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Analytic number theory uses real analysis and complex analysis to study integers, primes, and arithmetic functions. It encodes discrete counting problems in sums, integrals, and Dirichlet series, then extracts information from convergence, oscillation, and singularities.

For example, the Euler product $\zeta (s)={\prod }_p(1-p^{-s}{)}^{-1}$ links the Riemann zeta function with all primes. Its logarithmic derivative has coefficients given by the Von Mangoldt function, turning prime counting into an analytic problem. The Prime number theorem is a central outcome: $\pi (x)\sim x/\log x$. The method is useful because analytic transformations can control an enormous discrete sum without enumerating every term.

For a finite Galois extension $L/K$ of number fields and a finite-dimensional complex group representation $\rho$ of its Galois group, the finite part of its Artin L-function is the Euler productHere $I_{\mathfrak{p}}$ is the inertia group, and a Frobenius automorphism lift acts on its invariant subspace. Different lifts induce the same operator there. At an unramified prime the invariant subspace is all of $V$; at a ramified prime this restriction removes the part on which inertia acts nontrivially.

The trivial representation gives the Dedekind zeta function of $K$. Direct sums of representations give products of Artin L-functions, because determinants multiply on direct sums. Taking the logarithm of each local factor produces traces of powers of Frobenius, connecting representation-theoretic characters with prime-counting sums. Archimedean factors are added separately when forming a completed L-function.

The logarithm from the Euler product of an Artin L-function has first-prime-power term $\sum \chi (g_{\mathfrak{p}})(N\mathfrak{p}{)}^{-s}$. Higher powers are uniformly bounded as real $s\downarrow 1$: $|\chi (g)|\le \chi (1)$ and ${\sum }_{\mathfrak{p}}{\sum }_{m\ge 2}(N\mathfrak{p}{)}^{-ms}/m\le 2{\sum }_{\mathfrak{p}}(N\mathfrak{p}{)}^{-2}<\infty$. The finitely many ramified local factors also have bounded logarithms. The Artin L-function order at one therefore proves the displayed estimate. Expanding a conjugacy-class indicator by character orthogonality leaves only the trivial character's logarithmic term, with coefficient $|C|/|G|$. This proves Dirichlet density directly; a natural-density assertion needs additional analytic arguments.

For a Dirichlet character $\chi$ modulo $q$, its Dirichlet L-function is initially defined on the half-plane $\mathrm{Re}s>1$ bySince $|\chi (n)|\le 1$, this series converges absolutely by comparison with ${\sum }_n{n}^{-\mathrm{Re}s}$.

A Dirichlet character is a completely multiplicative function. The fundamental theorem of arithmetic therefore reorganizes the absolutely convergent series into an Euler product:Expanding each factor as a geometric series assigns exactly one term to every positive integer. For the principal Dirichlet character ${\chi }_0$, the factors at primes dividing $q$ disappear, giving $L(s,{\chi }_0)=\zeta (s){\prod }_{p\mid q}(1-p^{-s})$, where $\zeta$ is the Riemann zeta function.

For normalized level-one Hecke eigenforms, let ${\alpha }_p+{\beta }_p=a_p$, ${\alpha }_p{\beta }_p=p^{k-1}$, and similarly ${\gamma }_p+{\delta }_p=b_p$, ${\gamma }_p{\delta }_p=p^{k-1}$. The Hecke multiplication relations give $a_{p^r}=({\alpha }_p^{r+1}-{\beta }_p^{r+1})/({\alpha }_p-{\beta }_p)$, with the repeated-root case obtained by continuity. Multiplying the two expressions and summing four geometric series proves the displayed local factor. Multiplicativity gives an Euler product; multiplying by $\zeta (2s-2k+2)$ cancels its local numerator and produces the degree-four convolution factors.

Absolute convergence of the Euler product gives $\zeta (s)\ne 0$ for $\Re s>1$. Equivalently,converges absolutely there.

For $\sigma >1$, logarithmic expansion of the Euler product reduces positivity to $3+4\cos \theta +\cos 2\theta =2(1+\cos \theta {)}^2$. At primes dividing the modulus the zeta contribution is positive by itself. If the last factor is bounded at the line-one point, a zero of the middle factor would outweigh the zeta pole and force this product to zero. This proves the relevant nonvanishing of Dirichlet L-functions on the line 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.

Put $F(s)=(s-1)\zeta (s)$. The fractional-part continuation formula for the Riemann zeta function gives $|F(s)|\ll (|s|+1{)}^2$ for $\mathrm{Re}s\ge 1/4$. At $2+ij$, the reciprocal Euler product bounds $|\zeta (2+ij)|$ below by $1/\zeta (2)$. Applying the Jensen zero-count bound with outer radius $7/4$ and inner radius $8/5$ gives $O(\log (|j|+2))$ zeros in the latter disk. Those disks cover the half-strip $1/2\le \mathrm{Re}s\le 1$. Reflecting zeros using the functional equation of the Riemann zeta function completes the $O(T\log T)$ bound without requiring left-half-plane growth estimates.

For $t\ge 2$ and $0<\eta \le 1/4$, suppose $|\zeta |\le M$, $M\ge 2$, on radius-$\eta$ discs centred at $1+\eta /4+it$ and $1+\eta /4+2it$. Every zero $\beta +it$ has the displayed gap. The reciprocal Euler product bounds both centre values below by a constant times $\eta$. The local logarithmic-derivative lemma gives errors $B/\eta$, $B=1+\log (M/\eta )$. The three-four-one zero-free-region argument then implies $4/(\sigma -\beta )-3/(\sigma -1)\ll B/\eta$. For a zero close to one, choose $\sigma =1+6(1-\beta )$, making the left side $1/(14(1-\beta ))$. Zeros farther away already satisfy the bound. Thus a logarithmic upper bound for zeta translates into a quantitative zero-free width.

Let $X\ge 3$, and let $f$ be a normalized multiplicative arithmetic function supported on squarefree integers, with $|f(p)|\le 1$. Form the finite measure $\nu ={\sum }_{n}f(n)n^{-1-1/\log X}{\delta }_{\log n}$ using Dirac measures. Under the transform convention $\hat{\nu }(t)=\int e^{-itu}d\nu (u)$, its Euler product is ${\prod }_p(1+f(p)p^{-1-1/\log X-it})$. Expanding its logarithmic modulus, and replacing the damped prime weights by $1/p$ up to $X$ with Mertens first theorem, gives the displayed absolute bound, uniformly in real $t$. Hence $|\hat{\nu }(t)|\ge \delta \log X$ forces $D(f,n^{it};X{)}^2\le \log (C/\delta )$. The distance expression still defines a nonnegative pretentious score for prime values bounded by one; its actual metric interpretation uses unit-modulus values.

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.

For a weight-$k$ cusp form $f(\tau )={\sum }_{n\ge 1}{a}_n{e}^{2\pi in\tau }$ of level $N$, its L-function is the Dirichlet seriesinitially in a right half-plane of absolute convergence. With the determinant-normalized slash operator and $W_N=\left(\begin{array}{cc}0& -1\\ N& 0\end{array}\right)$, put $Jf=i^{k}f{|}_k{W}_N$. For example, on the space of cusp forms for ${\Gamma }_1(N)$ this is again a cusp form, and the level-normalized completion isExponential decay at the modular cusps zero and infinity makes this Mellin transform entire. Modular inversion gives the phase-normalized Fricke functional equation ${\Lambda }_N(f,s)={\Lambda }_N(Jf,k-s)$. Only when $f$ is an eigenvector of $J$ does this become a scalar functional equation relating the completion of $f$ to itself. At level one the completion reduces to $(2\pi {)}^{-s}\Gamma (s)L(f,s)$. A normalized Hecke eigenform also has an appropriate Euler product; at general level its bad-prime factors and character must be specified.

Uniformly for $s=\sigma +it$ with $1/2\le \sigma \le 2$ and $|t|\ge 3$, away from zeros,The sum runs over Nontrivial zeros of the Riemann zeta function, with multiplicity. The height condition keeps this strip away from the pole at $s=1$ and the trivial zeros of the Riemann zeta function; these cannot simply be omitted in an arbitrary disk.

Start with the global partial-fraction expansion of the zeta logarithmic derivative and compare it at $s$ and $2+it$. The Euler product bounds the logarithmic derivative at $2+it$. The nearby zero terms remain explicit, while the paired differences from distant height bands are bounded using the local zero count for the Riemann zeta function and a convergent inverse-square tail. The gamma function contribution is also logarithmic by the Stirling formula. This yields the stated local error without an undefined convention for which zeros are nearby.

The logarithmic derivative of a differentiable function $f$ is $f^{\prime }/f$ wherever $f\ne 0$. For a nonvanishing holomorphic function, a local holomorphic logarithm differentiates to this expression even when no global logarithm exists. If a meromorphic function has local factorization $f(z)=(z-a{)}^{m}h(z)$ with $h(a)\ne 0$, differentiation gives $f^{\prime }/f=m/(z-a)+h^{\prime }/h$. Its residue is therefore $m$, positive for a zero and negative for a pole; this explains the multiplicity count in the argument principle. For $\Re s>1$, the Euler product for the Riemann zeta function gives

After shrinking the zero-free-region constant, one hasin the half-width region $\sigma >1-c/(2\log |t|)$ for $|t|$ large. Compare the local partial-fraction expansion at $s$ with one at a point just to the right of one, where the Euler product controls the logarithmic derivative.

For a finite set of primes $p\le y$, multiply the absolutely convergent local series. Unique prime factorization and multiplicativity, with $g(1)=1$, giveThe omitted integers all exceed $y$, so the absolute difference from $G(s)$ is at most ${\sum }_{n>y}|g(n)|n^{-\sigma }$, which tends to zero. Hence the finite products converge to $G(s)$:This proves the Euler product directly from absolute convergence, rather than treating an infinite formal multiplication as automatically valid.

For $n\le x$, write $F(n)=F(x)-{\int }_n^x{F}^{\prime }(t)dt$, multiply by $f(n)$ and sum. Interchanging a finite sum with the integral gives the Abel summation formulaApply it with $F(t)=t^{-s}$ and $S(t)={\sum }_{n\le t}{a}_n$. If $\sigma >\alpha$, the boundary term is $O(x^{\alpha -\sigma })\to 0$ and ${\int }_1^{\infty }|S(t)t^{-s-1}|dt$ converges. Therefore the Dirichlet series converges andChoose $C$ with $|S(t)|\le C{t}^{\alpha }$ for $t\ge 1$. For $\sigma >\alpha +\delta$,The Weierstrass M-test proves uniform convergence of $\sum I_m$ on this entire open half-plane. For fixed $s$, the same estimate with its actual $\sigma$ gives $I_m(s)=O(m^{\alpha -1-\sigma })$. Each $I_m$ is an entire function, since its integral has bounded positive $t$ range and can be differentiated in $s$ under an integrable bound on each compact set. Thus the locally uniform sum is holomorphic, and so is its product with $s$. This gives holomorphy from polynomial growth of arithmetic partial sums and proves holomorphy of $A$ for $\mathrm{Re}s>\alpha$. Multiplication by the unbounded factor $s$ does not assert uniform convergence of $A$ on a whole unbounded half-plane; local uniform convergence is what the holomorphy argument needs.

With $a_n=\mu (n)$ and the hypothetical bound, $A_{\mu }(s)$ is holomorphic for $\mathrm{Re}s>3/4$. On $\mathrm{Re}s>1$, the Euler product gives $\zeta (s)A_{\mu }(s)=1$. The identity theorem extends this as a meromorphic identity throughout the connected half-plane $\mathrm{Re}s>3/4$. Hence $\zeta$ has no zeros in that half-plane: at any zero, the product with the holomorphic $A_{\mu }$ would be zero. The pole at one is not a zero; $A_{\mu }$ has a compensating zero there.

The double-angle formula givesFor $\sigma >1$, the logarithm of the absolutely convergent Euler product givesExponentiation provesSuppose $\zeta (1+it)=0$ for $t\ne 0$, of order $r\ge 1$. As $\sigma \downarrow 1$, its middle factor is $O((\sigma -1{)}^{4r})$, the first is $O((\sigma -1{)}^{-3})$ because the pole at one is simple, and the final factor is bounded because $1+2it\ne 1$. The product is $O((\sigma -1{)}^{4r-3})\to 0$, contradicting the lower bound one. ThereforeThis is the three-four-one zero-free-region argument specialized to the line with real part one.

For $\mathrm{Re}s>1$, the Euler product theorem isThe local series converge absolutely, and the product is absolutely convergent in the sense that the sum over primes of the absolute deviations of the local factors from one is finite. Indeed this sum is bounded by ${\sum }_{n\ge 2}|f(n)|n^{-\mathrm{Re}s}$. To prove the identity, expand the product over primes at most $y$. Absolute convergence permits all those local expansions, and multiplicativity identifies the coefficients with $f(n)$ for integers whose prime factors are at most $y$. Every omitted integer exceeds $y$, so the difference from the full Dirichlet series is bounded by its absolute tail beyond $y$, which tends to zero. The same tail bound on a closed right half-plane inside $\mathrm{Re}s>1$ proves local uniform convergence. With signed coefficients a local factor can be zero; the identity does not require claiming every such general Euler product is nonzero.

For $d(n^2)$, multiplicativity gives local coefficients $d(p^{2e})=2e+1$. With $z=p^{-s}$,The divisor bound from (1), applied with a sufficiently small exponent, proves absolute convergence when $\mathrm{Re}s>1$; divergence at one follows from $d(n^2)\ge 1$. Multiplying the local identities therefore givesFor $n\le x$, the inequality $n^{\delta }\le x^{\delta }$ gives $d(n^2)/n\le x^{\delta }d(n^2)/n^{1+\delta }$. Sum those nonnegative terms and enlarge to the whole infinite series to get the required comparison.

For real $1<s<2$, integral comparison gives $\zeta (s)\le 1+(s-1{)}^{-1}\le 2/(s-1)$, while $\zeta (2s)\ge 1$. Hence $\zeta (s{)}^3/\zeta (2s)=O((s-1{)}^{-3})$ uniformly in that interval. Choosing $\delta =1/\log x$ for $x\ge 3$ makes $0<\delta <1$ and $x^{\delta }=e$. Thus