# The prime number theorem

*Written by GPT-6.1 Sol (OpenAI) in Codex, Ultra setting, October 2026. Self-checked by the writing AI. Public domain (CC0).*

We now turn nonvanishing on the line one into a statement about the distribution of primes. The extra factor $1/(s+1)$ in a smoothed counting integral makes its vertical tails absolutely integrable. Removing the pole leaves a Fourier transform of an integrable function, which tends to zero. We can then remove the smoothing. The same integral proves cancellation of the Möbius function, but its unsmoothing requires a bound on its increments instead of monotonicity.

The elementary estimates and prime-counting equivalences are proved in Counting primes by elementary means, Theorems 2.1 and 5.2. We use the Euler series of $-\zeta'/\zeta$ and $1/\zeta$ from Dirichlet series and Euler products, §3, and the nonvanishing, pole neighbourhood and logarithmic bounds from Nonvanishing on the line one and a zero-free region, Theorems 1.1 and 4.1 and Lemma 3.1. We need only those boundary bounds for the first proof; its conclusion does not use the numerical width of the zero-free region. Sections 1–4 prove the smoothing and unsmoothing arguments in full. We also give the full analytic theorem of Newman and its alternative application in §5.

Write
$$
\psi(x)=\sum_{n\le x}\Lambda(n),\qquad
\theta(x)=\sum_{p\le x}\log p,\qquad
M(x)=\sum_{n\le x}\mu(n),
$$
and define $\psi_1(x)=\int_1^x\psi(u)\,du$ and $M_1(x)=\int_1^xM(u)\,du$ for $x\ge1$. The notation $\psi$ in this lesson always denotes Chebyshev's function. We write $\operatorname{li}(x)$ for the principal-value logarithmic integral and $\operatorname{Li}(x)=\int_2^xdu/\log u=\operatorname{li}(x)-\operatorname{li}(2)$.

The integration and complex-analysis tools used in this lesson are proved in Dirichlet series and Euler products, Appendix A, Lemmas A.1–A.4.

## 1. An integrable counting kernel

**Lemma 1.1 (the once-smoothed Mellin kernel).** For $y>0$ and $a>0$,
$$
\frac1{2\pi i}\int_{a-i\infty}^{a+i\infty}
\frac{y^s}{s(s+1)}\,ds=
\begin{cases}1-y^{-1},&y\ge1,\\0,&0<y\le1.\end{cases}
\tag{1.1}
$$
Both values at $y=1$ are zero, and the integral converges absolutely.

**Proof.** On the line, $|y^s|=y^a$ and $|s(s+1)|\asymp_a1+t^2$, proving convergence. For $y\ge1$, close the segment $a-iT$ to $a+iT$ by the left semicircle centred at $a$ with radius $T$. For $T$ sufficiently large its denominator has modulus at least a fixed multiple of $T^2$, while $|y^s|\le y^a$. The arc has length $\pi T$, so its integral tends to zero. Its enclosed poles at zero and minus one have residues 1 and $-y^{-1}$. The positively oriented residue formula gives (1.1), including $y=1$ by the same estimate. For $y<1$, close to the right: there are no poles, $|y^s|\le y^a$, and the same arc estimate gives zero. $\square$

The kernel vanishes continuously at the cutoff. It averages the usual sharp indicator rather than assigning a half-weight at a discontinuity.

**Proposition 1.2 (smoothed Perron identity).** Suppose $D(s)=\sum_{n\ge1}b(n)n^{-s}$ converges absolutely on $\Re s=a>1$. Then for $x\ge1$,
$$
\sum_{n\le x}b(n)(x-n)
=\frac1{2\pi i}\int_{a-i\infty}^{a+i\infty}
\frac{x^{s+1}D(s)}{s(s+1)}\,ds.
\tag{1.2}
$$
In particular it holds for $D=-\zeta'/\zeta$ and $D=1/\zeta$, giving $\psi_1$ and $M_1$ respectively.

**Proof.** The sum of the integrals of the absolute values is at most
$$
\frac{x^{a+1}}{2\pi}\sum_{n\ge1}\frac{|b(n)|}{n^a}
\int_{\mathbb R}\frac{dt}{|(a+it)(a+1+it)|}<\infty.
$$
Interchange sum and integral, and apply (1.1) with $y=x/n$. Its contribution is $x(1-n/x)$ when $n\le x$, zero otherwise. Integrating the finite counting sum over $1\le u\le x$ gives exactly $\sum_{n\le x}b(n)(x-n)$. The boundary $n=x$ contributes zero. $\square$

At fixed $x$ the denominator decays as $t^{-2}$. On a line of absolute convergence this suffices by boundedness of $D$; on the boundary line it also accommodates the $O(\log(|t|+2))$ bounds proved in the preceding lesson.

## 2. Removing the pole and reaching the boundary

**Lemma 2.1 (the Fourier limit needed here).** If $h:\mathbb R\to\mathbb C$ is continuous and absolutely integrable, then
$$
\int_{\mathbb R}h(t)e^{iut}\,dt\longrightarrow0
\quad\text{as }|u|\to\infty.
\tag{2.1}
$$

**Proof.** Given $\varepsilon>0$, choose $R$ so the integral of $|h|$ outside $[-R,R]$ is less than $\varepsilon$. Uniform continuity on that interval gives a finite step function $q$ with $\int_{-R}^R|h-q|<\varepsilon$. On a step interval $[b,d]$, the integral of $e^{iut}$ is $(e^{iud}-e^{iub})/(iu)$ when $u\ne0$. Thus the integral of $q e^{iut}$ tends to zero. The tail and approximation errors total less than $2\varepsilon$, independently of $u$. Let $\varepsilon$ decrease to zero. $\square$

This is the continuous, integrable case of the Riemann–Lebesgue lemma. It supplies the precise oscillatory limit used below.

**Theorem 2.2 (smoothed prime number theorem).** As $x\to\infty$,
$$
\psi_1(x)=\frac{x^2}{2}+o(x^2).
\tag{2.2}
$$

**Proof.** Put $A(s)=-\zeta'(s)/\zeta(s)$ and
$$
B(s)=A(s)-\frac1{s-1}.
$$
Since zeta has a simple pole at one, $A(s)=1/(s-1)+O(1)$ there and $B$ is holomorphic there. Nonvanishing on the rest of the line and the Euler product give a holomorphic extension of $B$ to an open neighbourhood of every point with $\Re s\ge1$. No uniform leftward neighbourhood is needed.

For $1\le\sigma\le2$ and $|t|\ge2$, the preceding lesson gives $B(s)=O(\log(|t|+2))$. On the remaining compact part of this strip, $B$ is bounded. In the smoothed integral take $a=2$ and first subtract the pole term. Direct residues give, for $x\ge1$,
$$
\frac1{2\pi i}\int_{2-i\infty}^{2+i\infty}
\frac{x^{s+1}}{(s-1)s(s+1)}\,ds
=\frac{x^2}{2}-x+\frac12=\frac{(x-1)^2}{2}.
\tag{2.3}
$$
To check this integral, close to the left as in Lemma 1.1. The arc is now $O_x(T^{-2})$, and the residues at $1,0,-1$ are $x^2/2,-x,1/2$.

For the regular part, shift a rectangle from $\sigma=2$ to $\sigma=1$. There are no poles. At height $T$, the horizontal integrals have modulus $O_x(\log(T+2)/T^2)$ and tend to zero; the vertical integrals converge absolutely. Therefore Proposition 1.2 becomes
$$
\psi_1(x)=\frac{(x-1)^2}{2}
+\frac{x^2}{2\pi}\int_{\mathbb R}
\frac{B(1+it)}{(1+it)(2+it)}e^{it\log x}\,dt.
\tag{2.4}
$$
The last kernel is continuous and integrable: it is bounded near zero and $O(\log(|t|+2)/t^2)$ at infinity. Lemma 2.1 sends its Fourier transform to zero as $\log x\to\infty$. Equation (2.4) proves (2.2). $\square$

Subtracting the entire pole transform before shifting avoids an integral through the pole at $s=1$. Its lower-order terms in (2.3) have no effect on the leading asymptotic, but keeping them makes the identity exact.

## 3. From an average to the number of primes

**Theorem 3.1 (prime number theorem).** The four equivalent asymptotics
$$
\psi(x)\sim x,\qquad \theta(x)\sim x,\qquad
\pi(x)\sim\frac{x}{\log x},\qquad
\pi(x)\sim\operatorname{li}(x)
\tag{3.1}
$$
hold. Equivalently $\pi(x)\sim\operatorname{Li}(x)$.

**Proof.** Fix $0<\varepsilon<1$. Monotonicity of $\psi$ gives
$$
\frac{\psi_1(x)-\psi_1((1-\varepsilon)x)}{\varepsilon x}
\le\psi(x)\le
\frac{\psi_1((1+\varepsilon)x)-\psi_1(x)}{\varepsilon x}.
\tag{3.2}
$$
Use Theorem 2.2 at each argument and divide by $x$. For fixed $\varepsilon$ the lower and upper bounds tend to $1-\varepsilon/2$ and $1+\varepsilon/2$. First let $x$ tend to infinity, then let $\varepsilon$ decrease to zero. This gives $\psi(x)/x\to1$.

The exact equivalences, including $\psi-\theta=O(\sqrt x)$, partial summation and $\operatorname{li}(x)\sim x/\log x$, were proved in lesson two, Theorems 2.1 and 5.2. They give the other three assertions. The constant difference between $\operatorname{Li}$ and $\operatorname{li}$ does not affect an asymptotic to an unbounded function. $\square$

The order of limits in (3.2) matters: the error $o(x^2)$ has no stated rate, so $\varepsilon$ is kept fixed until the first limit has been taken.

**Corollary 3.2 (the size of the $n$-th prime).** If $p_n$ is the $n$-th prime, then
$$
p_n\sim n\log n.
\tag{3.3}
$$

**Proof.** At $x=p_n$, $n=\pi(p_n)\sim p_n/\log p_n$. Taking logarithms gives
$$
\log n=\log p_n-\log\log p_n+o(1),
$$
so $\log n/\log p_n\to1$. Combining this with $p_n/(n\log p_n)\to1$ gives (3.3). The infinitely many primes and $p_n\to\infty$ follow already from the Euler-product lesson, or directly from (3.1). $\square$

**Example 3.3.** Enumerating the primes and their powers, with high-precision logarithms, gives

| $x$ | $\psi(x)$ | $\psi(x)/x$ |
|---:|---:|---:|
| $10^3$ | $996.680912247$ | $0.996680912247$ |
| $10^6$ | $999586.597496$ | $0.999586597496$ |

The computation is $\psi(x)=\sum_{p\le x}\lfloor\log x/\log p\rfloor\log p$, with the integer number of powers determined by integer multiplication. Finite examples illustrate the scale and do not supply the asymptotic proof.

## 4. Cancellation of the Möbius function

**Theorem 4.1.** We have $M(x)=o(x)$ and, as a convergent ordinary series,
$$
\sum_{n=1}^\infty\frac{\mu(n)}n=0.
\tag{4.1}
$$

**Proof.** The function $D(s)=1/\zeta(s)$ extends holomorphically across $s=1$ with a zero there, and is holomorphic at every other point of $\Re s\ge1$. The boundary bound in the preceding lesson gives $D(s)=O(\log(|t|+2))$ for $1\le\sigma\le2$, $|t|\ge2$. Shift the smoothed integral (1.2) from 2 to 1 just as in Theorem 2.2. This time there is no pole to subtract. We get
$$
M_1(x)=\frac{x^2}{2\pi}\int_{\mathbb R}
\frac{D(1+it)}{(1+it)(2+it)}e^{it\log x}\,dt=o(x^2)
\tag{4.2}
$$
by Lemma 2.1.

For $u\ge x$, the bound $|\mu(n)|\le1$ gives $|M(u)-M(x)|\le u-x+1$. With $h=\varepsilon x>0$,
$$
\left|M(x)-\frac{M_1(x+h)-M_1(x)}h\right|
\le\frac h2+1.
\tag{4.3}
$$
For fixed $\varepsilon$, the average in (4.3) is $o(x)$ by (4.2). Hence $\limsup|M(x)|/x\le\varepsilon/2$. Let $\varepsilon$ decrease to zero. This proves $M=o(x)$ without a false monotonicity assumption. The convergence and value in (4.1) follow from the floor identity and bounded-variation argument written in Solution 3 below. $\square$

An arbitrary $M(x)=o(x)$ estimate does not make $\int_1^\infty M(u)u^{-2}\,du$ absolutely convergent. That integral cannot on its own justify the final assertion. The arithmetic identity in Solution 3 supplies the missing information.

## 5. Newman's analytic theorem

There is also a boundary argument that requires no growth estimate at large imaginary parts. It uses boundedness of a function on the time axis and analytic continuation of its Laplace transform. The contour below proves the needed theorem in full. This method is due to Newman; [Zagier 1997] gives a short presentation of it.

**Theorem 5.1 (Newman's analytic theorem).** Let $f:[0,\infty)\to\mathbb C$ be locally integrable and bounded, with $|f(u)|\le H$. For $\Re z>0$ let
$$
G(z)=\int_0^\infty f(u)e^{-zu}\,du.
$$
Suppose $G$ extends holomorphically to an open set containing the closed half-plane $\Re z\ge0$. Then the improper integral of $f$ converges and
$$
\lim_{T\to\infty}\int_0^T f(u)\,du=G(0).
\tag{5.1}
$$
The continuation hypothesis is local at each boundary point; it does not assert a fixed-width strip to the left.

**Proof.** Put $G_T(z)=\int_0^T f(u)e^{-zu}\,du$, which is entire by differentiation under a finite integral. Fix $R>0$. Write $C_+$ for the right semicircle of radius $R$, oriented from $-iR$ to $iR$, and $C_-$ for the left semicircle oriented from $iR$ to $-iR$. Thus $C_++C_-$ is positively oriented.

Compactness of the segment $[-iR,iR]$ in the open continuation set gives $0<\delta<R/2$ such that the rectangle $[-\delta,0]+i[-R,R]$ lies in that set. Let $\Gamma_-$ run from $iR$ horizontally to $-\delta+iR$, vertically to $-\delta-iR$, and horizontally to $-iR$. Together with $C_+$ it encloses zero and lies, with its interior, in the domain of $G$. Use the kernel
$$
K_T(z)=\frac{e^{Tz}}z\left(1+\frac{z^2}{R^2}\right).
$$
Its sole pole inside either contour is at zero, with residue one. Cauchy's formula on the two respective contours yields
$$
\begin{split}
G(0)-G_T(0)=\frac1{2\pi i}\bigg(&
\int_{C_+}(G-G_T)K_T\,dz
+\int_{\Gamma_-}GK_T\,dz
-\int_{C_-}G_TK_T\,dz\bigg).
\end{split}
\tag{5.2}
$$
We estimate its three terms separately.

On $C_+$, away from the endpoints, $\sigma=\Re z>0$ and
$$
|G(z)-G_T(z)|\le H\int_T^\infty e^{-\sigma u}\,du
=\frac{H e^{-T\sigma}}\sigma.
$$
On $C_-$, where $\sigma<0$, the finite transform satisfies
$$
|G_T(z)|\le H\frac{e^{-T\sigma}-1}{-\sigma}
\le\frac{H e^{-T\sigma}}{-\sigma}.
$$
For $z=Re^{i\vartheta}$ the crucial cancellation is exact:
$$
\left|1+\frac{z^2}{R^2}\right|
=2|\cos\vartheta|=\frac{2|\sigma|}{R}.
\tag{5.3}
$$
Each arc integrand in (5.2) consequently has modulus at most $2H/R^2$. Each arc has length $\pi R$, so after the factor $1/(2\pi)$ each contributes at most $H/R$. The endpoints, where the kernel vanishes, have zero arc measure; no estimate dividing by zero is needed there.

The path $\Gamma_-$ is fixed as $T\to\infty$, avoids zero, and is compact in the holomorphic domain of $G$. Its other factors are bounded. On its interior $\Re z<0$, so $e^{Tz}\to0$, while $|e^{Tz}|\le1$ everywhere on the path. Dominated convergence gives $\int_{\Gamma_-}GK_T\,dz\to0$. Hence (5.2) implies
$$
\limsup_{T\to\infty}|G(0)-G_T(0)|\le\frac{2H}{R}.
\tag{5.4}
$$
Since $R$ was arbitrary, the left side is zero. This proves (5.1). $\square$

![The two contours in Newman's theorem](figures/NT-ZETA-09/newman_contours.png)

*Figure 1. The entire function $G_T$ uses the full circle on the left. The continued function $G$ uses the right semicircle and the narrow path $\Gamma_-$ on the right. Arrows show the positive orientations in (5.2); the factor in (5.3) vanishes at $\pm iR$. The drawing uses $R=3$ and $\delta=0.45$ to display the geometry. In the proof, $\delta$ is chosen separately for each $R$ from the actual continuation neighbourhood. No continuation of $G$ through the whole left semicircle is assumed. Original CC0 figure for Theorem 5.1, with reproducible source.*

**Corollary 5.2 (the prime number theorem by Newman).** Nonvanishing of $\zeta$ on $\Re s=1$, together with the elementary Chebyshev bound, implies $\psi(x)\sim x$.

**Proof.** Define
$$
q(u)=e^{-u}\psi(e^u),\qquad f(u)=q(u)-1.
$$
Chebyshev's bound makes $q$ and $f$ bounded; their step discontinuities cause no difficulty with local integrability. For $\Re z>0$, the substitution $x=e^u$ and termwise integration give
$$
\begin{split}
\int_0^\infty q(u)e^{-zu}\,du
&=\int_1^\infty\psi(x)x^{-z-2}\,dx\\
&=\sum_{n\ge1}\Lambda(n)\int_n^\infty x^{-z-2}\,dx
=\frac{A(1+z)}{1+z}.
\end{split}
\tag{5.5}
$$
Absolute interchange follows from $\sum\Lambda(n)n^{-1-\Re z}<\infty$; equivalently one can first apply the positive integral at the real part. Thus $G(z)=A(1+z)/(1+z)-1/z$ is the Laplace transform of $f$. At zero, writing $A(1+z)=1/z+B(1+z)$ gives
$$
G(z)=\frac{B(1+z)-1}{1+z},
\tag{5.6}
$$
so the apparent pole is removable. At every other point of $\Re z=0$, nonvanishing of zeta makes $A(1+z)$ holomorphic in a neighbourhood. The Euler product deals with $\Re z>0$. These neighbourhoods together form the open continuation set required in Theorem 5.1.

It follows that $\int_0^T(q(u)-1)\,du$ has a finite limit. For each fixed $a>0$, its differences at $u+a,u$ and at $u,u-a$ therefore tend to zero:
$$
\int_u^{u+a}q(v)\,dv\longrightarrow a,\qquad
\int_{u-a}^uq(v)\,dv\longrightarrow a.
\tag{5.7}
$$
For $0\le h\le a$, monotonicity of $\psi$ implies
$$
q(u+h)\ge e^{-h}q(u),\qquad
q(u-h)\le e^h q(u).
$$
Integrating these inequalities and using (5.7) gives
$$
\limsup_{u\to\infty}q(u)\le\frac{a}{1-e^{-a}},\qquad
\liminf_{u\to\infty}q(u)\ge\frac{a}{e^a-1}.
\tag{5.8}
$$
Let $a$ decrease to zero. Both bounding quantities tend to one, hence $q(u)\to1$, which is $\psi(x)/x\to1$. $\square$

This proof shows exactly where boundedness and the boundary continuation replace a quantitative bound in height. The Wiener–Ikehara theorem provides a broader Tauberian framework for prime counting. The elementary proofs of Selberg and Erdős offer another route. We mention those methods for orientation and use neither as a proof input here.

## Exercises

1. **Easy: unsmoothing a monotone count.** Deduce $\psi(x)\sim x$ from $\psi_1(x)\sim x^2/2$. Keep track of the order of the two limits.
2. **Medium: inverting prime counting.** Deduce $p_n\sim n\log n$ from $\pi(x)\sim x/\log x$ without assuming the size of $\log p_n$ in advance.
3. **Medium: the Möbius series at one.** Starting from $M(x)=o(x)$ and $\sum_{n\le x}\mu(n)\lfloor x/n\rfloor=1$, prove that $\sum_{n\le x}\mu(n)/n\to0$. Explain why an unquantified partial-summation integral does not suffice.
4. **Hard: Newman's theorem and prime counting.** Prove the analytic theorem for a bounded locally integrable function whose Laplace transform continues holomorphically across every point of the imaginary axis. Apply it to $f(u)=e^{-u}\psi(e^u)-1$ and deduce the prime number theorem.
5. **Medium: a prime in a multiplicative interval.** For each fixed $\varepsilon>0$, prove that $(x,(1+\varepsilon)x]$ contains a prime for all sufficiently large $x$.

## Solutions

**Solution 1.** On $[(1-\varepsilon)x,x]$, the monotone function $\psi(u)$ is at most $\psi(x)$; on $[x,(1+\varepsilon)x]$ it is at least $\psi(x)$. These integral comparisons give (3.2). For fixed $\varepsilon$, substitution of $\psi_1(y)=y^2/2+o(y^2)$ gives
$$
1-\frac\varepsilon2\le\liminf\frac{\psi(x)}x
\le\limsup\frac{\psi(x)}x\le1+\frac\varepsilon2.
$$
Letting $\varepsilon\downarrow0$ proves the assertion. Choosing a shrinking $\varepsilon(x)$ before controlling the remainder would be unjustified.

**Solution 2.** Substitution of $p_n$ in the prime number theorem gives $p_n/(n\log p_n)\to1$. Its logarithm gives $\log n=\log p_n-\log\log p_n+o(1)$. Since $p_n\to\infty$, divide this last equality by $\log p_n$: the quotient $\log\log p_n/\log p_n$ tends to zero, so $\log n/\log p_n\to1$. Multiplication of the two known ratios now gives $p_n/(n\log n)\to1$.

**Solution 3.** The finite floor identity follows directly from $\mu*1$:
$$
\sum_{n\le x}\mu(n)\left\lfloor\frac xn\right\rfloor
=\sum_{m\le x}\sum_{n\mid m}\mu(n)=1\quad(x\ge1).
$$
Consequently, if $S(x)=\sum_{n\le x}\mu(n)/n$, then
$$
xS(x)=1+\sum_{n\le x}\mu(n)\{x/n\}.
\tag{6.1}
$$
Fix $0<\varepsilon<1$. The portion $n\le\varepsilon x$ on the right has modulus at most $\varepsilon x$. For the remaining portion, put
$$
E_x=\sup_{\varepsilon x\le u\le x}|M(u)|=o(x)
$$
with $\varepsilon$ fixed. This follows from $M(u)/u\to0$ uniformly once $u\ge\varepsilon x\to\infty$.

Partition $(\varepsilon x,x]$ at the points $x/k$. It has at most $1+\lceil1/\varepsilon\rceil$ pieces of the form $(b,d]$ lying in $(x/(k+1),x/k]$, possibly truncated at the lower end. On a piece, $\{x/n\}=r(n)=x/n-k$, with $0\le r\le1$ and $r$ decreasing. Ordinary partial summation on that piece is the exact identity
$$
\sum_{b<n\le d}\mu(n)r(n)
=M(d)r(d)-M(b)r(b)-\int_b^d M(u)r'(u)\,du.
\tag{6.2}
$$
Here $r(b)$ is the smooth branch's value, even when the lower endpoint is a partition point; that point is excluded from the sum and belongs to the preceding piece. The endpoint values have modulus at most one, and $\int_b^d|r'(u)|\,du\le1$. Thus each piece costs at most $3E_x$. The whole long portion is $o(x)$ for fixed $\varepsilon$.

Dividing (6.1) by $x$ gives $\limsup|S(x)|\le\varepsilon$. Let $\varepsilon\downarrow0$. This proves the convergence of the ordinary partial sums to zero, as asserted in Theorem 4.1. An arbitrary bound $M=o(x)$ alone leaves a possible nonintegrable $o(1)/u$ in $\int M(u)u^{-2}\,du$; the finite floor identity is what closes the argument.

**Solution 4.** Theorem 5.1 supplies the complete analytic proof, including the contour identity (5.2), the two arc bounds $H/R$ from the exact cancellation (5.3), and the vanishing narrow-path integral. These give $\limsup|G(0)-\int_0^Tf|\le2H/R$ for every $R$, and then convergence by $R\to\infty$. The construction of $\Gamma_-$ uses only compactness of a finite segment of the continuation domain, so the stated local continuation hypothesis suffices.

For the required arithmetic application, Chebyshev's estimate bounds $f$. Its transform is $A(1+z)/(1+z)-1/z$ by (5.5); its value at zero is removable by (5.6), and all other imaginary-axis points are regular by zeta's nonvanishing. The analytic theorem gives convergence of the integral of $q-1$. Its differences over each fixed forward and backward interval converge to zero. Monotonicity then bounds $q(u)$ by the forward integral divided by $1-e^{-a}$ and from below by the backward integral divided by $e^a-1$. These yield (5.8). Taking $a\downarrow0$ proves $q(u)\to1$, hence the prime number theorem. No growth assumption on $G$ at large imaginary parts was inserted.

**Solution 5.** For fixed $\varepsilon>0$, $\log((1+\varepsilon)x)=\log x+O_\varepsilon(1)$. The prime number theorem at the two arguments gives
$$
\pi((1+\varepsilon)x)-\pi(x)
=\bigl(\varepsilon+o_\varepsilon(1)\bigr)\frac{x}{\log x}.
$$
This is positive for sufficiently large $x$. Since it counts precisely the primes in the specified half-open interval, at least one prime lies there. Fixed $\varepsilon$ is essential to this argument; it asserts no uniform shrinking-interval estimate.

## Freely readable sources

- D. Koukoulopoulos, [*The Distribution of Prime Numbers*](https://dms.umontreal.ca/~koukoulo/documents/publications/primes.pdf), freely readable author preliminary version, 367 pages. Chapter 8, Theorem 8.1 and its proof on.
- D. Zagier, [*Newman’s Short Proof of the Prime Number Theorem*](https://people.mpim-bonn.mpg.de/zagier/files/doi/10.2307/2975232/fulltext.pdf) (1997), complete author-hosted article;. Printed pages 707–708 give Newman’s analytic theorem and its proof.
- E. Bombieri, [*Problems of the Millennium: the Riemann Hypothesis*](https://www.claymath.org/wp-content/uploads/2022/05/riemann.pdf), freely readable official Clay Mathematics Institute paper.