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

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 $\Re s=1/2$.

Use the Jacobi theta function $\Theta (t)={\sum }_{n\in \mathbb{Z}}{e}^{-\pi n^{2}t}$, $t>0$. Its transformation law follows directly from the Gaussian Fourier transform. With the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(u)e^{-2\pi i\xi u}du$,For example, differentiating the transform and using integration by parts gives ${\hat{f}}^{\prime }(\xi )=-(2\pi \xi /t)\hat{f}(\xi )$; its value at zero is the Gaussian integral $t^{-1/2}$. Periodize the Gaussian. Its Fourier coefficients are these transforms at integer frequencies, and both series converge absolutely, so evaluating its Fourier series at zero proves the Poisson summation formula here:

For $\Re s>1$, integrating the exponentially convergent series termwise gives the Mellin representation of the completed Riemann zeta functionSplit at one and substitute $u=1/t$ in the lower part, using the transformation just proved. The contribution of $t^{-1/2}-1$ is $1/(s-1)-1/s$. We obtain the pole-subtracted theta integral for the completed zeta functionThe integral is entire: $\Theta (t)-1$ decays exponentially, uniformly dominating the integrand and all its $s$ derivatives on compact sets. Both the rational term and the integral are invariant under $s\mapsto 1-s$. This proves the meromorphic identityEquivalently the Riemann xi functionis entire and satisfies $\xi (s)=\xi (1-s)$. The reflection and duplication formulas for the Gamma function turn the same identity intounderstood by meromorphic continuation at removable exceptional points. This establishes the functional equation of the Riemann zeta function, rather than assuming it.

If $N(T)$ counts zeros $\rho =\beta +i\gamma$ in the critical strip with $0<\gamma \le T$, including zero multiplicity, the Riemann–von Mangoldt formula isAt a zero ordinate one may instead use the conventional half-weight boundary count; that only alters the stated error. In particular $N(T)\sim T\log T/(2\pi )$.

To invert without assuming simple zeros, fix $0<\epsilon <1$ and set $T_n^{\pm }=(1\pm \epsilon )2\pi n/\log n$. Then $\log T_n^{\pm }/\log n\to 1$, soFor large $n$, $N(T_n^{-})<n<N(T_n^{+})$, and hence $T_n^{-}<{\gamma }_n\le T_n^{+}$. Let $\epsilon \downarrow 0$. The asymptotic inversion of the zeta zero count givesAs in the zero-counting formula, the sequence lists ordinates with zero multiplicity; “increasing” means nondecreasing here. The argument does not require the Riemann hypothesis or simplicity of zeros.

For $\mathrm{Re}s>1$, define the Riemann zeta function by the absolutely convergent Dirichlet seriesThe Euler product follows from unique prime factorization and absolute convergence. The counting-function integral givesBecause the fractional part is bounded, the last integral, and its derivatives with respect to $s$ on compact subsets, converge uniformly for $\mathrm{Re}s>0$. It is a holomorphic function there. This proves the Meromorphic continuation of the Riemann zeta function to the right half-plane, with a simple pole at one of residue one and no other singularity in that half-plane.

The Gamma function isThe Gamma function recurrence extends it meromorphically; its poles are at the nonpositive integers and it has no zeros. The functional equation of the Riemann zeta function can be stated asor, equivalently,These are identities of meromorphic functions; the second also extends the Riemann zeta function to the remaining half-plane. The critical strip is $0<\mathrm{Re}s<1$; its boundary lines will also be treated below, rather than included among possible exceptional zeros.

For $\mathrm{Re}s>1$, the absolutely convergent reciprocal Euler product gives $\zeta (s)\ne 0$. For $\mathrm{Re}s<0$, the factors $2^s$, ${\pi }^{s-1}$, $\Gamma (1-s)$ and $\zeta (1-s)$ in the second functional equation of the Riemann zeta function are finite and nonzero. Hence the only zeros there are the zeros of the sine factor:Each is simple. At zero, the sine zero cancels the pole of $\zeta (1-s)$; using its residue one gives $\zeta (0)=-1/2$, not zero. Nonvanishing on the rest of $\mathrm{Re}s=0$ follows from the boundary-line proof below and the functional equation of the Riemann zeta function. Together these facts prove that the trivial zeros of the Riemann zeta function are the only zeros outside the open critical strip.

Here is a Jensen disk proof of the zeta zero-count bound which avoids any unproved left-half-plane growth estimate. Set $F(s)=(s-1)\zeta (s)$, a holomorphic function throughout $\mathrm{Re}s>0$, including at one. The integral continuation formula givesFor each integer $j$, apply Jensen's formula in the disk with centre $2+ij$, outer radius $R=7/4$ and inner radius $r=8/5$. The outer disk stays in $\mathrm{Re}s\ge 1/4$, so its maximum modulus is $O((|j|+2{)}^2)$. At its centre,since $|1/\zeta (2+ij)|\le {\sum }_n|\mu (n)|n^{-2}\le \zeta (2)$. Therefore the number of zeros in its inner disk, counted with multiplicity, is at mostThe rectangle $1/2\le \mathrm{Re}s\le 1$, $|\mathrm{Im}s-j|\le 1/2$ fits in the inner disk because its farthest point has distance $\sqrt{(3/2{)}^2+(1/2{)}^2}=\sqrt{5/2}<8/5$. Summing over $O(T)$ such disks counts $O(T\log T)$ zeros in the right half of the critical strip up to height $T$. The first functional equation of the Riemann zeta function bijects zeros, with multiplicities, in the left half with their reflections $s\mapsto 1-s$ in the right half; its Gamma function factors are finite and nonzero in the strip. ThusThe pole of zeta function at one does not count as a zero: $F(1)=1$.

Finally, for real $\sigma >1$ and $t\ne 0$, the Euler-product logarithms and the nonnegative trigonometric polynomialgive the three-four-one product proof of zeta boundary nonvanishing:If $\zeta (1+it)$ had a zero of order $a\ge 1$, its factor would be $O((\sigma -1{)}^{4a})$ as $\sigma \downarrow 1$. The real zeta function factor has pole order three, and the factor at $1+2it$ stays bounded because $t\ne 0$. The product would tend to zero as $O((\sigma -1{)}^{4a-3})$, contradicting that it is at least one. ThereforeAt $t=0$ there is a pole, not a zero. Applying the sine-form functional equation of the Riemann zeta function at $s=it\ne 0$ now proves nonvanishing on the remaining imaginary axis, completing the earlier assertion about all zeros outside the critical strip.

We use the Van der Corput sum-integral lemma. Put $e(u)=e^{2\pi iu}$ and $I_h={\int }_a^{b}e(f(x)-hx)dx$. The Fourier series of the periodization of $1_{[a,b]}(x)e(f(x))$ giveswhere integer endpoints have half weight. This is the Dirichlet-Jordan convergence theorem for a piecewise smooth, or more generally bounded-variation, periodic function. Here $f$ is $C^1$, so the periodized function has bounded variation. Changing to the requested endpoint convention costs at most one.

Write $u=f^{\prime }$. For $h\ne 0$, $|u-h|\ge |h|-\delta >0$. Since $u$ is continuous and monotone, the reciprocal has bounded variation, and integration by parts in the Riemann-Stieltjes sense yieldsThe variation of the reciprocal is at most $2\delta /(h^2-{\delta }^2)$. Summing over $h\ne 0$ gives $O((1-\delta {)}^{-1})$, separating $|h|=1$ and using convergence of ${\sum }_{h\ge 2}{h}^{-2}$. For each endpoint, useThe symmetric partial sums of the first term are a constant multiple of ${\sum }_{h=1}^H\sin (2\pi hx)/h$, uniformly bounded in $H$ and $x$; this standard Fourier series bound follows by splitting at $h\asymp 1/\Vert x\Vert$ and applying Abel summation to the remaining sine sum. The second term is absolutely summable with bound $O((1-\delta {)}^{-1})$. The same bound therefore holds for the whole sum of the $h\ne 0$ integrals. Since $I_0$ is the ordinary integral,No second derivative is required; monotonicity supplies the needed variation estimate.

For the Hardy-Littlewood approximation to the Riemann zeta function, take $f(w)=-t\log w/(2\pi )$. On $w\ge x\ge |t|/\pi$, $|f^{\prime }(w)|\le 1/2$ and $f^{\prime }$ is monotone. The proved lemma says that the difference between the partial sum of $w^{-it}$ and its integral over $(x,Y]$ is $O(1)$ uniformly in $Y$. Weighted Abel summation with the decreasing weight $w^{-\sigma }$ then makes the weighted difference $O(x^{-\sigma })$, since its total variation on $[x,\infty )$ is $x^{-\sigma }$. Initially for $\sigma >1$, the tail integral is $x^{1-s}/(s-1)$. The bounded primitive of the discrepancy gives a locally uniformly convergent weighted discrepancy integral for every $\sigma >0$, continuing the identity to that region. Thus, away from the pole,If $t=0$ the ordinary sum-integral comparison supplies the same estimate. At $s=1$ the formula is understood meromorphically. It approximates the Riemann zeta function by a finite Dirichlet polynomial, transfers exponential sum estimates to bounds in the critical strip, yields elementary near-one bounds for $\zeta$ and its derivative, and supports estimates for the mean value of Dirichlet polynomials and numerical calculations.