Riemann zeta function ($\zeta (s)$)

$\zeta (s)={\sum }_{n\ge 1}{n}^{-s}$ for $\Re s>1$; its Euler product is ${\prod }_p(1-p^{-s}{)}^{-1}$.

For $\sigma >1$, the logarithm of the displayed product is a prime-power sum with coefficients $3+4\cos \theta +\cos 2\theta =2(1+\cos \theta {)}^2\ge 0$. A zero of order $a\ge 1$ at $1+it$, with $t\ne 0$, would make the product vanish like $O((\sigma -1{)}^{4a-3})$: the real factor has a pole of order three, and the third factor is bounded. This contradicts its lower bound one. Hence the Riemann zeta function has no zeros on its line of real part one; its pole at one is not a zero.

The hypothesis asks for growth smaller than every fixed positive power on the critical line. The implied constant may depend on the chosen exponent. The Riemann hypothesis implies the Lindelöf hypothesis through a smoothed explicit formula and strip convexity.

For $T/2\le \Im s\le T$ and $0<\Re s=\sigma \le 1$, the Hardy-Littlewood approximation to the Riemann zeta function truncates zeta at $T$ with error $O(T^{-\sigma })$. Multiply the finite sums and use $a_n={\sum }_{m\mid n,m\le M,n/m\le T}\mu (m)$. All coefficients with $1<n\le \min (M,T)$ vanish by the Möbius divisor-sum identity. The total error is bounded using ${\sum }_{m\le M}{m}^{-\sigma }\ll M^{1-\sigma }\log M$.

For large positive $t$ and $0<\sigma \le 1$, a fixed sufficiently large $C$ gives the displayed upper bound; to the right of one use $|\zeta (\sigma +it)|\ll {\log }^{2/3}t$. Its nonlinear dependence on $1-\sigma$ permits a wider zero-free region of the Riemann zeta function through the Landau zero-free-region theorem. Its proof uses estimates for exponential sums; its use as an upper-bound input is separate from a proof of a zero-free region.

For $s=\sigma +it$, $\sigma >0$, and $x\ge |t|/\pi$, the displayed approximation holds away from the pole. Apply the Van der Corput sum-integral lemma to $-t\log w/(2\pi )$ on $w\ge x$, where its derivative has modulus at most one half. Abel summation with $w^{-\sigma }$ converts the bounded unweighted discrepancy to $O(x^{-\sigma })$. Locally uniform convergence of this weighted discrepancy extends the identity from $\sigma >1$ to $\sigma >0$. It turns estimates for finite exponential sums into estimates for the Riemann zeta function.

Abel summation applied to the counting function $\lfloor w\rfloor$ gives the displayed identity for $\Re s>1$ and every $x>0$. Since $0\le \{w\}<1$, the integral is locally uniformly convergent and holomorphic for $\Re s>0$. This supplies a meromorphic continuation of the Riemann zeta function to that half-plane, with its sole pole at one and residue one. Keeping the fractional endpoint term makes the formula valid at noninteger cutoffs.

The completed zeta function $\Lambda (s)={\pi }^{-s/2}\Gamma (s/2)\zeta (s)$ satisfies $\Lambda (s)=\Lambda (1-s)$ and has poles at zero and one. It is also sometimes denoted $\xi (s)$; this differs from the entire xi function obtained by multiplying by $s(s-1)/2$. The distinction matters in the constant term of a nonholomorphic Eisenstein series.

Removing the completion's poles produces an entire function with $\xi (0)=\xi (1)=1/2$ and $\xi (s)=\xi (1-s)$. Its zeros are the nontrivial zeros of the Riemann zeta function. Its finite order allows Hadamard factorization, while the xi modulus criterion for the Riemann hypothesis characterizes their horizontal location.

Differentiate the genus-one Hadamard factorization. Its complex difference terms converge as $O(|\rho {|}^{-2})$, and the individual real sums converge by the absolute convergence of the real xi logarithmic derivative. The remaining constant is $\Re B+\sum \Re (1/\rho )$. On the critical line the xi logarithmic derivative has zero real part, while the kernel terms cancel in pairs reflected across that line. Hence this constant vanishes and gives the displayed formula away from zeros.

The Jensen zero-count bound gives $O(T\log T)$ zeros up to ordinate $T$. Since their real parts lie in $(0,1)$, a fixed-point real logarithmic derivative summand is $O(|\Im \rho {|}^{-2})$. Dyadic ordinate bands then contribute $O(j{2}^{-j})$. This proves absolute convergence of the real sum, while the unpaired complex sum of reciprocals need not converge absolutely.

The trivial zeros of the Riemann zeta function are $s=-2,-4,-6,\dots$. The remaining zeros are the Nontrivial zeros of the Riemann zeta function.

The Riemann zeta function satisfies ${\int }_0^T|\zeta (1/2+it){|}^{4}dt\ll T(\log T{)}^4$ for $T\ge 2$. Squaring a short Dirichlet polynomial produces the divisor function; the divisor-square summatory bound controls the off-diagonal and diagonal terms.

For every integer $n\ge 1$, the Riemann zeta function satisfies $\zeta (1-n)=(-1{)}^{n-1}{B}_n/n$. One proof applies meromorphic continuation of a Mellin transform from an asymptotic expansion to $(e^y-1{)}^{-1}$, whose transform is the Bose integral $\Gamma (s)\zeta (s)$, and divides residues by the residues of the Gamma function. The formula includes $\zeta (0)=-1/2$.

For $\Re s>0$,The integral defines a holomorphic function in that half-plane, so this continues $\zeta$ meromorphically across $\Re s=1$, with a simple pole of residue one at $s=1$.

The Basel problem asks for the exact value of $\zeta (2)$. Its answer is

The completed zeta functionis an entire function satisfying $\xi (s)=\xi (1-s)$. Thus every nontrivial zero $\rho$ is accompanied by $1-\rho$, $\overline{\rho }$, and $1-\overline{\rho }$.

For $\Theta (u)={\sum }_{n\in \mathbb{Z}}{e}^{-\pi n^{2}u}$ and initially $\Re s>1$,The Poisson summation formula gives $\Theta (u)=u^{-1/2}\Theta (1/u)$. Splitting the integral at one therefore continues it meromorphically and makes its invariance under $s\mapsto 1-s$ visible.

Let $J(s)={\int }_1^{\infty }(\theta (t)-1)t^{s/2-1}dt$, with $\theta (t)={\sum }_{n\in \mathbb{Z}}{e}^{-\pi n^{2}t}$. Exponential decay makes $J$ an entire function. The Jacobi theta function transformation and a substitution in the Mellin transform giveThis continues the completed Riemann zeta function meromorphically, with residues $1$ and $-1$ at $1$ and $0$, and makes the functional equation of the Riemann zeta function immediate. Dividing by the Gamma function removes the apparent pole at zero from $\zeta (s)$.

Writingthe Stirling formula gives, uniformly on every fixed vertical strip and for large $|t|$,

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

A nontrivial zero of the Riemann zeta function lies in the critical strip $0<\Re s<1$. The functional equation of the Riemann zeta function reflects such zeros across the critical line $\Re s=1/2$.

The critical strip for the Riemann zeta function is $0<\Re s<1$.

The critical line is the symmetry line $\Re s=1/2$ inside the critical strip.

The Riemann hypothesis asserts that every Nontrivial zero of the Riemann zeta function lies on the critical line $\Re s=1/2$.

The subpower zeta bound to the right of the critical line gives an arbitrarily small exponent at $1/2+\delta$. The functional equation gives exponent $\delta +\epsilon$ at $1/2-\delta$. The Phragmén–Lindelöf principle gives exponent $\delta /2+\epsilon$ at the midpoint. Letting these fixed positive parameters be arbitrarily small proves the Lindelöf hypothesis.

Under Riemann hypothesis, integrate a smoothed explicit formula from $1/2+\delta$ to two, taking $x=\log |t|$. The prime contribution is $O_{\delta }(x^{1-2\delta })$, while the local zero count bounds the zero contribution by $O_{\delta }(x^{-\delta }\log |t|/\log x)$. Both are $o(\log |t|)$ for fixed $0<\delta <1/4$, yielding the displayed bound; other fixed positive shifts follow by standard strip bounds and the Euler product.

Riemann hypothesis is equivalent to monotonicity of $\sigma \mapsto |\xi (\sigma +it)|$ for every fixed $t$ on $\sigma \ge 1/2$. Under Riemann hypothesis the real logarithmic derivative of Riemann xi is positive for $\sigma >1/2$. Conversely a zero right of that line would force the nonnegative monotone modulus to vanish on an interval, contrary to the identity theorem. Functional symmetry rules out zeros to the left.

The Riemann hypothesis is equivalent tofor every $ϵ>0$. In fact, the hypothesis and the Riemann–von Mangoldt explicit formula give the stronger $O(x^{1/2}(\log x{)}^2)$ bound. Conversely, the stated error continues $-{\zeta }^{\prime }/\zeta -s/(s-1)$ holomorphically to $\Re s>1/2$, excluding zeros there; the functional equation supplies the other half.

The classical zero-free region asserts that for some $c>0$,Together with bounds for the logarithmic derivative, it permits contour arguments with exponentially small errors in $\sqrt{\log x}$.

For sufficiently large $|t|$, the Riemann zeta function has no zeros in the displayed region, for a fixed positive $c$. To derive it from the Richert bound for the Riemann zeta function, take $\eta \asymp (\log \log t/\log t{)}^{2/3}$. The logarithm of the maximum on the two discs in the Landau zero-free-region theorem is $O(\log \log t)$, as is $\log (1/\eta )$. Dividing $\eta$ by this logarithmic factor gives the displayed width.

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.

The logarithm of the Euler product and $3+4\cos \theta +\cos 2\theta \ge 0$ give $\zeta (\sigma {)}^3|\zeta (\sigma +it){|}^4|\zeta (\sigma +2it)|\ge 1$. At $\sigma =1+a/{\log }^9|t|$, upper bounds $\zeta (\sigma )\ll {\log }^9|t|/a$ and $|\zeta (\sigma +2it)|\ll \log |t|$ imply the lower bound $|\zeta |\gg a^{3/4}{\log }^{-7}|t|$. The Cauchy estimate for derivatives gives $|{\zeta }^{\prime }|\ll {\log }^2|t|$ nearby, allowing a sufficiently small leftward displacement of order ${\log }^{-9}|t|$. This elementary region is weaker than the classical logarithmic region but gives an explicit reciprocal bound by simple estimates.

For $F=-{\zeta }^{\prime }/\zeta$ and $\sigma >1$, positivity ofterm by term in the Euler product givesComparing this with the pole of $F$ at one and the local poles of ${\zeta }^{\prime }/\zeta$ at zeros proves the classical zero-free region.

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.

If $N(T)$ counts the nontrivial zeta zeros $\rho =\beta +i\gamma$ with $0<\gamma \le T$, including multiplicity, thenIn particular,

The number of Nontrivial zeros of the Riemann zeta function in a height interval of length one is $O(\log (|t|+3))$, including multiplicity and either endpoint. Subtract the Riemann–von Mangoldt formula at the endpoints and use conjugation at negative heights. Summing these counts with reciprocal-height weights gives ${\sum }_{|\Im \rho |\le T}1/|\rho |\ll {\log }^2(T+3)$.

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.

Evaluate the real logarithmic derivative of Riemann xi at real part two, where every summand is positive and comparable to this kernel. The gamma logarithmic derivative and the bounded zeta logarithmic derivative give the logarithmic bound. It implies the local unit-interval count, and conversely such local counts imply this smoothed bound by summing the decaying tails.

The logarithmic derivative of a nonzero differentiable function $f$ is $f^{\prime }/f$. For $\Re s>1$, the Euler product for the Riemann zeta function gives

If $f$ is holomorphic near the closed radius-$R$ disc, $f(z_0)\ne 0$, and $|f|\le M$, the displayed estimate holds away from zeros for $|z-z_0|\le R/3$, counting zeros with multiplicity. Factoring local zeros isolates their poles; the remaining logarithm is controlled by disc estimates. If all zeros lie to the left of a point $z$ in real part, their real contributions are nonnegative. This is the disc estimate used by the Landau zero-free-region theorem; a version with general radius ratios is Lemma 24.17 in Montgomery and Vaughan.

Uniformly in a fixed bounded-width disk around height $t$,This follows by taking the logarithmic derivative of the Hadamard product for the completed zeta function and estimating the distant zeros.

The prime number theorem states that $\pi (x)\sim x/\log x$, equivalently ${\sum }_{n\le x}\Lambda (n)\sim x$.

Suppose $A>0$ and every sufficiently high Nontrivial zero of the Riemann zeta function satisfies the displayed gap, with bounded heights also separated from one. The truncated explicit formula for the second Chebyshev function and the local zero count for the Riemann zeta function give errors $x{\log }^{2}x\exp (-c\log x/(\log T{)}^A)$ and $x{\log }^{2}x/T$. Balance them by $\log T\asymp (\log x{)}^{1/(A+1)}$. This explains how the width of a zero-free region of the Riemann zeta function determines the exponential scale in the Prime number theorem error.

The classical zero-free region and a truncated Perron contour givefor some constant $c>0$.

For a smooth compactly supported $F:(0,\infty )\to \mathbb{R}$ whose Mellin transform has suitable vertical decay, Mellin inversion andgiveMoving the contour through the pole at $s=1$ and into the classical zero-free region gives main term $X\tilde{F}(1)$ and error $O(X^{1-c/\log \log X})$ when $\tilde{F}(\sigma +it)\ll e^{-|t{|}^{1/2}}$.

If, for some real $u$,then partial summation continues $-{\zeta }^{\prime }(s-iu)/\zeta (s-iu)$ meromorphically to $\Re s>1/2$ with no pole except at $1+iu$. The zeta function has no zero to the right of the critical line, and its functional equation then implies the Riemann hypothesis.

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

The identity $\zeta (s)=\eta (s)/(1-2^{1-s})$ continues the Riemann zeta function meromorphically to $\Re s>0$. Apparent singularities at nonreal zeros of $1-2^{1-s}$ are removable, as one sees by replacing $2$ with an integer $k$ for which $1-k^{1-s}\ne 0$. At $s=1$, $\eta (1)=\log 2$ and $1-2^{1-s}\sim (s-1)\log 2$, so $\zeta$ has a simple pole of residue one.