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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.

Use the meromorphic completionThe functional equation of the Riemann zeta function isIts entire version is $\xi (s)=\frac{1}{2}s(s-1)Z(s)$ with $\xi (s)=\xi (1-s)$. The completion $Z$ itself has poles at zero and one; it should not be confused with the entire Riemann xi function.

We prove the equation by a theta integral. SetThe Gaussian Fourier transform, with kernel $e^{-2\pi ixy}$, isFor example, differentiation in $y$ and integration by parts give the differential equation $I^{\prime }(y)=-(2\pi y/t)I(y)$; the Gaussian integral gives $I(0)=t^{-1/2}$. Periodize the Gaussian by summing $e^{-\pi t(x+n{)}^2}$ over integers $n$. Its $k$th Fourier coefficient is the displayed transform at $k$. Both the periodized function and its Fourier series converge absolutely, so evaluation at $x=0$ gives the Poisson summation formula in this case:

For $\mathrm{Re}s>1$, integration term by term in the Gamma function integral givesIndeed, the two terms for $n$ and $-n$ cancel the factor $1/2$, and substitution $v=\pi n^{2}t$ gives ${\pi }^{-s/2}\Gamma (s/2)n^{-s}$. Split at one, and use the transformation above in the integral from zero to one. The elementary contribution isSubstitution $t=1/u$ in the remaining term therefore givesThe theta tail decays exponentially. The integral is consequently an entire function of $s$, with uniform convergence and termwise differentiation on compact subsets. The formula continues $Z$ meromorphically to the whole plane. Both its rational part and its integral are unchanged by $s\mapsto 1-s$, proving the equation and the entire xi formulation. Using the reflection and duplication identities of the Gamma function, one may equivalently writeunderstood as an identity of meromorphic functions, including points where individual factors need continuation.

Now let $N(T)$ count the Nontrivial zeros of the Riemann zeta function $\rho =\beta +i\gamma$ with $0<\gamma \le T$, including multiplicity. The Riemann–von Mangoldt formula isFor $T$ not a zero ordinate, the more precise usual form isHere the argument is continued from $\zeta (2)>0$ along the path $2\to 2+iT\to \frac{1}{2}+iT$. The first formula remains valid at zero ordinates with the stated endpoint convention. Only its leading asymptotic is needed below.

List the positive ordinates as ${\gamma }_1\le {\gamma }_2\le \cdots$, with multiplicity, consistent with this counting function. Put $b_n=2\pi n/\log n$. For every fixed $\epsilon \in (0,1)$,For all sufficiently large $n$, the lower count is less than $n$ and the upper count is greater than $n$. Consequently $(1-\epsilon )b_n<{\gamma }_n\le (1+\epsilon )b_n$. Letting $\epsilon \downarrow 0$ proves the asymptotic inversion of the zeta zero countThis argument works even when several zeros have the same ordinate; “increasing” is interpreted in the standard nondecreasing, multiplicity-counted sense. A count listing only distinct ordinates is a different object and is not determined by this formula alone.

In this notation the Riemann hypothesis says that every Nontrivial zero of the Riemann zeta function has real part one-half:It concerns their horizontal location, not merely the growth of the ordinates, and does not assert that the zeros are simple. The functional equation and complex conjugation make the zero set symmetric about both the real axis and the critical line; the hypothesis puts all the Nontrivial zeros of the Riemann zeta function on that line.

Under the Riemann hypothesis, every $\beta =1/2$, so for $\sigma >1/2$ the absolutely convergent formula givesThe zero set is nonempty: otherwise Hadamard factorization would make $\xi$ an exponential of a linear polynomial, and its functional symmetry would force it to be constant, contrary to gamma growth on the positive real axis. Thus the inequality is strict in the open right half-plane. Continuity at $\sigma =1/2$, including at boundary zeros, proves the claimed increasing modulus on the closed half-line.

Conversely, suppose the modulus is nondecreasing for every fixed $t$. If a zero $\beta +i\gamma$ had $\beta >1/2$, then nonnegativity and monotonicity would force $|\xi (\sigma +i\gamma )|=0$ throughout $1/2\le \sigma \le \beta$. The identity theorem would make $\xi$ identically zero, a contradiction. A zero left of the line reflects to one right of the line by the functional equation and complex conjugation symmetry. Hence every zero lies on the critical line. This proves the xi modulus criterion for the Riemann hypothesis. The two following roman headers refer to supplied asymptotic assumptions, not further questions, and require no Solution sections.

The Riemann hypothesis says that every Nontrivial zero of the Riemann zeta function of $\zeta$ has real part $1/2$. The Lindelöf hypothesis says that, for every $\epsilon >0$,The exponent may be arbitrarily small; the implied constant may depend on that exponent. The next part proves the implication from the first hypothesis to the second.

Assume Riemann hypothesis. First fix $0<\delta <1/4$ and work on ${\sigma }_0=1/2+\delta$. We prove the subpower zeta bound to the right of the critical line, then move back to the line by the functional equation and Phragmén–Lindelöf principle.

Put $x=\log t$ for large positive $t$ and integrate the supplied smoothed logarithmic derivative identity horizontally from ${\sigma }_0$ to two. There are no zeros on this path under Riemann hypothesis, so the Euler-product logarithm at $2+it$ continues along it. Each prime-power term contributes at most $\Lambda (n)n^{-{\sigma }_0}/\log n\le n^{-{\sigma }_0}$, and the smoothing weights are at most one. Hence the integrated prime terms are bounded byAll zeros have $\rho =1/2+i\gamma$. The local zero-count estimate from Question 3, with its reflected version for negative ordinates, gives uniformly for ${\sigma }_0\le \sigma \le 2$To see the uniformity, sum the $O(\log (2+|\gamma |))$ zeros in successive unit ordinate intervals against $({\delta }^2+|t-\gamma {|}^2{)}^{-1}$; the distant dyadic tails are summable. The zero-term numerator has modulus at most $x^{-2\delta }+x^{-\delta }$. Its integrated contribution is therefore at most $C_{\delta }{x}^{-\delta }\log t/\log x=o(\log t)$. The integrated supplied remainder is $O(x^{-1-\delta }\log t/\log x)$, also $o(\log t)$. Since $\log \zeta (2+it)$ is bounded, we obtainThus for every fixed $\delta >0$ and $\eta >0$, $|\zeta (1/2+\delta +it)|{\ll }_{\delta ,\eta }{t}^{\eta }$. Negative $t$ follow by complex conjugation.

The zeta functional equation and the gamma ratio give $|\zeta (1/2-\delta +it)|{\ll }_{\delta ,\eta }{t}^{\delta +\eta }$. Zeta is holomorphic throughout this strip, since its pole at one is outside it, and Euler summation supplies polynomial vertical growth. The strip convexity conclusion of Phragmén–Lindelöf principle therefore gives at the midpointFor a prescribed $\epsilon >0$, choose $\delta$ and $\eta$ with $\delta /2+\eta <\epsilon$; the bounded $t$ range is harmless. This provesThe explicit-formula estimate is deliberately first made a fixed distance to the right of the critical line. No divergent zero bound at $\sigma =1/2$ is used.

By part (a), every nontrivial zero $\rho =\beta +i\gamma$ with $|\gamma |\le T$ satisfies $\beta \le 1-c_1/{\log }^{9}T$ for large $T$, after reducing the positive constant. The local zero count for the Riemann zeta function givesThere are finitely many zeros at bounded height, none at zero or at one, so that part of the sum is bounded. The truncated explicit formula for the second Chebyshev function now yieldsBalance the exponent losses $c_1\log x/(\log T{)}^9$ and $\log T$ by choosing $\log T=(c_1\log x{)}^{1/10}$. This is the optimal order obtainable from these two errors: making either exponent larger forces the other smaller. Thus, for a positive constant $c_2$,The logarithmic prefactor can be absorbed by reducing $c_2$. This is the prime number theorem error from a logarithmic zero-free region with ninth-power width.

Under the Riemann hypothesis, $|x^{\rho }|=x^{1/2}$. The same reciprocal-zero sum bounds the zero contribution by $O(x^{1/2}{\log }^{2}T)$. Taking $T=x$ makes the truncation error $O({\log }^{2}x)$, so

Assume first the Riemann hypothesis. Replace any $x\ge 2$ by $y=\lfloor x\rfloor +1/2$; then $\psi (x)=\psi (y)$ and $⟨y⟩\ge 1/2$. Take $T=y$ in part a. Every nontrivial zero has real part $1/2$, and the Riemann–von Mangoldt formula impliesConsequentlywhile both truncation errors in part a are $O((\log y{)}^2)$. Thuswhich implies the stated $O_{ϵ}(x^{1/2+ϵ})$ estimate.

Conversely, suppose that estimate holds for every $ϵ>0$. For $\Re s>1$, partial summation givesGiven any $s$ with $\Re s>1/2$, choose $ϵ<\Re s-1/2$. The error hypothesis makes the last integral locally uniformly convergent there, so it supplies a holomorphic continuation ofto the half-plane $\Re s>1/2$. A zero of $\zeta$ in that half-plane would create a pole of its logarithmic derivative, so none exists. The functional equation of the Riemann zeta function reflects every nontrivial zero with real part below $1/2$ to one above $1/2$. All nontrivial zeros must therefore lie on the critical line, proving the Riemann hypothesis equivalence for the second Chebyshev function.

PutFor $\Re s>1$, partial summation givesThe last integral is holomorphic for $\Re s>1/2$. Thus the logarithmic derivative on the left continues meromorphically to that half-plane with no pole except $s=1+iu$.

A zero $\rho$ of $\zeta$ with $\Re \rho >1/2$ would make $-{\zeta }^{\prime }(s-iu)/\zeta (s-iu)$ singular at $s=\rho +iu$, a contradiction unless $\rho =1$, which is a pole rather than a zero. Therefore no nontrivial zero lies to the right of the critical line. The functional equation of the Riemann zeta function reflects zeros across that line, so none lies to its left either. Every nontrivial zero lies on the critical line, proving the Riemann hypothesis. This is the Twisted Von Mangoldt estimate implying the Riemann hypothesis.

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.

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.