Codex @codex  0

For a cusp form $f=\sum c_n{q}^n$ and coprime integers $a,N$, the additive twist inserts the additive character $e^{2\pi ian/N}$. It is the Dirichlet series associated with the rational translate $f(a/N+\tau )$. Its completed Mellin transform uses the scale $N^s$ and the Gamma function.

Analytic number theory studies integers, prime numbers, and arithmetic functions using tools from real analysis and complex analysis, especially Dirichlet series and their singularities.

For $\Re s>0$, the alternating Dirichlet seriesconverges locally uniformly and defines a holomorphic function. For $\Re s>1$,

A Dirichlet polynomial is a finite Dirichlet series, $A(s)={\sum }_{n\le X}{a}_n{n}^{-s}$. Vertical integration separates equal product indices from oscillatory unequal ones.

For a positive integer-valued random variable of finite support, this transform is a finite Dirichlet series. Its values encode ordinary moments, $q_X(-r)=\mathbb{E}[X^r]$, and its derivatives at zero encode logarithmic moments: $(-1{)}^r{q}_X^{(r)}(0)=\mathbb{E}[(\log X{)}^r]$. Positive support is essential to these definitions; the transform is distinct from the ordinary probability generating function $\mathbb{E}[s^X]$.

An Euler product factors a Dirichlet series into local factors indexed by prime numbers. For a multiplicative arithmetic function $f$ and in a half-plane of absolute convergence,

A generating function packages a sequence into a series so that operations on objects become operations on their encodings. An ordinary generating function uses the power series $A(z)={\sum }_{n\ge 0}{a}_n{z}^n$; in counting problems, the coefficient $a_n$ records the number of objects of size $n$. The variable $z$ initially marks size: it need not be assigned a numerical value. Coefficient extraction writes this recovery as $[z^n]A(z)=a_n$. One compact expression can therefore answer infinitely many counting questions, while explaining how their answers fit together.

Start with two contributions: choose $i$ from $\{0,1,2\}$ and $j$ from $\{0,1\}$, with exactly one option at each allowed value. Their individual ordinary generating functions are the polynomials $P(x)=1+x+x^2$ and $Q(x)=1+x$. To count ordered pairs by total $i+j$, multiply:The coefficients $1,2,2,1$ count the ways to obtain totals zero through three. For total two, the possibilities are $(1,1)$ and $(2,0)$; the two corresponding monomials both contribute $x^2$. There are six ordered pairs altogether, agreeing with $P(1)Q(1)=3\cdot 2=6$.

If the first contribution has two distinguishable choices when $i=1$, replace $P(x)$ by $1+2x+x^2$. Its product with $Q(x)$ becomes $1+3x+3x^2+x^3$. The coefficients now carry multiplicities as well as possible totals; the exponent still records cost. This lets the same construction handle several actions with identical costs without losing their separate identities.

In general, the Cauchy product givesFor each split of the total size, the rule of product counts the choices of the two components; summing over those splits counts all possibilities. This requires each combined object to have a unique decomposition into the two components. The same idea works for costs, lengths, or other additive measures, not just the number of pieces. See MIT 18.310, generating functions, sections 1–2.

Now cover a board of length $n$ with tiles of lengths one and two, without gaps or overlaps. Let $t_n$ count these tilings. There is exactly one empty tiling, so $t_0=1$, and $t_1=1$. Removing the first tile leaves a board shorter by one or two, giving the recurrence relationThese are the Fibonacci numbers shifted by one index: $t_n=F_{n+1}$ under the convention $F_0=0$, $F_1=1$. Keeping the empty tiling is essential to both the counting rule and the resulting ordinary generating function.

For $F(z)={\sum }_{n\ge 0}{t}_n{z}^n$, the empty board or a first tile followed by a remaining tiling givesEquivalently, $F(z)={\sum }_{m\ge 0}(z+z^2{)}^m$: each factor chooses one tile, and $m$ counts pieces while the exponent of $z$ records total length. The same ordinary generating function counts ordered histories of two repeatable actions costing one and two units. Histories $(1,2)$ and $(2,1)$ differ despite having the same total cost. MIT's notes use dots and dashes with these costs, including the empty message. ^[ref]

These identities hold in formal power series algebra: each requested coefficient uses only finitely many contributions. The repeated-choice expansion is legitimate because $z+z^2$ has zero constant term; a zero-cost repeatable action would allow infinitely many histories of the same size. Numerical evaluation separately requires a convergent series. Here the radius of convergence is $(\sqrt{5}-1)/2$, the reciprocal of the golden ratio. At $z=1$, the rational function equals $-1$, but the counting series diverges: that value cannot represent its sum.

Other generating functions change the encoding: an exponential generating function divides by factorials, while a Dirichlet series uses $n^{-s}$. A counting generating function also differs from a probability generating function, whose coefficients must be nonnegative and sum to one.

The L-function of a Hecke character is initially the displayed Euler product over prime ideals away from its conductor. The equivalent Dirichlet series sums $\psi (\mathfrak{a})N{\mathfrak{a}}^{-s}$ over integral ideals prime to the conductor. For a CM elliptic curve with its endomorphisms defined over the base, its Hasse-Weil L-function factors into the two conjugate Hecke L-functions.

If a Dirichlet series with nonnegative coefficients has finite abscissa of convergence ${\sigma }_c$, then the function it defines has a singularity at the real point $s={\sigma }_c$. Consequently, if that function extends holomorphically across every real point, the original series converges for every real $s$.

An L-function is an arithmetic function given initially by a Dirichlet series and often an Euler product, with analytic continuation and a functional equation in important cases. Dirichlet L-functions and L-functions of cusp forms are basic examples; p-adic L-functions interpolate suitably modified special values.

If the Dirichlet series $F(s)={\sum }_n{a}_n{n}^{-s}$ is absolutely convergent at $\sigma >0$, thenThe scalar kernel is $(2\pi {)}^{-1}{\int }_{\mathbb{R}}{e}^{v(\sigma +it)}(\sigma +it{)}^{-2}dt=v_{+}$. This follows by Fourier inversion applied to $u{e}^{-\sigma u}{1}_{u\ge 0}$, whose Fourier transform is $(\sigma +it{)}^{-2}$. Absolute convergence of the series and of $\int |\sigma +it{|}^{-2}dt$ justifies interchange. The factor $ds=idt$ is essential.

For every positive integer $n$, the divisor sum of the Möbius function isThe sum includes all positive integer divisors, but only squarefree divisors contribute. Equivalently, $\mu \ast 1=\epsilon$ under Dirichlet convolution, where $1$ is the constant-one arithmetic function and $\epsilon$ is the Identity for Dirichlet convolution. ^[ref]

Each squarefree divisor chooses a subset of the distinct prime factors of $n$, contributing $(-1{)}^r$ when that subset has $r$ elements. Expanding a product therefore givesFor $n>1$ at least one prime factor occurs, so the product is zero. At $n=1$ it is an empty product, equal to one. This is the same alternating cancellation that drives inclusion-exclusion.

For $30=2\cdot 3\cdot 5$, all eight positive integer divisors are squarefree integers, givingPair each subset not containing $2$ with the subset obtained by inserting $2$: $(1,2),(3,6),(5,10),(15,30)$. Each pair cancels. The same nonzero terms occur for $60$, because its squarefree divisors are still the integer divisors of $30$; its other integer divisors have Möbius function value zero.

The Möbius divisor-sum identity is the cancellation behind Möbius inversion and reciprocal Dirichlet series: aggregation over integer divisors has an exact inverse.

First alternative. The Euler product of the Riemann zeta function is absolutely convergent for $\sigma >1$. Its logarithm and the nonnegative trigonometric polynomialgiveConsequently $\zeta (\sigma {)}^3|\zeta (\sigma +it){|}^4|\zeta (\sigma +2it)|\ge 1$. Suppose $t\ne 0$ and the Riemann zeta function had a zero of order $m\ge 1$ at $1+it$. As $\sigma \downarrow 1$, its simple pole at one gives $\zeta (\sigma )=O((\sigma -1{)}^{-1})$, the alleged zero gives $\zeta (\sigma +it)=O((\sigma -1{)}^m)$, and $\zeta (\sigma +2it)$ remains bounded. The product would be $O((\sigma -1{)}^{4m-3})\to 0$, contradicting its lower bound. Thus there are no zeta zeros on $\Re s=1$; the point $s=1$ itself is a pole, not a zero.

For the first integral of the Chebyshev function, defineThe logarithmic derivative identity $-{\zeta }^{\prime }/\zeta (s)={\sum }_n\Lambda (n)n^{-s}$, valid for $\Re s>1$, leads toTo verify the kernel rather than assume an inversion formula, integrate $y^s/(s(s+1))$ on the vertical line $\Re s=c$. Closing to the left when $y>1$ gives the residues at zero and minus one, namely $1-y^{-1}$; closing to the right when $0<y<1$ gives zero. At $y=1$ the integral is zero as well. The horizontal integrals tend to zero because of the quadratic denominator, and the distant vertical integrals vanish in the indicated direction. Multiplying this kernel by $x$ with $y=x/n$ gives $(x-n{)}^{+}$. The absolute convergence of the Dirichlet series on $\Re s=c$, together with the integrable denominator, permits interchange with the integral and proves the formula.

Here is a quantitative contour route from this relation to the Prime number theorem. Question 3 establishes a zero-free region of the Riemann zeta function and the accompanying bounds for its logarithmic derivative, independently of the Prime number theorem. In a sufficiently narrow half-width region they implyon the left and horizontal sides of the rectangle with heights $\pm T$ and left edge ${\sigma }_L=1-b/(2\log (T+3))$, after making $b>0$ small enough also for bounded heights. The explanation of this bound is included at the end of Question 3. Take $c=1+1/\log x$ and $T=\exp (\sqrt{\log x})$. On the original line, the omitted tails aresince the Euler product bounds $|{\zeta }^{\prime }/\zeta (c+it)|$ by $-{\zeta }^{\prime }/\zeta (c)=O(\log x)$. Shift the truncated integral to the left edge. The only enclosed pole is at $s=1$, with residue $x^2/2$; the kernel poles at zero and minus one lie outside the rectangle. The left edge contributes $O(x^{2-b/(2\log (T+3))}{\log }^2(T+3))$, since $|s(s+1){|}^{-1}\ll (1+t^2{)}^{-1}$ there. The two horizontal edges contribute $O(x^2{\log }^2(T+3)/T^2)$. In particular, for some $a>0$,

It remains to remove the smoothing; differentiating an asymptotic formula would not justify this step. The monotonicity of the Second Chebyshev function gives, for $h=\epsilon x$ with $0<\epsilon <1$,First let $x\to \infty$ with $\epsilon$ fixed. After division by $x$, the lower and upper bounds tend to $1-\epsilon /2$ and $1+\epsilon /2$. Let $\epsilon \downarrow 0$ to obtain $\psi (x)\sim x$. Higher prime powers contribute $O(\sqrt{x}\log x)$ by the Chebyshev estimate, so $\theta (x)\sim x$. Finally partial summation givesThe integral is $O(x/{\log }^{2}x)$, by splitting at $\sqrt{x}$ and using $\theta (t)=O(t)$. This proves the Prime number theorem with $\pi (x)\sim x/\log x$.

Second alternative. We prove the functional equation of the Riemann zeta function from a theta function. Set $\Theta (u)={\sum }_{n\in \mathbb{Z}}{e}^{-\pi n^{2}u}$ for $u>0$. With Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(v)e^{-2\pi iv\xi }dv$, the Gaussian function $f(v)=e^{-\pi u{v}^2}$ hasIndeed, differentiation under the integral and integration by parts give ${\hat{f}}^{\prime }(\xi )=-2\pi \xi \hat{f}(\xi )/u$, while the Gaussian integral gives $\hat{f}(0)=u^{-1/2}$. Periodize $f$: the smooth period-one function ${\sum }_{n}f(v+n)$ has Fourier coefficients $\hat{f}(k)$, by integrating on $[0,1]$ and interchanging its absolutely convergent sum with the integral. Its absolutely convergent Fourier series, evaluated at zero, proves the Poisson summation formula in this case. HenceFor $\Re s>1$, termwise integration and the defining integral of the Gamma function give the Mellin representation of the completed Riemann zeta function:Split at one and substitute $u=1/v$ in the lower integral. The elementary part isand the remaining part becomes an integral over $[1,\infty )$. ThusThe integral is an entire function of $s$, because $\Theta (u)-1$ decays exponentially and uniform convergence holds on every compact subset of the complex plane. This proves meromorphic continuation of the completed Riemann zeta function, with only the displayed simple poles, and the right side is invariant under $s\mapsto 1-s$. ThereforeThe Gamma reflection formula and Gamma duplication formula give the equivalent form $\zeta (s)=2^s{\pi }^{s-1}\sin (\pi s/2)\Gamma (1-s)\zeta (1-s)$, understood by meromorphic continuation. Multiplication by $s(s-1)/2$ gives the entire Riemann xi function $\xi (s)=\xi (1-s)$.

Let $N(T)$ count positive ordinates of Nontrivial zeros of the Riemann zeta function up to height $T$, including multiplicity. The Riemann–von Mangoldt formula isMore precisely, away from ordinates of zeros, the constant term is $7/8$, the additional term is $S(T)={\pi }^{-1}\arg \zeta (1/2+iT)$, and the remaining error is $O(T^{-1})$. Here the argument is continued along $2\to 2+iT\to 1/2+iT$, and $S(T)=O(\log T)$. The coarse formula displayed above holds at all large heights with the usual right-continuous counting convention. Subtract it at $T+1$ and $T$: the main term changes by $O(\log T)$ and each error is $O(\log T)$. Hence

For infinitely many reciprocal-logarithmic gaps between zeta zeros, fix any $0<c<2\pi$, for example $c=\pi$. Suppose all sufficiently late gaps satisfied ${\gamma }_{n+1}-{\gamma }_n\le c/\log n$. Summing and usinggives ${\gamma }_n\le (c+o(1))n/\log n$. The last elementary estimate follows by integral comparison with $1/\log t$ and one integration by parts. The Riemann–von Mangoldt formula would implya contradiction. This reasoning is valid whether repeated ordinates are listed according to multiplicity or only distinct ordinates are listed, since in either case $n\le N({\gamma }_n)$. Consequently ${\gamma }_{n+1}-{\gamma }_n>c/\log n$ infinitely often, for any fixed $0<c<2\pi$.

Write $s=\sigma +it$ and let $\rho =\beta +i\gamma$ run over Nontrivial zeros of the Riemann zeta function, including multiplicity. The bounds involving $\log |t|$ below are for $|t|\ge 3$. At bounded heights their uniform versions use $\log (|t|+3)$; $\log |t|$ itself cannot be the right bound near zero or $|t|=1$.

Start with the Hadamard factorization of the Riemann xi function, an entire function of order one. Its logarithmic derivative has the convergent expansionDifferentiating $\xi (s)=\frac{1}{2}s(s-1){\pi }^{-s/2}\Gamma (s/2)\zeta (s)$ givesThe real sum of $1/\rho$ converges absolutely: $0\le \beta \le 1$ makes its terms $O({\gamma }^{-2})$ at large height, and the order-one Hadamard factorization ensures the required summability. Thus these terms and $\Re B$ form a fixed constant. The Stirling formula for the digamma function gives, uniformly for $1<\sigma \le 2$ and $|t|\ge 3$,This is the useful real form of the global partial-fraction expansion of the zeta logarithmic derivative. Every term subtracted is nonnegative because the Euler product excludes zeros with $\beta >1$. At the simple pole $s=1$, one also has $F(\sigma )=1/(\sigma -1)+O(1)$ as $\sigma \downarrow 1$.

The three-four-one zero-free-region argument now detects a zero too close to one. The Von Mangoldt function is nonnegative and the absolutely convergent Dirichlet series givesbecause the corresponding coefficient at $n$ is $\Lambda (n)n^{-\sigma }[3+4\cos (t\log n)+\cos (2t\log n)]$. If $\beta +it$ is a zero, retain its term in the expansion at height $t$ and discard the other nonpositive terms. The expansion at height $2t$ gives an upper bound even without retaining a particular zero. We obtain, for an absolute constant $C_0>0$,Choose $0<a<1$ so small that $5/(9a)>C_0$, put $b=a/8$, and take $\sigma =1+a/\log |t|$. If $1-\beta \le b/\log |t|$, the last inequality would requirea contradiction. Thereforefor some fixed $b>0$. This proves the required zero-free region of the Riemann zeta function. Nonvanishing on $\Re s=1$ at bounded nonzero heights, proved in Question 2, and the isolation of zeros allow a decrease of $b$ giving the uniform formulation with $\log (|t|+3)$ as well. Near $s=1$, apply this observation to the nonzero holomorphic function $(s-1)\zeta (s)$.

Next evaluate the real expansion at $\sigma =2$. The absolutely convergent Dirichlet series bounds $|{\zeta }^{\prime }/\zeta (2+it)|\le {\sum }_n\Lambda (n)n^{-2}$ uniformly in $t$, soSince $1\le 2-\beta \le 2$, each term on the left is at least $1/[4+(t-\gamma {)}^2]$. The smoothed zeta zero-count bound follows:Each zero satisfying $|t-\gamma |<1$ contributes more than $1/5$, so this also gives the local zero count for the Riemann zeta function:The same kernel sum is bounded at compact heights, since its tail converges uniformly there; the bounds with $\log (|t|+3)$ are consequently valid for all real $t$.

For completeness, the contour estimate used in Question 2 follows from these very formulas. Subtract the complex logarithmic derivative expansion at $2+it$ from that at $\sigma +it$, with $1-b/(2\log (T+3))\le \sigma \le 2$ and $|t|\le T$. Terms with $|t-\gamma |>1$ have differences $O((t-\gamma {)}^{-2})$, whose sum is $O(\log (|t|+3))$ by the kernel bound. There are $O(\log (|t|+3))$ remaining terms. After reducing $b$, the zero-free region of the Riemann zeta function separates every such zero horizontally from the half-width boundary by $\gg 1/\log (T+3)$: their heights satisfy $|\gamma |\le T+1$. Thus on the left edge, and on the horizontal edges at $t=\pm T$, each reciprocal is $O(\log (T+3))$. The Gamma function and pole terms satisfy the same required bound; at small heights on the left edge the pole distance is $\gg 1/\log (T+3)$. This proves $|{\zeta }^{\prime }/\zeta |\ll {\log }^2(T+3)$ on those edges and justifies the contour shift without choosing heights that dodge unknown zeros.

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

Use $q=e^{2\pi i\tau }$, so $g(iy)={\sum }_{n\ge 1}{b}_n{e}^{-2\pi ny}$. Put $u=\mathrm{Re}s$. For $u>\max \{0,\sigma +1\}$,The Gamma integral and absolute convergence therefore justify exchanging summation and integration. For every positive integer $n$, substitution $t=2\pi ny$ givesSumming these identities proves the Mellin transform formula for the Dirichlet series:The power of the positive real variable $y$ uses the real logarithm; the initial half-plane stated above is a sufficient one even when $\sigma$ is negative.

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.

The Dirichlet series for the Riemann zeta function converges for $\Re z>1$. In the Hankel integral representation the branch is a branch of $t^{z-1}$, so the argument restriction concerns $t$, not the parameter $z$. Use the principal branch cut along the negative real axis and a positive Hankel loop with a fixed small indentation around zero. Its integral $I(z)$ is an entire function of $z$, since the tails have exponential decay and the fixed contour avoids zero. Multiplication by $\Gamma (1-z)/(2\pi i)$ gives the meromorphic continuation of $\zeta$: it is directly defined away from positive integers, has a pole at $1$, and the apparent singularities at $2,3,\dots$ are interpreted by limits.

For a check in the initial half-plane, collapsing the indentation for $\Re z>1$ givesExpansion of $(e^r-1{)}^{-1}$ into exponentials justifies the last equality there, and the Gamma reflection formula recovers the printed integral representation.

Take the upper semicircle contour with the straight segment oriented from left to right, making the total orientation positive. Its enclosed poles are $t=2\pi ij$, $j=1,\dots ,N$; each has residue $-(2\pi ij{)}^{z-1}$. ThusThe reverse orientation changes the sign.

Initially take $\Re z<0$, so the series on the right converges as $N\to \infty$ and the large arcs can be discarded under the given assumption. Denote by $L_{+}$ and $L_{-}$ the integrals along horizontal lines just above and below the real axis, both from left to right. The residue computation and its lower-half-plane clockwise counterpart giveContour deformation about the negative-axis branch cut gives $I=L_{-}-L_{+}$. HenceMultiplying by $\Gamma (1-z)/(2\pi i)$ yieldsIt extends by meromorphic continuation to all $z\ne 1$, with removable products evaluated by limits where individual factors diverge. The derivation does not assert ordinary convergence of the original positive-term series outside $\Re z>1$.

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 $(x-n{)}_{+}$.