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First continue the Riemann zeta function into $\Re s>0$. Summing the integral of the integer-part function gives, initially for $\Re s>1$,The last integral defines a holomorphic function for $\Re s>0$, so this continuation has only a simple pole at one, of residue one.

For $\sigma >1$, logarithmic expansion of the Euler product and $3+4\cos v+\cos 2v=2(1+\cos v{)}^2\ge 0$ giveThis proves the three-four-one product proof of zeta boundary nonvanishing inequalityIf $t\ne 0$ and $\zeta (1+it)=0$ to order $r\ge 1$, the middle factor is $O((\sigma -1{)}^{4r})$, the first is $O((\sigma -1{)}^{-3})$, and the last remains bounded. Their product tends to zero, a contradiction. At $t=0$ there is a pole, not a zero. The zeta function has no zero on $\Re s=1$.

Put $I(x)={\int }_0^x\psi (u)du={\sum }_{n\le x}(x-n)\Lambda (n)$, where $\Lambda$ is the Von Mangoldt function. Logarithmic differentiation of the Euler product gives $-{\zeta }^{\prime }(s)/\zeta (s)={\sum }_n\Lambda (n)n^{-s}$ for $\Re s>1$. The first integral of the Chebyshev function is consequentlyTo check its kernel, evaluate the contour integral of $y^s/(s(s+1))$ by closing to the left for $y>1$ and to the right for $0<y<1$. The residues at zero and minus one give $1-y^{-1}$ in the first case and zero in the second. Multiplication by $x$ and substitution $y=x/n$ therefore produce $(x-n{)}_{+}$. On the line $c>1$, the Dirichlet series and vertical integral have absolute convergence, justifying their interchange; at $x=n$ the kernel is zero on either side.

Here is the smoothing step that establishes the requested asymptotic. Shift the line of this contour integral to the left. The pole of $-{\zeta }^{\prime }/\zeta$ at one has residue $+1$; at each Nontrivial zero of the Riemann zeta function $\rho$ of zero multiplicity $m_{\rho }$ its residue is $-m_{\rho }$. The denominators also contribute at zero and minus one, and the trivial zeros of the Riemann zeta function occur at $-2,-4,\dots$. Summing these residues yields the smoothed explicit formula for the Chebyshev functionZeros are repeated with zero multiplicity. For the limits of the contour integral, choose horizontal heights avoiding zero ordinates, where the logarithmic derivative on each fixed strip is $O((\log T{)}^2)$. The factor $s(s+1)$ makes their integrals tend to zero. On left edges chosen a fixed distance from the trivial zeros, the functional equation of the Riemann zeta function controls the logarithmic derivative by $O(\log (|s|+2))$; for fixed $x>1$ the remaining vertical integral vanishes as that edge moves left. This is why the extra integration of $\psi$ is useful.

The Riemann–von Mangoldt formula, stated in Question 3, gives $N(T)=O(T\log T)$, henceIndeed each dyadic height block contributes $O((\log T)/T)$. The zero-free line just proved, the nonvanishing Euler product on its right, and the functional equation of the Riemann zeta function put every Nontrivial zero of the Riemann zeta function in $0<\Re \rho <1$. Therefore each $x^{\rho -1}$ tends to zero, with absolute value at most one for $x\ge 1$. Divide the explicit formula by $x^2$ and use dominated convergence on the zero terms with absolute convergence. The remaining displayed terms are $o(x^2)$, and we obtain

Monotonicity of $\psi$ removes the smoothing. For $0<h<1$,After division by $x$, the limits of the two bounds are $1-h/2$ and $1+h/2$. Let $h\downarrow 0$ to get $\psi (x)\sim x$. Question 1 gives $\psi (x)-\vartheta (x)=O(\sqrt{x}\log x)=o(x)$, so $\vartheta (x)\sim x$. Partial summation now givesThe integral is $O(x/(\log x{)}^2)$, by splitting at $\sqrt{x}$ and using $\vartheta (t)=O(t)$. Inverting this at $x=p_n$, we have $n\sim p_n/\log p_n$ and hence $\log n\sim \log p_n$. Thus the $n$th prime satisfies

Fix a modulus $q$ and a residue class $a$ with $(a,q)=1$. The Dirichlet characters modulo $q$ are the one-dimensional characters of a representation of $(\mathbb{Z}/q\mathbb{Z}{)}^{\times }$, extended by zero on nonunits. Let ${\chi }_0$ be the principal Dirichlet character.

For every nonprincipal Dirichlet character the partial sums ${\sum }_{n\le u}\chi (n)$ are bounded: the sum over a full period is zero, and a remaining incomplete period has bounded length. Partial summation therefore continues its Dirichlet L-function holomorphically to $\Re s>0$ byThis shows in particular that it has no pole at one. By contrasthas a simple pole there. We use the stipulated nonvanishing $L(1,\chi )\ne 0$ for every real nonprincipal Dirichlet character, and establish the remaining nonvanishing as follows.

For real $\sigma >1$, take the product over all Dirichlet characters. The Euler product logarithms and Orthogonality of Dirichlet characters giveThus this product is real and at least one. If a nonreal Dirichlet character had $L(1,\chi )=0$, complex conjugation would also give $L(1,\overline{\chi })=0$. These are distinct factors. Two zeros and only one possible simple pole force the whole product to tend to zero as $\sigma \downarrow 1$, contradicting the lower bound. Hence all nonprincipal Dirichlet characters have $L(1,\chi )\ne 0$, under the given real-character assumption.

For $\sigma >1$, absolute convergence of each Euler product giveswith bounded remainder as $\sigma \downarrow 1$: the terms of prime-power exponent at least two have total modulus at most ${\sum }_{n\ge 2}1/(n(n-1))$. Since a nonprincipal $L$ defines a holomorphic function that is nonzero near one, its Euler-product logarithm differs from a local holomorphic logarithm by a fixed integer multiple of $2\pi i$ on a short real interval. It is therefore bounded there. For the principal Dirichlet character the pole instead givesApply the Dirichlet character orthogonality relation at each prime. The prime-character sum near one yieldsThis tends to infinity. A finite set of primes would have a bounded sum as $\sigma \downarrow 1$, so every coprime residue class contains infinitely many primes. This proves the Dirichlet theorem on primes in arithmetic progressions from the specified assumption. In fact it proves divergence of the corresponding reciprocal-prime series and gives Dirichlet density $1/\phi (q)$.

Finally, for real $s>1$ each prime satisfiesThe integrands are nonnegative, so Tonelli theorem permits summation inside the integral. At a fixed $t$, exactly $\pi (t,q,a)$ primes from the residue class have entered. ThereforeBoth sides are finite, since they are bounded respectively by ${\sum }_{n\ge 2}{n}^{-s}$ and $s{\int }_1^{\infty }{t}^{-s}dt$. The values of a counting function at its isolated jump endpoints do not change this integral.

A precise endpoint version of the Perron formula is as follows. If $F(s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$ converges absolutely on $\Re s=c>0$, then, for $x>0$,When $x$ is not an integer, the left side is simply the sum over $n\le x$.

Prove first the scalar kernel formula, with symmetric vertical limits:For $y>1$, close the contour to the left and apply the residue theorem at $s=0$; the exponential $e^{s\log y}$ makes the closing integral vanish. For $y<1$, close to the right, where there is no enclosed pole and the exponential decreases. The usual semicircle estimate, or integration by parts on the closing arcs, justifies these limits. At $y=1$ the truncated integral is ${\pi }^{-1}\arctan (T/c)$, tending to $1/2$.

One also needs a uniform bound to pass through the Dirichlet series. Pairing the two halves of the vertical integral gives, with $u=\log y$,The cosine term is absolutely bounded. The sine term is uniformly bounded by integration by parts against the bounded sine integral ${\int }_0^v\sin (tu)dt/t$, since $t^2/(c^2+t^2)$ is increasing from zero to one. Thus $|K_T(y)|\le C_c{y}^c$, independently of $T,y$.

For finite $T$, absolute convergence permits termwise integration, giving ${\sum }_n{a}_n{K}_T(x/n)$. The uniform bound dominates this by $C_c{x}^c{\sum }_n|a_n|n^{-c}<\infty$. The dominated convergence theorem and the scalar kernel formula now prove the boxed Perron formula, including its half-weight endpoint convention.

For the link with primes, the logarithmic derivative of the Euler product givesApplying Perron formula therefore expresses the second Chebyshev function ${\psi }_0(x)$, with half weight at a prime-power endpoint, as the vertical integral of $-{\zeta }^{\prime }/\zeta (s)x^s/s$. Shift this contour left, choosing heights away from zeros and using the partial-fraction expansion to control the integrand. The pole at one contributes $x$, a nontrivial zero $\rho$ of multiplicity $m$ contributes $-m{x}^{\rho }/\rho$, the trivial zeros contribute ${\sum }_{k\ge 1}{x}^{-2k}/(2k)$, and the pole from $1/s$ contributes $-{\zeta }^{\prime }(0)/\zeta (0)=-\log (2\pi )$. Thus the resulting Riemann–von Mangoldt explicit formula iswith zeros summed symmetrically in height. A truncated contour gives the corresponding truncated formula and remainder. The pole at one gives the main term for primes; zeros farther left give smaller oscillating corrections. Zero-free regions therefore translate into error bounds for $\psi (x)$ and, by partial summation, for the prime-counting function.

With the normalization of the preceding completion, the super-completion is the Riemann xi functionThe pole-subtracted theta integral for the completed zeta function gives the particularly useful expressionIt proves directly that $\xi$ is an entire function, with $\xi (0)=\xi (1)=1/2$ and $\xi (s)=\xi (1-s)$. Here “integral function” means entire, not integer-valued.

To prove the order-one growth of the Riemann xi function, put $M(R)={\max }_{|s|\le R}|\xi (s)|$. For $R\ge 3$, the exponentially decreasing theta integrand bounds both $J(s)$ and $J(1-s)$ byMultiplication by the quadratic factor in $s$ and the Stirling formula imply $\log M(R)=O(R\log R)$. Hence the order of an entire function is at most one. On the positive real axis, $\zeta (R)=1+O(2^{-R})$, and the defining formula and Stirling formula giveThis lower bound proves that the order is at least one. Consequently $\xi$ is entire of order exactly one, generally not of finite exponential type.

Its zeros are precisely the Nontrivial zeros of the Riemann zeta function, counted with multiplicity. The values at zero and one are nonzero; the Gamma function has no zeros; and the completed functional equation together with the Euler product shows that there are no further completed zeros outside the closed critical strip. By Hadamard factorization for entire functions of order one,The genus-one factors give ${\sum }_{\rho }|\rho {|}^{-2}<\infty$, and the logarithmic derivative converges locally normally away from the zeros:Differentiate the defining product of $s(s-1)$, the gamma factor, and $\zeta (s)$. Rearranging proves the global partial-fraction expansion of the zeta logarithmic derivative:The summands must be kept together: each is $O_s(|\rho {|}^{-2})$ for large $|\rho |$, whereas separating the two complex reciprocal sums need not preserve absolute convergence. The gamma term accounts for the trivial zeros as well as cancellation at zero.

Put $F(s)=-{\zeta }^{\prime }(s)/\zeta (s)$. The Euler product gives $F(s)={\sum }_{n\ge 1}\Lambda (n)n^{-s}$ for $\Re s>1$. The elementary nonnegative polynomialtherefore gives the three-four-one zero-free-region argument:The coefficients $\Lambda (n)$ are nonnegative; applying the identity with $u=t\log n$ justifies the inequality term by term.

Take real parts in the global partial-fraction expansion of the zeta logarithmic derivative. For $1<\sigma \le 2$, the digamma function bounds yieldAll terms in the zero sum are nonnegative, because the Euler product and completed functional equation locate the nontrivial zeros in the closed critical strip. The constant terms are bounded: in particular ${\sum }_{\rho }\Re (1/\rho )={\sum }_{\rho }\beta /|\rho {|}^2$ converges. Near the real pole, $F(\sigma )=1/(\sigma -1)+O(1)$.

First exclude zeros on the line one. If $1+it$, $t\ne 0$, were a zero of multiplicity $m\ge 1$, the selected zero term would giveInserted into the positivity inequality, this yields $0\le (3-4m)/(\sigma -1)+O_t(1)$, impossible as $\sigma \downarrow 1$. At $t=0$, zeta has a pole, with a punctured neighbourhood containing no zeros. Thus the line one has no zeros.

Now let $\rho =\beta +i\gamma$ with $|\gamma |\ge 1$ and put $L=\log (|\gamma |+2)$. Select this zero in the formula at $t=\gamma$ and discard every other nonnegative zero kernel. The pole terms at heights $\gamma ,2\gamma$ are bounded. ConsequentlyChoose a fixed positive $a$ small enough that $C_{1}a\le 1/2$ and $1+a/L\le 2$, and set $\sigma =1+a/L$, $h=1-\beta$. Rearranging givesThis is the required logarithmic gap at large height.

For $|\gamma |\le 1$, zeros with $\beta \le 1/2$ already have a fixed gap. The remaining compact region contains finitely many zeros, none on the line one; the pole at one cannot be an accumulation point of zeros. Hence those zeros also have a fixed positive distance from that line. Reduce the absolute constant to cover this bounded-height region and to make the preceding weak inequality strict. The zero-free region of the Riemann zeta function is thereforeThis proof has established boundary nonvanishing as well as the quantitative gap, rather than assuming the former implicitly.

The global partial-fraction expansion of the zeta logarithmic derivative, with zeros counted with multiplicity and paired summands retained, isFor a nonprincipal primitive Dirichlet character $\chi$ of Dirichlet conductor $q$, put $a=0$ or $1$ according as $\chi (-1)=1$ or $-1$. Its completed Dirichlet L-function isThe global partial-fraction expansion of a Dirichlet L-function logarithmic derivative iswhere $B_{\chi }={\mathcal{L}}_{\chi }^{\prime }(0)/{\mathcal{L}}_{\chi }(0)$ and the zeros are the nontrivial zeros of the completion. These expansions follow from genus-one Hadamard factorization; the combined terms converge locally normally. The primitive principal Dirichlet character has Dirichlet conductor one and gives the zeta formula. If a Dirichlet character modulo $Q$ is induced by a primitive Dirichlet character ${\chi }^{\ast }$ of Dirichlet conductor $q$, its missing Euler factors addto the primitive logarithmic derivative. This correction is $O(\log Q)$ on the real interval $1<\sigma \le 2$.

The form needed here is Landau theorem for two real Dirichlet characters: there is an absolute $b>0$ such that, if ${\chi }_1,{\chi }_2$ are real primitive nonprincipal Dirichlet characters of Dirichlet conductors $q_1,q_2$ and their product is nonprincipal, thencounting multiplicity. In particular, two distinct primitive real Dirichlet characters cannot each have a zero in that interval.

For the proof, induce the Dirichlet characters to a common modulus and considerThe last factor can be imprimitive but is nonprincipal. For $\sigma >1$, logarithmic differentiation of the Euler products givesThis positivity is the reason to include all four factors: each real Dirichlet character takes values in $\{0,1,-1\}$, including at ramified primes.

The real partial-fraction identity for a primitive completion has no residual real constant. Indeed, after taking real parts in its Hadamard derivative, the constant is determined to be zero by the functional-equation symmetry of its modulus across the critical line; the zero kernels there cancel in reflected pairs. ConsequentlyFor $1<\sigma \le 2$ all the zero kernels are nonnegative, and the gamma term is bounded. The missing Euler factors have the previously bounded correction. Combining the four logarithmic derivatives, and retaining any two candidate real zeros ${\beta }_1,{\beta }_2$ of $L({\chi }_1)L({\chi }_2)$, therefore givesZeros in any other factors only decrease this upper bound. The sole pole term comes from zeta.

Put $H=\log (q_1{q}_2)$ and choose $a>0$ so small that $Ca<1/10$. If both $1-{\beta }_j<a/(4H)$, take $\sigma =1+a/H$. The upper bound is then at mosta contradiction. Taking $b=a/4$ proves the theorem, including the multiplicity assertion.

A Siegel zero is a possible exceptional real zero of the $L$-function of a real nonprincipal primitive Dirichlet character, in the very narrow zero-free-region exception near one. Fix a sufficiently small absolute threshold $\eta >0$ and call a zero exceptional here whenTake $\eta <b/4$ and also smaller than the constant in the classical zero-free region for Dirichlet L-functions. Such a zero, if present, is real and simple; its existence is not asserted. This choice fixes the constants consistently for the requested separation conclusions.

For two different real primitive Dirichlet characters of the same modulus $q$, their product cannot be principal: otherwise they agree on all units and hence are the same Dirichlet character. Their two exceptional zeros would both lie above $1-b/\log (q^2)$, contradicting Landau's theorem. Thus at most one real primitive Dirichlet character modulo $q$ has an exceptional zero.

For distinct primitive Dirichlet conductors $q_1<q_2$, the product of their Dirichlet characters is again nonprincipal; otherwise their common induced Dirichlet character would have two different primitive Dirichlet conductors. Suppose $q_2\le q_1^2$. ThenAgain both zeros violate the two-zero prohibition. Therefore the sparse conductors of exceptional real Dirichlet zeros obeyThe quadratic spacing uses a fixed sufficiently small exceptional-zero constant; it does not claim this conclusion for an arbitrarily enlarged neighbourhood of one.

Let $T(N)$ count pairs $n,n+2$ both prime with $n+2\le N$. The Selberg sieve replaces primality by the necessary condition that the polynomial $F(n)=n(n+2)$ have no small prime divisor. Its strength is a nonnegative quadratic majorant with optimizable divisor weights; its limitation is that a sifted integer need not be prime.

For a prime $p$, let $\nu (p)$ count the forbidden residues of $n$ modulo $p$. We have $\nu (2)=1$, since the two roots coincide modulo two, and $\nu (p)=2$ for every odd prime, where the roots are zero and $-2$. The Chinese remainder theorem makes $\nu (d)$ multiplicative on squarefree integers. With $g(d)=\nu (d)/d$,These local densities constitute the sieve distribution for the twin-prime problem. Since ${\sum }_{p\le z}g(p)=2\log \log z+O(1)$, it is a dimension-two sieve, which is why the expected upper bound has two logarithms in the denominator.

Let $P(R)={\prod }_{p\le R}p$ and choose real Selberg sieve weights ${\lambda }_d$ supported on squarefree $d\le R$, with ${\lambda }_1=1$. If $\gcd (F(n),P(R))=1$, the only supported divisor of $F(n)$ is one. HenceExpanding and using the distribution of the polynomial roots givesHere $[d,e]$ denotes the least common multiple. The divisor-square support reaches $R^2$, although each individual weight is supported only to $R$; keeping these two levels distinct is essential for estimating the remainder.

The Selberg sieve diagonalization solves the main-term minimization exactly. On squarefree integers putMultiplicativity givesAlso, Möbius inversion gives ${\sum }_{r\le R}\mu (r)Y_r={\lambda }_1=1$. Therefore the Cauchy-Schwarz inequality proveswith equality at $Y_r=\mu (r)w(r)/G(R)$. Inverting the divisor sums gives the optimizing weightsIn particular ${\lambda }_1=1$ and $Q(\lambda )=1/G(R)$.

These weights have $|{\lambda }_d|\le 1$, which supplies a simple rigorous remainder estimate. Indeed, multiply the inner sum by ${\prod }_{p\mid d}(1-g(p){)}^{-1}={\sum }_{e\mid d}w(e)$. The terms are $w(em)$ for distinct squarefree integers $em\le R$ with $e\mid d$ and $(m,d)=1$, hence form a subset of the sum defining $G(R)$. Since $\nu ([d,e])\le \nu (d)\nu (e)$ and $\nu (d)\le \tau (d)$,Thus

For this sieve $w(2)=1$ and $w(p)=2/(p-2)$ for odd primes. Here is a direct proof that $G(R)\asymp {\log }^{2}R$, sufficient for the upper bound. PutThe Mertens theorem for reciprocal primes implies $W(z)\asymp {\log }^{2}z$, by taking logarithms and using the convergence of ${\sum }_{p}g(p{)}^2$. Give each divisor $d\mid P(z)$ probability $w(d)/W(z)$. Its expected logarithm isby the Mertens first theorem. With $z=R^{1/8}$ this is at most $\frac{1}{3}\log R$ for large $R$. The Markov inequality shows that at least two thirds of the total product weight has $d\le R$. These terms occur in $G(R)$, so $G(R)\gg W(R^{1/8})\gg {\log }^{2}R$. Conversely $G(R)\le W(R)\ll {\log }^{2}R$, proving the claim without treating an untruncated Euler product as a truncated divisor sum.

Choose, for example, $R=N^{1/4}$. Any twin-prime pair with smaller member greater than $R$ survives this sieve; pairs with smaller member at most $R$ contribute at most $O(R)$. ConsequentlyThe chosen level is deliberately enough to prove the correct logarithmic order; sharper level and remainder analyses improve the numerical upper-bound constant.

The local correction factor also explains the conjectured main-term constant. Relative to two independent primality conditions, the singular series isThe twin-prime constant $C_2$ is positive and convergent because its logarithm has an absolutely convergent tail. The factor two at $p=2$ records that only one parity is allowed, while each odd-prime factor records the two distinct forbidden residues. The Hardy–Littlewood twin-prime asymptotic conjecture predictsThis is a conjecture, not a consequence of the upper-bound calculation.

The parity problem in sieve theory is the obstruction to reversing the argument. The nonnegative square majorizes all survivors, including composites with sufficiently large prime factors; small-divisor information does not isolate the desired prime-factor parity or force both factors to be prime. Therefore the Selberg sieve establishes the correct upper-bound scale and the arithmetic local factors, but by itself does not prove a positive lower bound for twin primes, their infinitude, or the conjectured asymptotic.

For $\sigma >1$, the absolutely convergent logarithm of the Euler product for the Riemann zeta function isThe elementary identity $3+4\cos v+\cos 2v=2(1+\cos v{)}^2\ge 0$ therefore givesThus the expression inside the logarithm is at least one. The Meromorphic continuation of the Riemann zeta function to the right half-plane has only a simple pole at one, of residue one. It follows, for example, fromwhose integral is holomorphic there.

Suppose $t\ne 0$ and $\zeta$ has a zero of order $m\ge 1$ at $1+it$. As $\sigma \downarrow 1$, the real factor is $O((\sigma -1{)}^{-3})$, the middle factor is $O((\sigma -1{)}^{4m})$, and the factor at $1+2it$ is bounded. Their product would tend to zero, contradicting its lower bound one. HenceAt $t=0$ there is a pole, not a zero; the statement about the line does not assert a finite value for $\zeta (1)$. This proves the three-four-one product proof of zeta boundary nonvanishing.

To relate the Second Chebyshev function to the logarithmic derivative, differentiate the Euler product in its half-plane of absolute convergence:Since $\psi (u)={\sum }_{n\le u}\Lambda (n)$, its first integral of the Chebyshev function isFor any fixed $c>1$, its Mellin inversion formula isFor justification, the elementary contour kernel is $\frac{1}{2\pi i}{\int }_{(c)}{y}^s/(s(s+1))ds=1-y^{-1}$ for $y>1$ and zero for $0<y\le 1$. Close left or right and take the residues at zero and minus one; at $y=1$ the continuous value is zero. Absolute convergence on $\mathrm{Re}s=c$ allows termwise integration of the Dirichlet series, and multiplying this kernel by $x$ gives $(x-n{)}_{+}$.

Here is how the relation yields the asymptotic. Move the contour to the left, using the functional equation of the Riemann zeta function to control the left-hand side. The residue at $s=1$ is $x^2/2$, and a Nontrivial zero of the Riemann zeta function $\rho$ contributes $-x^{\rho +1}/(\rho (\rho +1))$, multiplied by its multiplicity. The kernel poles at zero and minus one contribute $-x\log (2\pi )$ and ${\zeta }^{\prime }(-1)/\zeta (-1)$; a trivial zero of the Riemann zeta function $-2k$ contributes $-x^{1-2k}/(2k(2k-1))$. Thus the smoothed explicit formula for the Chebyshev function isFor the contour argument, use heights avoiding the zero ordinates and then let those heights increase. The local zero bound from the Riemann–von Mangoldt formula, together with the Local partial-fraction expansion of the Riemann zeta logarithmic derivative, permits heights with logarithmic-derivative bound $O({\log }^{2}T)$ in each fixed vertical strip. The horizontal integrals then vanish because the denominator is of size $T^2$. After that, send the left edge through negative odd integers to minus infinity; the functional equation of the Riemann zeta function bounds the logarithmic derivative there by a logarithm, and $x^{s+1}$ makes the left integral vanish for fixed $x>1$. This explains why this smoothing allows a convergent contour calculation without a quantitative zero-free region.

In fact the Riemann–von Mangoldt formula, stated in the next solution, gives $N(T)=O(T\log T)$ and henceThe Nontrivial zeros of the Riemann zeta function satisfy $0<\mathrm{Re}\rho <1$, using the Euler product, the functional equation of the Riemann zeta function, and the nonvanishing just proved. For each such zero, $x^{\rho -1}\to 0$ as $x\to \infty$, whereas $|x^{\rho -1}|\le 1$ for $x\ge 1$. The dominated convergence theorem therefore makes the absolutely convergent zero sum, divided by $x^2$, tend to zero. All the other displayed terms are $o(x^2)$. We obtain

It remains to unsmooth; differentiating an asymptotic without justification would not suffice. Monotonicity of $\psi$ gives, for fixed $0<h<1$,Divide by $x$ and use the integral asymptotic. The lower and upper limits lie between $1-h/2$ and $1+h/2$. Letting $h\downarrow 0$ proves $\psi (x)\sim x$. Higher prime powers contribute $O(\sqrt{x}\log x)$, as in the first solution, so $\vartheta (x)\sim x$. Finally partial summation givesThe integral is $O(x/(\log x{)}^2)$, by $\vartheta (t)=O(t)$ and splitting at $\sqrt{x}$. Hencewhich is the Prime number theorem.

Fix integers $q\ge 1$ and $a$ with $(a,q)=1$. We prove the Dirichlet theorem on primes in arithmetic progressions by showing that the prime Dirichlet sum in this residue class diverges as its real exponent decreases to one.

First justify the required Dirichlet L-function facts. If $\chi$ is a nonprincipal Dirichlet character modulo $q$, its sum over a full period is zero: choose a unit $b$ with $\chi (b)\ne 1$, and multiplication by $b$ permutes the residue classes, forcing the sum to equal $\chi (b)$ times itself. Therefore the partial sums $C_{\chi }(u)={\sum }_{n\le u}\chi (n)$ are bounded by $q$. Partial summation giveswhich supplies holomorphic continuation near one, with locally uniform convergence. For the principal Dirichlet character ${\chi }_0$,has a simple pole at one, with residue $\phi (q)/q$.

The assumption in the question gives nonvanishing at one for real nonprincipal characters. We must also prove it for nonreal characters. For a nonreal $\chi$, its square ${\chi }^2$ is nonprincipal. For real $\sigma >1$, logarithms of the Euler products giveFor a prime not dividing $q$, its logarithmic coefficient is $3+4\mathrm{Re}\chi (p{)}^k+\mathrm{Re}\chi (p{)}^{2k}=2(1+\cos \theta {)}^2\ge 0$, where $\chi (p{)}^k=e^{i\theta }$. Primes dividing $q$ contribute only the positive zeta term. If $L(1,\chi )$ vanished to order $m\ge 1$, the product would be $O((\sigma -1{)}^{4m-3})$ and tend to zero: the squared-character factor is bounded because ${\chi }^2$ is nonprincipal. This is a contradiction. HenceThis step uses the given real-character hypothesis and the Euler product positivity for L-function nonvanishing for all remaining characters.