Prime number theorem with the de la Vallée Poussin error term

We prove the classical quantitative prime number theorem: there is an absolute constant $c>0$ for which

$\displaystyle \psi(x):=\sum_{n\le x}\Lambda(n)=x+O(x\exp(-c\sqrt{\log x})).$

Consequently,

$\displaystyle \vartheta(x):=\sum_{p\le x}\log p=x+O(x\exp(-c\sqrt{\log x})),$

and

$\displaystyle \pi(x)=\text{Li}(x)+O(\frac{x}{\log x}\exp(-c\sqrt{\log x})).$

The main issue is not to prove merely that $\psi(x)\sim x$ , but to understand why the error has the particular shape $\exp(-c\sqrt{\log x})$ . The answer is a balance between two costs in a contour shift. A zero-free region lets us move a Perron contour to the left of the line $\text{Re}(s)=1$ , gaining a factor from the power $x^s$ . But the shift is only available up to a finite height $T$ , and truncating the contour costs a factor involving $T^{-1}$ . The de la Vallée Poussin zero-free region permits a horizontal displacement of order $1/\log T$ . Thus the saving is roughly $\exp(-c\log x/\log T)$ , while the truncation cost is roughly $\exp(-\log T)$ . Balancing these forces $\log T\asymp\sqrt{\log x}$ , and that is where the square root comes from.

The proof has four linked stages. The Euler product turns primes into the Dirichlet coefficients of $-\zeta'/\zeta$ . The completed zeta function then turns the nontrivial zeros of $\zeta$ into a positive potential. The elementary inequality $3+4\cos\theta+\cos 2\theta\ge0$ uses the positivity of the von Mangoldt coefficients to prevent this potential from placing a zero too close to $\text{Re}(s)=1$ . Finally, a smoothed Perron formula converts the zero-free region back into an estimate for primes. All implied constants below are absolute.

Primes and Riemann Zeta function

Define

$\displaystyle \Lambda(n):=\begin{cases}\log p,&n=p^k\text{ for some prime }p\text{ and some integer }k\ge1,\\ 0,&\text{otherwise}.\end{cases}$

The function $\psi(x)=\sum_{n\le x}\Lambda(n)$ counts prime powers, weighting $p^k$ by $\log p$ . This is exactly the weight forced by logarithmic differentiation of the Euler product. For $\text{Re}(s)>1$ ,

$\displaystyle \zeta(s)=\prod_p(1-p^{-s})^{-1},\qquad -\frac{\zeta'}{\zeta}(s)=\sum_{n=1}^{\infty}\frac{\Lambda(n)}{n^s}.$

Indeed, differentiating the logarithm of the Euler product gives $\sum_p(\log p)/(p^s-1)$ . Expanding $(p^s-1)^{-1}$ as the geometric series $\sum_{k\ge1}p^{-ks}$ produces one term $(\log p)p^{-ks}$ for every prime power $p^k$ . Thus the problem of estimating primes is transformed into the problem of estimating the logarithmic derivative of $\zeta(s)$ near the boundary of its initial half-plane of convergence.

We shall also need the elementary continuation of $\zeta$ to $\text{Re}(s)>0$ . Let $A(u)=\lfloor u\rfloor$ . Summation by parts gives, initially for $\text{Re}(s)>1$ ,

$\displaystyle \zeta(s)=s\int_1^{\infty}A(u)u^{-s-1}du=\frac{s}{s-1}-s\int_1^{\infty}{u}u^{-s-1}du.$

The second integral converges already when $\text{Re}(s)>0$ , since $0\le{u}<1$ . Hence $\zeta$ continues meromorphically to this larger half-plane, with a unique singularity there: a simple pole at $s=1$ of residue $1$ . In particular,

$\displaystyle -\frac{\zeta'}{\zeta}(s)=\frac{1}{s-1}+O(1)$

near $s=1$ . This pole is the analytic origin of the main term $x$ .

The nontrivial zeros of $\zeta$ are most useful after adjoining the Gamma factor that arises from Poisson summation. Let

$\displaystyle \theta(u):=\sum_{n\in\mathbb Z}\exp(-\pi n^2u),\qquad u>0.$

The Fourier transform of a Gaussian is another Gaussian, and Poisson summation gives the exact relation

$\displaystyle \theta(u)=u^{-1/2}\theta(1/u).$

For $\text{Re}(s)>1$ , Mellin integration term by term gives

$\displaystyle \pi^{-s/2}\Gamma\left(\frac{s}{2}\right)\zeta(s)=\frac12\int_0^{\infty}(\theta(u)-1)u^{s/2-1}du.$

Split this integral at $u=1$ . On the interval $0<u<1$ , insert the theta transformation and then substitute $u=1/v$ . One obtains

$\displaystyle \pi^{-s/2}\Gamma\left(\frac{s}{2}\right)\zeta(s)=\frac{1}{s-1}-\frac{1}{s}+F(s),$

where

$\displaystyle F(s):=\frac12\int_1^{\infty}(\theta(u)-1)\left(u^{s/2-1}+u^{-(s+1)/2}\right)du.$

The function $\theta(u)-1$ decays exponentially as $u\to\infty$ , so this integral converges locally uniformly for every complex $s$ . Hence $F$ is entire. The displayed expression is also unchanged when $s$ is replaced by $1-s$ . Therefore the completed zeta function

$\displaystyle \xi(s):=\frac12s(s-1)\pi^{-s/2}\Gamma\left(\frac{s}{2}\right)\zeta(s)$

is entire and satisfies the functional equation

$\displaystyle \xi(s)=\xi(1-s).$

The zeros of $\xi$ are exactly the nontrivial zeros $\rho=\beta+i\gamma$ of $\zeta$ . The Euler product shows that $\zeta(s)\ne0$ for $\text{Re}(s)>1$ ; the functional equation reflects zeros across $\text{Re}(s)=1/2$ . Hence every nontrivial zero lies in the strip $0\le\beta\le1$ .

We also need a mild estimate for the Gamma factor. The classical series

$\displaystyle \frac{\Gamma'}{\Gamma}(z)=-\gamma+\sum_{m=0}^{\infty}\left(\frac{1}{m+1}-\frac{1}{m+z}\right)$

shows that

$\displaystyle \frac{\Gamma'}{\Gamma}(z)\ll\log(|z|+2),\qquad \text{Re}(z)\ge\frac12.$

To see this directly, split the sum at $m\asymp |z|+2$ . The initial portion is a harmonic sum of size $O(\log(|z|+2))$ , while the remainder is bounded because each summand is $O(|z|/m^2)$ .

The crucial gain from $\xi$ is that its logarithmic derivative records the zeros with positive weights to the right of the critical strip. The required form follows from the Hadamard factorization of the entire function $\xi$ . Its order is at most one; more concretely,

$\displaystyle \log|\xi(s)|\ll |s|\log(|s|+3).$

The canonical product therefore has genus one, and, with zeros counted with multiplicity,

$\displaystyle \frac{\xi'}{\xi}(s)=B+\sum_{\rho}\left(\frac{1}{s-\rho}+\frac{1}{\rho}\right).$

The sum is understood in the usual symmetric sense. Using the functional equation and the symmetry $\rho\leftrightarrow1-\overline\rho$ of the zero set, the real part simplifies to

$\displaystyle \text{Re}\frac{\xi'}{\xi}(\sigma+it)=\sum_{\rho}\frac{\sigma-\beta}{(\sigma-\beta)^2+(t-\gamma)^2},\qquad \sigma>1.$

Every summand is nonnegative. One may picture a zero $\rho$ as producing a positive peak centred at height $\gamma$ ; the closer $\sigma+it$ is to that zero, the larger its contribution.

Differentiating the definition of $\xi$ now gives

$\displaystyle \frac{\xi'}{\xi}(s)=\frac{1}{s}+\frac{1}{s-1}-\frac12\log\pi+\frac12\frac{\Gamma'}{\Gamma}\left(\frac{s}{2}\right)+\frac{\zeta'}{\zeta}(s).$

Combining this with the Gamma estimate and the positive zero formula yields, uniformly for $1<\sigma\le2$ ,

$\displaystyle -\text{Re}\frac{\zeta'}{\zeta}(\sigma+it)=\frac{\sigma-1}{(\sigma-1)^2+t^2}-\sum_{\rho}\frac{\sigma-\beta}{(\sigma-\beta)^2+(t-\gamma)^2}+O(\log(|t|+3)).$

The sign in front of the zero sum is the feature to remember. A zero lying close to $\sigma+it$ makes $-\zeta'/\zeta$ very negative. We will contradict such a large negative contribution using a nonnegative trigonometric polynomial and the nonnegative coefficients $\Lambda(n)$ .

Trigonometric Identity and Zero Free region

The whole zero-free argument rests on the elementary identity

$\displaystyle 3+4\cos\theta+\cos 2\theta=2(1+\cos\theta)^2\ge0.$

For $\sigma>1$ , the Dirichlet series for $-\zeta'/\zeta$ gives

$\displaystyle \begin{aligned} &-3\frac{\zeta'}{\zeta}(\sigma)-4\text{Re}\frac{\zeta'}{\zeta}(\sigma+it)-\text{Re}\frac{\zeta'}{\zeta}(\sigma+2it)\\ &=\sum_{n=1}^{\infty}\frac{\Lambda(n)}{n^{\sigma}}\left(3+4\cos(t\log n)+\cos(2t\log n)\right)\ge0. \end{aligned}$

This is not a formal trick. It compares the logarithmic derivative at three heights, $0,t,2t$ . If there were a zero close to $1+it$ , the middle term would become strongly negative. The coefficient $4$ in front of that term is deliberately larger than the coefficient $3$ in front of the pole at $s=1$ . That small numerical advantage is what leaves a positive gap between zeros and the line $\text{Re}(s)=1$ .

Let $\rho_0=\beta_0+i\gamma_0$ be a zero, write $\ell:=\log(|\gamma_0|+3)$ , and put $\sigma=1+\delta$ . The pole estimate at $1+\delta$ gives $-\zeta'/\zeta(1+\delta)\le\delta^{-1}+O(1)$ . At the point $1+\delta+i\gamma_0$ , the single zero $\rho_0$ contributes $-1/(\delta+1-\beta_0)$ , and every other zero contributes with the same nonpositive sign. At height $2\gamma_0$ , we may simply discard the nonpositive zero contributions. Substitution into the positive trigonometric inequality gives

$\displaystyle 0\le\frac{3}{\delta}-\frac{4}{\delta+1-\beta_0}+C\ell.$

Choose $\delta=(2C\ell)^{-1}$ , enlarging $C$ once if necessary. Then $3/\delta+C\ell=7C\ell$ , so

$\displaystyle \delta+1-\beta_0\ge\frac{4}{7C\ell}.$

Subtracting $\delta$ leaves

$\displaystyle 1-\beta_0\ge\frac{1}{14C\log(|\gamma_0|+3)}.$

Thus there is an absolute constant $a>0$ for which

$\displaystyle \zeta(s)\ne0\quad\text{whenever}\quad \text{Re}(s)\ge1-\frac{a}{\log(|\text{Im}(s)|+3)}.$

This is the de la Vallée Poussin zero-free region. It says that a zero at height $T$ must stay at least on the order of $1/\log T$ away from the line $\text{Re}(s)=1$ . The reciprocal logarithm may look weak, but it is already strong enough to give an exponentially decaying error in the prime number theorem.

Before shifting a contour, we need a uniform bound for $\zeta'/\zeta$ in the zero-free region. The positive potential formula first gives a local zero count. At $s=2+it$ ,

$\displaystyle \text{Re}\frac{\xi'}{\xi}(2+it)=\sum_{\rho}\frac{2-\beta}{(2-\beta)^2+(t-\gamma)^2}.$

Every zero with $|\gamma-t|\le1$ contributes at least $1/5$ , because $1\le2-\beta\le2$ . On the other hand, the explicit formula for $\xi'/\xi$ , the Gamma estimate, and the absolute convergence of $\zeta'/\zeta$ on $\text{Re}(s)=2$ show that the left side is $O(\log(|t|+3))$ . Hence

$\displaystyle |\{\rho:|\gamma-t|\le1\}| \ll\log(|t|+3).$

This tells us that only logarithmically many zero peaks can be near any given height. To convert this into a partial-fraction estimate, compare $\xi'/\xi(s)$ with $\xi'/\xi(2+it)$ , where $s=\sigma+it$ and $1/2\le\sigma\le2$ . For zeros with $|\gamma-t|>1$ ,

$\displaystyle \left|\frac{1}{s-\rho}-\frac{1}{2+it-\rho}\right|\ll\frac{1}{|t-\gamma|^2}.$

Group the zeros into strips $m<|t-\gamma|\le m+1$ . The local zero count bounds the number in each strip by $O(\log(|t|+m+3))$ , and summing the resulting convergent series gives

$\displaystyle \frac{\zeta'}{\zeta}(s)=\sum_{|\gamma-t|\le1}\frac{1}{s-\rho}+O\left(\log(|t|+3)+\frac{1}{|s-1|}\right).$

Now fix $T\ge3$ and put

$\displaystyle \sigma_0:=1-\frac{a}{2\log T}.$

If $|t|\le T$ and $\sigma_0\le\sigma\le2$ , then the zero-free region keeps every zero with $|\gamma-t|\le1$ at horizontal distance $\gg1/\log T$ from $s$ . There are only $O(\log T)$ such zeros, so their total contribution is $O((\log T)^2)$ . The remaining terms are smaller. Consequently,

$\displaystyle \frac{\zeta'}{\zeta}(\sigma+it)\ll(\log T)^2,\quad \sigma_0\le\sigma\le2,\quad |t|\le T.$

This is the estimate that makes the contour shift quantitative.

Perron’s Formula

A direct Perron formula for $\psi(x)$ has a kernel $1/s$ , which decays only like $1/|t|$ and makes truncation awkward. We therefore integrate once and work instead with

$\displaystyle \psi_1(x):=\sum_{n\le x}(x-n)\Lambda(n).$

The new kernel is $1/(s(s+1))$ , which decays like $1/t^2$ . The elementary Mellin calculation behind the formula is

$\displaystyle \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{y^s}{s(s+1)},ds=\begin{cases}1-y^{-1},& y>1, \\ 0, & 0<y<1 \end{cases}$

For $y>1$ , shift the contour to the left and collect the residues at $s=0$ and $s=-1$ ; for $0<y<1$ , shift it to the right, where there are no poles. Insert the Dirichlet series for $-\zeta'/\zeta$ and apply this kernel with $y=x/n$ . The result is

$\displaystyle \psi_1(x)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}-\frac{\zeta'}{\zeta}(s)\frac{x^{s+1}}{s(s+1)}ds,\qquad c>1.$

The pole of $-\zeta'/\zeta$ at $s=1$ has residue $1$ . Hence, when the contour is moved to the left, it contributes the main term $x^2/2$ .

Let $L:=\log x$ , take $c=1+1/L$ , and choose $\displaystyle \log T=\sqrt{\frac{aL}{2}},\quad \sigma_0=1-\frac{a}{2\log T}.$

For bounded $x$ , the desired estimate can be absorbed into the constant, so we may assume that $T$ is large. Shift the contour through the rectangle with vertices $c\pm iT$ and $\sigma_0\pm iT$ . The zero-free region shows that no zero of $\zeta$ is crossed. The only singularity crossed is the pole at $s=1$ , so

$\displaystyle \psi_1(x)=\frac{x^2}{2}+E_{\text{vert}}+E_{\text{hor}}+E_{\text{tail}}.$

On the new vertical side, the logarithmic derivative estimate and the integrability of $(1+t^2)^{-1}$ give

$\displaystyle E_{\text{vert}}\ll x^{\sigma_0+1}(\log T)^2=x^2(\log T)^2\exp\left(-\frac{a\log x}{2\log T}\right).$

On the two horizontal sides, $|s(s+1)|\gg T^2$ , so $\displaystyle E_{\text{hor}}\ll \frac{x^2(\log T)^2}{T^2}.$

Finally, on the original line $\text{Re}(s)=c$ , the absolutely convergent Dirichlet series gives $|\zeta'/\zeta(c+it)|\ll(c-1)^{-1}\ll L$ . The omitted tails above height $T$ therefore contribute $\displaystyle E_{\text{tail}}\ll\frac{x^2L}{T}.$

With our choice of $T$ , both $a\log x/(2\log T)$ and $\log T$ equal $\sqrt{a\log x/2}$ . The logarithmic factors can be absorbed by slightly weakening the constant in the exponential. Thus, for some absolute $b>0$ ,

$\displaystyle \psi_1(x)=\frac{x^2}{2}+O\left(x^2\exp\left(-b\sqrt{\log x}\right)\right).$

Removing Smoothing

The final step uses only positivity. Let $0<h\le x/2$ . Every term with $n\le x-h$ contributes exactly $h\Lambda(n)$ to $\psi_1(x)-\psi_1(x-h)$ , while terms with $x-h<n\le x$ make an additional nonnegative contribution. The same observation on the interval to the right of $x$ gives

$\displaystyle \frac{\psi_1(x)-\psi_1(x-h)}{h}\le\psi(x)\le\frac{\psi_1(x+h)-\psi_1(x)}{h}.$

Insert the asymptotic formula for $\psi_1$ . The main terms on the right and left are respectively $x+h/2$ and $x-h/2$ . Choose

$\displaystyle h=x\exp\left(-\frac{b}{2}\sqrt{\log x}\right).$

The smoothing error, after division by $h$ , has the same order as $h$ . Hence, after decreasing the positive constant once more,

$\displaystyle \psi(x)=x+O\left(x\exp\left(-c\sqrt{\log x}\right)\right).$

This is the de la Vallée Poussin form of the prime number theorem for the Chebyshev function.

The smoothing is a convenience rather than a necessity. One can begin directly from the truncated sharp Perron integral

$\displaystyle I(x,T):=\frac{1}{2\pi i}\int_{c-iT}^{c+iT}-\frac{\zeta'(s)}{\zeta(s)}\frac{x^s}{s},ds,\qquad c=1+\frac{1}{\log x}.$

The infinite version of this integral inverts the discontinuous cutoff $\displaystyle \mathbf 1_{n\le x}$ : it equals $1$ when $n<x$ , $0$ when $n>x$ , and $1/2$ at $n=x$ . At finite height $T$ , however, the cutoff is blurred in the short window $\displaystyle |n-x|\ll x/T.$ The standard truncated Perron estimate therefore gives

$\displaystyle \psi(x)=I(x,T)+O\left(\frac{x(\log x)^2}{T}+\log x\right).$

The term $\displaystyle x(\log x)^2/T$ is the price of using a sharp cutoff: the contour cannot distinguish perfectly between integers just below and just above $x$ .

The contour shift is otherwise the same. Move the segment to $\displaystyle \text{Re}(s)=1-\frac{2}{a\log T},$ where $a>0$ comes from the zero-free region. The pole at $s=1$ contributes $x$ . Since the kernel is now only $1/s$ , rather than $1/(s(s+1))$ , the vertical integral has one extra logarithmic loss and the horizontal segments have size about $\displaystyle x(\log T)^2/T.$ One obtains

$\displaystyle \psi(x)=x+O\left(x\exp\left(-\frac{2\log x}{a\log T}\right)(\log T)^3+\frac{x(\log x)^2}{T}\right).$

Choosing $\displaystyle \log T\asymp\sqrt{\log x}$ balances the two exponential savings and yields again

$\displaystyle \psi(x)=x+O\left(x\exp(-c\sqrt{\log x})\right).$

Thus smoothing does not improve the final de la Vallée Poussin error term. Its role is expository and technical: replacing the discontinuous cutoff by $\displaystyle (x-n)_+$ upgrades $\displaystyle 1/s$ to $\displaystyle 1/(s(s+1)),$ removes the boundary layer near $n=x$ , and makes the contour estimates absolutely convergent.

Prime Number Theorem

The difference $\psi(x)-\vartheta(x)$ counts only proper prime powers. Without using any prime number theorem, one has

$\displaystyle 0\le\psi(x)-\vartheta(x)=\sum_{{p^k\le x, ~ k\ge2}}\log p\ll\sqrt{x}(\log x)^2.$

Indeed, for each $k\ge2$ , there are at most $x^{1/k}$ possible primes $p$ , each weighted by at most $\log x$ , and the sum over $k$ is dominated by its first term up to a logarithmic factor. This is negligible compared with $x\exp(-c\sqrt{\log x})$ . Therefore

$\displaystyle \vartheta(x)=x+O\left(x\exp\left(-c\sqrt{\log x}\right)\right).$

Finally, partial summation gives

$\displaystyle \pi(x)=\frac{\vartheta(x)}{\log x}+\int_2^x\frac{\vartheta(t)}{t(\log t)^2}dt.$

Substituting the main term $\vartheta(t)=t$ gives $\text{Li}(x)+O(1)$ , because the derivative of $x/\log x+\int_2^xdt/(\log t)^2$ is $1/\log x$ . For the error term, split the integral at $\sqrt{x}$ . The lower portion is far smaller than the claimed final error. On the interval $[\sqrt{x},x]$ , one has $\sqrt{\log t}\ge\sqrt{\log x}/\sqrt2$ , so the exponential factor is uniformly bounded by a slightly weaker exponential in $\sqrt{\log x}$ . Consequently, after one final reduction of $c$ ,

$\displaystyle \pi(x)=\text{Li}(x)+O\left(\frac{x}{\log x}\exp\left(-c\sqrt{\log x}\right)\right).$

It is useful to see the proof as a single chain rather than as a collection of unrelated estimates. The Euler product says that weighted prime powers are the coefficients of $-\zeta'/\zeta$ . Completion and factorization say that zeros of $\zeta$ appear as a positive potential in $\text{Re}(\xi'/\xi)$ . The polynomial $3+4\cos\theta+\cos2\theta$ converts the positivity of the coefficients $\Lambda(n)$ into an inequality that cannot coexist with a zero extremely close to one. The zero-free region then permits a contour shift; the shifted contour produces a saving, and the finite truncation produces a competing loss. The two costs balance at height $\log T\asymp\sqrt{\log x}$ , giving the classical error term. The square root is therefore not an incidental artifact of the proof. It is the numerical fingerprint of a reciprocal-logarithmic zero-free region. A stronger zero-free region would move the contour farther left and improve the error term; a weaker one would give less saving. The prime number theorem with de la Vallée Poussin’s error term is exactly what this classical zero-free region is designed to yield.

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