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Here $\psi ={\Gamma }^{\prime }/\Gamma$ is the digamma function, $B=\log 2+\frac{1}{2}\log \pi -1-\gamma /2$, and $\gamma$ is the Euler--Mascheroni constant. The sum runs over Nontrivial zeros of the Riemann zeta function with multiplicities. Each displayed paired summand is $O_s(|\rho {|}^{-2})$, so it converges locally uniformly away from zeros. Differentiate the genus-one Hadamard factorization of the Riemann xi function and use the Gamma function recurrence. Unlike an unpaired reciprocal-zero sum, this expression does not require an unstated summation convention.

The pole-subtracted theta integral for the completed zeta function bounds the Riemann xi function on $|s|\le R$ by a polynomial in $R$ times ${\int }_1^{\infty }{e}^{-\pi t}{t}^{(R+1)/2}dt$. The Stirling formula gives logarithmic growth $O(R\log R)$. On the positive real axis, the definition and $\zeta (R)\ge 1$ give $\log \xi (R)=\frac{R}{2}\log R+O(R)$. Thus its order of an entire function is exactly one.

// Target: analytic-number-theory.bigb

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.

Use $\Xi (u)={\pi }^{-u/2}\Gamma (u/2)\zeta (u)$ for the completed Riemann zeta function, without the polynomial factor used to define the entire Riemann xi function. For complex $\nu$ putThis is the Modified Bessel function of the second kind. The substitution $t=e^v$ writes it as $\frac{1}{2}{\int }_{\mathbb{R}}{e}^{-Y\cosh v+\nu v}dv$, proving that it is entire in $\nu$, that $K_{\nu }=K_{-\nu }$, and that it decays exponentially as $Y\to \infty$, uniformly for $\nu$ in a compact set.

For $\Re s>1$ all the lattice sums and the integrals below converge absolutely. The $m=0$ terms in the completed nonholomorphic Eisenstein series contribute $\Xi (2s)y^s$. For $m\ne 0$, use the Gamma integral and then Poisson summation in $n$:Here the Poisson identity is ${\sum }_n{e}^{-\pi u(n+mx{)}^2}=u^{-1/2}{\sum }_h{e}^{-\pi h^2/u}{e}^{2\pi ihmx}$. At $h=0$, evaluating the Gamma integral and summing over $m$ gives $\Xi (2s-1)y^{1-s}$. At $h\ne 0$, the substitution $u=|h|t/(|m|y)$ and the integral representation of the modified Bessel function of the second kind giveCollect the frequency $r=mh\ne 0$. There are two signed choices of $m$ for each positive divisor of $|r|$. Thus, with ${\sigma }_a(n)={\sum }_{d\mid n}{d}^a$, the full Fourier expansion isThe factor $2$ is important: the lattice sum includes both signs, while the defining factor $1/2$ was already used in its Mellin integral. This expansion uses exactly the completion in the question.

We can deduce both of the offered conclusions. First recall, with a short proof, the needed functional equation of the Riemann zeta function. Put $\Theta (t)={\sum }_{n\in \mathbb{Z}}{e}^{-\pi n^{2}t}$. Poisson summation gives $\Theta (t)=t^{-1/2}\Theta (1/t)$. Splitting the Mellin integral of $\Theta -1$ at one yieldsThe integral is entire and invariant under $u\mapsto 1-u$. Hence $\Xi (u)=\Xi (1-u)$, with only simple poles at $u=0,1$, of residues $-1,1$. This is the pole-bearing completion, not the entire xi function.

For fixed $y>0$, the nonconstant part of the Eisenstein expansion is entire in $s$ and locally normally convergent: the divisor sums grow at most polynomially on a compact set of $s$, whereas $K_{s-1/2}(2\pi |r|y)$ decays exponentially in $|r|$. The same argument gives local uniformity for $z$ in compact subsets of the complex upper half-plane. Thus only the constant terms can have poles. Their apparent poles at $s=1/2$ cancel: $\Xi (2s)$ has principal part $1/(2(s-1/2))$, while $\Xi (2s-1)$ has its negative, and both multiply $y^{1/2}$ there. The remaining poles are simple, at $s=0,1$, with residues $-1/2,1/2$.

Under $s\mapsto 1-s$, the two constant terms interchange by the zeta functional equation. The nonconstant terms are unchanged because $K_{\nu }=K_{-\nu }$ andwhich follows by replacing a divisor $d$ by $|r|/d$. Therefore

For the Kronecker limit formula, evaluate the finite Laurent coefficient at $s=1$. Write ${\gamma }_E$ for the Euler--Mascheroni constant. From $\zeta (u)=1/(u-1)+{\gamma }_E+O(u-1)$ and ${\Gamma }^{\prime }(1/2)/\Gamma (1/2)=-{\gamma }_E-2\log 2$ we obtainThe gamma derivative follows by differentiating the Gamma duplication formula and using ${\Gamma }^{\prime }(1)=-{\gamma }_E$. Multiplication by $y^{1-s}$ adds $-\frac{1}{2}\log y$ to the constant term; the other constant term contributes $\Xi (2)y=\pi y/6$.

The half-order Modified Bessel function of the second kind is $K_{1/2}(Y)=\sqrt{\pi /(2Y)}{e}^{-Y}$. For example, in its even integral representation set $v/2$ as variable and then $u=\sinh (v/2)$: $\cosh v=1+2u^2$ and the $\cosh (v/2)$ factor converts the integral to a Gaussian. At $s=1$ the nonconstant Eisenstein coefficients are consequently ${\sigma }_{-1}(|r|)e^{-2\pi |r|y}$. On the other hand, the absolutely convergent logarithmic series givesAdding this identity to its conjugate identifies the nonconstant sum as $-2\log |{\prod }_{n\ge 1}(1-q^n)|$. With the analytic Dedekind eta function product $\eta (z)=q^{1/24}{\prod }_{n\ge 1}(1-q^n)$, the resulting answer isThis derivation of the Kronecker limit formula uses the analytic eta product only; it does not assume its modular transformation. The lattice series is invariant for $\Re s>1$ by an integral change of the pair $(m,n)$ and $\Im (\gamma z)=y/|cz+d{|}^2$. Analytic continuation preserves this invariance, so its finite Laurent coefficient is invariant too. This supplies the independent weight-two transformation argument used in Question 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.

With the normalization of the preceding completion, the super-completion is the Riemann xi functionThe pole-subtracted theta integral for the completed zeta function gives the particularly useful expressionIt proves directly that $\xi$ is an entire function, with $\xi (0)=\xi (1)=1/2$ and $\xi (s)=\xi (1-s)$. Here “integral function” means entire, not integer-valued.

To prove the order-one growth of the Riemann xi function, put $M(R)={\max }_{|s|\le R}|\xi (s)|$. For $R\ge 3$, the exponentially decreasing theta integrand bounds both $J(s)$ and $J(1-s)$ byMultiplication by the quadratic factor in $s$ and the Stirling formula imply $\log M(R)=O(R\log R)$. Hence the order of an entire function is at most one. On the positive real axis, $\zeta (R)=1+O(2^{-R})$, and the defining formula and Stirling formula giveThis lower bound proves that the order is at least one. Consequently $\xi$ is entire of order exactly one, generally not of finite exponential type.

Its zeros are precisely the Nontrivial zeros of the Riemann zeta function, counted with multiplicity. The values at zero and one are nonzero; the Gamma function has no zeros; and the completed functional equation together with the Euler product shows that there are no further completed zeros outside the closed critical strip. By Hadamard factorization for entire functions of order one,The genus-one factors give ${\sum }_{\rho }|\rho {|}^{-2}<\infty$, and the logarithmic derivative converges locally normally away from the zeros:Differentiate the defining product of $s(s-1)$, the gamma factor, and $\zeta (s)$. Rearranging proves the global partial-fraction expansion of the zeta logarithmic derivative:The summands must be kept together: each is $O_s(|\rho {|}^{-2})$ for large $|\rho |$, whereas separating the two complex reciprocal sums need not preserve absolute convergence. The gamma term accounts for the trivial zeros as well as cancellation at zero.

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.

Use the Fourier transform convention $\hat{g}(\xi )={\int }_{\mathbb{R}}g(x)e^{-2\pi ix\xi }dx$. For $t>0$, completing the square in the Gaussian integral givesBoth functions are rapidly decreasing, so the Poisson summation formula applies:In terms of the Jacobi theta function, this is $\vartheta (i/t)=\sqrt{t}\vartheta (it)$. The theta series converges locally uniformly in the complex upper half-plane, defining a holomorphic function. The branch of $\sqrt{-i\tau }$ with positive value at $\tau =it$ is holomorphic there, because $-i\tau$ lies in the right half-plane. Both sides of the proposed inversion identity are holomorphic; agreement on the positive imaginary axis and the identity theorem giveEquivalently, the complex Gaussian has transform $(-i\tau {)}^{-1/2}\exp (-\pi i{\xi }^2/\tau )$, which produces the same identity directly by Poisson summation.

Put $\psi (t)=(\vartheta (it)-1)/2={\sum }_{n\ge 1}{e}^{-\pi n^{2}t}$. For $\mathrm{Re}s>1$, integrate term by term in its Mellin transform:The theta function inversion givesSplit the integral at one. In the term containing $\psi (1/t)$ put $t=1/u$. The elementary remainder integral isThus the pole-subtracted theta integral for the completed zeta function isThe integral is entire by exponential decay. This supplies a meromorphic continuation with simple poles at zero and one, of residues $-1$ and $1$. The expression is unchanged by $s\mapsto 1-s$, proving the functional equation of the Riemann zeta functionThe reflection and duplication identities $\Gamma (z)\Gamma (1-z)=\pi /\sin \pi z$ and $\Gamma (z)\Gamma (z+1/2)=2^{1-2z}\sqrt{\pi }\Gamma (2z)$ convert it toMultiplying $Z$ by $s(s-1)/2$ removes its two poles and yields the Riemann xi function, which is an entire function, satisfying $\xi (s)=\xi (1-s)$.

Use $e(t)=e^{2\pi it}$ and the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e(-x\xi )dx$, initially for an integrable function. For the interval indicator function, direct integration givesThe integral bound is one, and the displayed numerator has modulus at most two. By the convolution theorem, the Fourier transform of the $m$-fold convolution is its $m$th power. Thus the Fourier decay of repeated interval convolutions givesAt zero, interpret the second term of the minimum as infinity.