)\Psi(s):

| n n+11n s1x sdx| = | n n+1 n xst s+1dtdx| n n+1 n x|s||t s+1|dtdx = n n+1 n x|s|t Re(s)+1dtdx n n+1 n x|s|n Re(s)+1dtdx |s|n Re(s)+1.\array{ \left| \int_n^{n+1} \frac1{n^s} - \frac1{x^s}\; d x \right| & = & \left| \int_n^{n+1} \int_n^x \; \frac{s}{t^{s+1}}\; d t\; d x \right|\\ & \leq & \int_n^{n+1} \int_n^x \; \frac{{|s|}}{{|t^{s+1}|}}\;d t \; d x \\ & = & \int_n^{n+1} \int_n^x \; \frac{{|s|}}{t^{Re(s) + 1}}\; d t\; d x \\ & \leq & \int_n^{n+1} \int_n^x \; \frac{{|s|}}{n^{Re(s) + 1}}\;d t \; d x \\ & \leq & \frac{{|s|}}{n^{Re(s) + 1}}. }

Since n=1 1n Re(s)+1\sum_{n=1}^\infty \frac1{n^{Re(s) + 1}} converges over the region Re(s)>0Re(s) \gt 0, it follows that the series for Φ(s)\Phi(s) converges absolutely, and Ψ(s)\Psi(s) is a holomorphic function, in that region.

Over the region Re(s)>1Re(s) \gt 1, we have

Ψ(s) = n=1 1n s n n+11x sdx = n=1 1n s n=1 n n+11x sdx = ζ(s) 1 1x sdx = ζ(s)1s1\array{ \Psi(s) & = & \sum_{n=1}^\infty \frac1{n^s} - \int_n^{n+1} \frac1{x^s}\; d x \\ & = & \sum_{n=1}^\infty \frac1{n^s} - \sum_{n=1}^\infty \int_n^{n+1} \frac1{x^s}\; d x \\ & = & \zeta(s) - \int_1^\infty \frac1{x^s}\; d x \\ & = & \zeta(s) - \frac1{s-1} }

as asserted.

The zeta function has no zeroes on the line Re(s)=1Re(s) = 1

Now we relate θ(x)\theta(x) to ζ(s)\zeta(s) by introducing an auxiliary function Φ\Phi that is connected with the logarithmic derivative of ζ(s)\zeta(s). For Re(s)>1Re(s) \gt 1 put

Φ(s)= plog(p)p s.\Phi(s) = \sum_p \frac{\log(p)}{p^s}.

Over Re(s)>1Re(s) \gt 1, we calculate

logζ(s)= plog(1p s)\log \zeta(s) = -\sum_p \log(1 - p^{-s})

which upon differentiating yields

ζ (s)ζ(s) = plog(p)p s1p s = plog(p)p s plog(p)p sp s1p s = Φ(s) plog(p)p s(p s1)\array{ \frac{\zeta^'(s)}{\zeta(s)} & = & -\sum_p \frac{\log(p) p^{-s}}{1 - p^{-s}} \\ & = & -\sum_p \log(p) p^{-s} -\sum_p \frac{\log(p) p^{-s}p^{-s}}{1 - p^{-s}} \\ & = & -\Phi(s) - \sum_p \frac{\log(p)}{p^s(p^s - 1)} }

with the last sum converging over the region Re(s)>12Re(s) \gt \frac1{2}. From this and the analytic continuation of ζ(s)\zeta(s) (Proposition ), we see that Φ(s)\Phi(s) admits a meromorphic continuation over Re(s)>12Re(s) \gt \frac1{2}, and holomorphic in that region except possibly where ζ(s)\zeta(s) has zeroes, or at the simple pole s=1s = 1.


Φ(s)1s1\Phi(s) - \frac1{s-1} is holomorphic over the region Re(s)1Re(s) \geq 1, i.e., ζ(s)\zeta(s) has no zeroes over Re(s)1Re(s) \geq 1.


The case Re(s)>1Re(s) \gt 1 is easily handled by the product formula (2).

For a fixed t>0t \gt 0, let μ\mu be the order of the zero of ζ(1+it)\zeta(1 + i t) (if ζ(1+it)0\zeta(1 + i t) \neq 0 then of course μ=0\mu = 0); it is the same as the order of the zero at s=1its = 1 - i t. Let ν\nu be the order of the zero at 1±2it1 \pm 2i t. We then have

  1. limϵ0ϵΦ(1+ϵ)=1\underset{\epsilon \searrow 0}{\lim} \epsilon \cdot \Phi(1 + \epsilon) = 1 (since ζ(s)\zeta(s) has a pole of order 11 at s=1s = 1), and similarly

  2. limϵ0ϵΦ(1+ϵ±it)=μ\underset{\epsilon \searrow 0}{\lim} \epsilon \cdot \Phi(1 + \epsilon \pm i t) = -\mu,

  3. limϵ0ϵΦ(1+ϵ±2it)=ν\underset{\epsilon \searrow 0}{\lim} \epsilon \cdot \Phi(1 + \epsilon \pm 2i t) = -\nu.

Observe that for ϵ>0\epsilon \gt 0,

0 ϵ plog(p)p 1+ϵ(p it/2+p it/2) 4 = ϵ plog(p)p 1+ϵ k=2 2(4r+2)p irt = ϵ r=2 2(4r+2)Φ(1+ϵ+irt)\array{ 0 & \leq & \epsilon \cdot \sum_p \frac{\log(p)}{p^{1 + \epsilon}} (p^{i t/2} + p^{-i t/2})^4 \\ & = & \epsilon \cdot \sum_p \frac{\log(p)}{p^{1 + \epsilon}} \sum_{k = -2}^2 \binom{4}{r + 2} p^{-i r t} \\ & = & \epsilon \cdot \sum_{r = -2}^2 \binom{4}{r+2} \Phi(1 + \epsilon + i r t) }

from which, in conjunction with 1., 2., 3. above, we deduce 068μ2ν0 \leq 6 - 8\mu - 2\nu, so that μ=0\mu = 0.

The function θ(x)\theta(x) is related to the function Φ(s)\Phi(s) by way of a Riemann-Stieltjes integral:

Φ(s)= plog(p)p s= 1 dθ(x)x s\Phi(s) = \sum_p \frac{\log(p)}{p^s} = \int_1^\infty \frac{d \theta(x)}{x^s}

for Re(s)>1Re(s) \gt 1. An integration by parts yields

1 Mdθ(x)x s=[θ(M)M sθ(1)]+ 1 Mθ(x)sx s+1dx\int_1^M \frac{d \theta(x)}{x^s} = \left[\frac{\theta(M)}{M^s} - \theta(1)\right] + \int_1^M \theta(x) \frac{s}{x^{s+1}}\; d x

where limMθ(M)M s=0\underset{M \to \infty}{\lim} \frac{\theta(M)}{M^s} = 0 by Lemma . Thus

(3)Φ(s)=s 1 θ(x)x s+1dx=s 0 e stθ(e t)dt. \Phi(s) = s \int_1^\infty \frac{\theta(x)}{x^{s+1}}\; d x = s \int_0^\infty e^{-s t} \theta(e^t)\; d t.

Reduction to a Tauberian theorem

The asymptotic result θ(x)x\theta(x) \sim x will be established by appeal to the following result:


Suppose that the integral

1 θ(x)xx 2dx\int_1^\infty \; \frac{\theta(x) - x}{x^2}\; d x

converges. Then θ(x)x\theta(x) \sim x.


Suppose that for some λ>1\lambda \gt 1 there exist arbitrarily large xx such that λxθ(x)\lambda x \leq \theta(x). The monotonicity of θ\theta implies the first inequality that follows:

x λxθ(t)ttdtt x λxλxttdtt= 1 λλxtxtxdtt= 1 λλtt 2dt>0.\int_x^{\lambda x} \; \frac{\theta(t) - t}{t}\; \frac{d t}{t} \geq \int_x^{\lambda x} \; \frac{\lambda x - t}{t}\; \frac{d t}{t} = \int_1^\lambda \; \frac{\lambda x - t x}{t x} \; \frac{d t}{t} = \int_1^\lambda \; \frac{\lambda - t}{t^2}\; d t \gt 0.

This is a contradiction, because convergence of 1 θ(t)tt 2dt\int_1^\infty \; \frac{\theta(t) - t}{t^2}\; d t implies that x λxθ(t)tt 2dt\int_x^{\lambda x} \; \frac{\theta(t) - t}{t^2}\; d t can be made arbitrarily small by taking xx large.

Similarly, if we suppose that for some λ<1\lambda \lt 1 there exist arbitrarily large xx such that θ(x)λx\theta(x) \leq \lambda x, then

λx xθ(t)ttdtt λx xλxttdtt= λ 1λxtxtxdtt= λ 1λtt 2dt<0\int_{\lambda x}^x \; \frac{\theta(t) - t}{t}\; \frac{d t}{t} \leq \int_{\lambda x}^x \; \frac{\lambda x - t}{t}\; \frac{d t}{t} = \int_\lambda^1 \; \frac{\lambda x - t x}{t x} \; \frac{d t}{t} = \int_\lambda^1 \; \frac{\lambda - t}{t^2}\; d t \lt 0

which again leads to a contradiction.

Thus we focus on convergence of

(4) 1 θ(x)xxdxx= 0 θ(e t)e te tdt \int_1^\infty \; \frac{\theta(x) - x}{x}\; \frac{d x}{x} = \int_0^\infty \; \frac{\theta(e^t) - e^t}{e^t}\; d t

For Re(s)>1Re(s) \gt 1 we have, from (3),

Φ(s)s1s1 = 0 e stθ(e t)e (1s)tdt = 0 e st(θ(e t)e t)dt\array{ \frac{\Phi(s)}{s} - \frac1{s-1} & = & \int_0^\infty \; e^{-s t}\theta(e^t) - e^{(1-s)t}\; dt \\ & = & \int_0^\infty \; e^{-s t} (\theta(e^t) - e^t)\; d t }

or what is the same, for the region Re(s)>0Re(s) \gt 0,

(5)Φ(s+1)s+11s= 0 e (s+1)t(θ(e t)e t)dt= 0 e st(θ(e t)e te t)dt \frac{\Phi(s+1)}{s+1} - \frac1{s} = \int_0^\infty \; e^{-(s+1) t} (\theta(e^t) - e^t)\; d t = \int_0^\infty \; e^{-s t}\left( \frac{\theta(e^t) - e^t}{e^t} \right)\; d t

where the left side is in fact holomorphic over Re(s)0Re(s) \geq 0, hence takes a finite value at s=0s = 0, say LL.

The Tauberian theorem we will establish permits the conclusion

L=lims0 0 e st(θ(e t)e te t)dt= 0 θ(e t)e te tdtL = \underset{s \searrow 0}{\lim} \int_0^\infty \; e^{-s t}\left( \frac{\theta(e^t) - e^t}{e^t} \right)\; d t = \int_0^\infty \; \frac{\theta(e^t) - e^t}{e^t} \; d t

which in turn will yield the Prime Number Theorem:


The integral 0 θ(e t)e te tdt\int_0^\infty \; \frac{\theta(e^t) - e^t}{e^t} \; d t converges.

(The prime number theorem follows from this by (4), Theorem , and Lemma .)

Proof of the Tauberian theorem

Here is the statement of the particular Tauberian theorem used (recalling that “Tauberian” signifies a family of theorems that roughly speaking give conditions for convergence of series or integral expressions occurring at natural boundaries of regions where series or integral representations are defined). This particular form was given by D.J. Newman, and allows a dramatic shortening of the proof of the prime number theorem, as shown convincingly by Zagier, who as we said compressed it down to under three pages.


Let f(t)f(t) be a function that is bounded and locally integrable over the domain t0t \geq 0, and suppose that its Laplace transform, defined for Re(s)>0Re(s) \gt 0 by the equation

g(s)= 0 e stf(t)dt,g(s) = \int_0^\infty \; e^{-s t} f(t)\; d t,

extends to a function gg that is holomorphic over the region Re(s)0Re(s) \geq 0. Then the integral 0 f(t)dt\int_0^\infty\; f(t)\; d t converges and equals g(0)g(0).

The proof below involves some fairly refined complex analysis. It may be well to recall that we say a function gg is holomorphic at a point z 0z_0 if gg is defined over a small neighborhood {z:|zz 0|<δ}\{z: {|z-z_0|} \lt \delta\} of the point and in that neighborhood is expressible as a convergent Taylor series

g(z)= n=0a n(zz 0) nn!.g(z) = \sum_{n=0} \frac{a_n (z - z_0)^n}{n!}.

Then we say gg is holomorphic over a (not necessarily open) region if it is holomorphic at every point in the region.

In the case of interest f(t)=θ(e t)e te tf(t) = \frac{\theta(e^t) - e^t}{e^t}, the order estimate Lemma guarantees that ff is bounded and locally integrable, and we have already observed that its Laplace transform, which is

g(s)=Φ(s+1)s+11sg(s) = \frac{\Phi(s+1)}{s+1} - \frac1{s}

by (5), is holomorphic in the region Re(s)0Re(s) \geq 0 by Proposition . Thus all the hypotheses of the theorem are satisfied in this case, and its conclusion in this case is Theorem . So all that remains is to prove Theorem .


For T0T \geq 0, put

g T(s)= 0 Te stf(t)dt.g_T(s) = \int_0^T \; e^{-s t} f(t)\; d t.

Clearly g Tg_T is an entire (i.e., everywhere holomorphic) function. We are trying to show that limTg T(0)=g(0)\underset{T \to \infty}{\lim} g_T(0) = g(0).

A straightforward application of Cauchy’s theorem yields

g T(0)g(0)=12πi C(g T(z)g(z))e zTdzzg_T(0) - g(0) = \frac1{2\pi i} \int_C \; \left(g_T(z) - g(z)\right) e^{z T}\; \frac{d z}{z}

for contours CC having 00 as an interior point, such that g Tgg_T - g is holomorphic on CC and its interior. We will choose CC to be the boundary of the closed region D={z:|z|RandRe(z)δ}D = \{z: {|z|} \leq R\; and\; Re(z) \geq -\delta\} where for a given RR, we choose δ\delta small enough that g Tgg_T - g is holomorphic in DD (this uses compactness of the interval I={it:RtR}I = \{i t: -R \leq t \leq R\} and holomorphicity of g Tgg_T - g along II).

The idea would be to estimate the contribution of the integral over C +C_+, the part of CC where the real part is nonnegative, and over C C_- where it is nonpositive, hoping to show these contributions are small when RR is large. This almost works, except that we don’t have good control over the size of the contributions near the area where Re(z)=0Re(z) = 0. To offset this, a mollifying factor (1+z 2R 2)(1 + \frac{z^2}{R^2}) is introduced in the integrand (“Carleman’s formula”): this is close to zero near Re(z)=0Re(z) = 0 and has the desired mitigating effect.

Thus we instead use

g T(0)g(0)=12πi C(g T(z)g(z))e zT(1+z 2R 2)dzzg_T(0) - g(0) = \frac1{2\pi i} \int_C \; \left(g_T(z) - g(z)\right) e^{z T}\; \left(1 + \frac{z^2}{R^2}\right)\; \frac{d z}{z}

Suppose now that |f(t)|{|f(t)|} is globally bounded by BB. For Re(z)>0Re(z) \gt 0 we have

|g T(z)g(z)| T |f(t)e zt|dtB T e Re(z)tdt=Be Re(z)TRe(z){|g_T(z) - g(z)|} \leq \int_T^\infty\; {|f(t)e^{-z t}|}\; d t \leq B \int_T^\infty e^{-Re(z)t}\; d t = B \frac{e^{-Re(z)T}}{Re(z)}

and we also have, over the circle |z|=R{|z|} = R,

|e zT(1+z 2/R 2)(1/z)| = e Re(z)T|(R 2z 1+z)/R 2| = e Re(z)T|(|z| 2z 1+z)/R 2| = e Re(z)T|(z¯+z)/R 2| = e Re(z)T|2Re(z)|/R 2\array{ {|e^{z T}(1 + z^2/R^2)(1/z)|} & = & e^{Re(z)T} \cdot {|(R^2 z^{-1} + z)/R^2|} \\ & = & e^{Re(z)T} \cdot {|({|z|}^2 z^{-1} + z)/R^2|} \\ & = & e^{Re(z)T} \cdot {|(\bar{z} + z)/R^2|} \\ & = & e^{Re(z)T} \cdot {|2Re(z)|}/R^2 }

so that putting the factors together, the integral over C +C_+ is bounded in absolute value by

2B/R 2 C +|dz|=2πB/R2B/R^2 \int_{C_+}\; {|d z|} = 2\pi B/R

(and of course this hasn’t taken into account the factor 12πi\frac1{2\pi i} outside the integral). In any case, with the Carleman factor, the contribution from the C +C_+ part is B/RB/R, which is small if RR is large.

Now we bound the contribution from C C_- by considering g Tg_T and gg separately. First let us bound

12πi C g T(z)e zT(1+z 2R 2)dzz.\frac1{2\pi i} \int_{C_-} \; g_T(z) e^{z T}\; \left(1 + \frac{z^2}{R^2}\right)\; \frac{d z}{z}.

For this, we can take advantage of the fact that g Tg_T is entire, and (courtesy of Cauchy’s theorem) replace C C_- by the semicircle where |z|=R,Re(z)<0{|z|} = R, Re(z) \lt 0. The estimates are similar in form to those above, starting with

|g T(z)| 0 T|f(t)||e zt|dtB 0 Te Re(z)tdtB Te Re(z)tdt=Be Re(z)TRe(z).{|g_T(z)|} \leq \int_0^T {|f(t)|}\cdot {|e^{-z t}|}\; d t \leq B \int_0^T e^{-Re(z)t}\; d t \leq B \int_{-\infty}^T e^{-Re(z)t}\; d t = B \frac{e^{-Re(z)T}}{-Re(z)}.

Bear in mind here that we are now considering Re(z)<0Re(z) \lt 0 and Re(z)=|Re(z)|-Re(z) = {|Re(z)|}; the bound produced tends to be very large. But, large as it may be, it is offset by the bound on the product of the remaining factors in the integrand, which is e Re(z)T|2Re(z)|/R 2e^{Re(z)T} \cdot {|2Re(z)|}/R^2. We conclude as before that over C C_- the overall contribution from g Tg_T has upper bound B/RB/R.

Finally, we bound

12πi C g(z)e zT(1+z 2R 2)dzz.\frac1{2\pi i} \int_{C_-} \; g(z) e^{z T}\; \left(1 + \frac{z^2}{R^2}\right)\; \frac{d z}{z}.

Holding RR fixed, the expression g(z)(1+z 2R 2)1zg(z)\cdot \left(1 + \frac{z^2}{R^2}\right) \cdot \frac1{z} taken over C C_- is bounded in absolute value by a number KK. Whereas along Re(z)=δRe(z) = -\delta, by taking TT very large we can make |e zT|{|e^{z T}|} as small as we like, so that considering gg separately, the contribution over C C_- can be discarded as vanishingly small (in other words, first pick RR, then pick δ\delta, and finally pick TT, to get this contribution to be as small as we like). This completes the proof.


  • Donald J. Newman, Simple analytic proof of the prime number theorem, Amer. Math. Monthly 87 (1980), 693-696.

  • Don Zagier, Newman’s short proof of the prime number theorem, Amer. Math. Monthly 104 (1997), 705-708. (pdf)

  • Don Zagier, A one-sentence proof that every prime p1(mod4)p \equiv 1 (\mod 4) is a sum of two squares, Amer. Math. Monthly, Vol. 97 No. 2 (1990), p. 144. (link)

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