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For a primitive Dirichlet character whose conductor of a Dirichlet character is $q>1$ and whose character parity is $a$, Poisson summation gives ${\theta }_{\chi }(x)={\epsilon }_{\chi }{x}^{-a-1/2}{\theta }_{\overline{\chi }}(1/x)$, where ${\epsilon }_{\chi }=\tau (\chi )/(i^a\sqrt{q})$. Its rapid decay at both ends yields the entire Mellin representation of the completed Dirichlet L-function. The case of conductor of a Dirichlet character equal to one uses the ordinary Jacobi theta function and a subtracted constant term.

Take the Gaussian function $f_a(x)=e^{-\pi a{x}^2}$ with $a>0$. Every derivative is a polynomial times the same Gaussian function, so $f_a$ is a Schwartz function. The Fourier transform of a Gaussian gives, in the cycles convention,Thus the Poisson summation formula gives the exact identityThe Gaussian Fourier transform is given in scaled form in Exercise 41 of Terence Tao's 245C, Notes 2: The Fourier transform.^[ref] The identity is also the imaginary-axis transformation of the Jacobi theta function. At $a=1$ the Gaussian function is its own Fourier transform; at reciprocal values of $a$ its spatial and spectral widths exchange.

For a short derivation of the Gaussian Fourier transform, write $H(\xi )={\hat{f}}_a(\xi )$. Differentiation under the integral and integration by parts, using $f_a^{\prime }(x)=-2\pi ax{f}_a(x)$, giveThe Gaussian integral gives $H(0)=a^{-1/2}$, so this ordinary differential equation yields the claimed formula. The decay of the Gaussian function justifies these steps and removes the boundary term.^[ref]

For $a=0.04$, the physical Gaussian function has width parameter $1/\sqrt{a}=5$, while its Fourier transform has width parameter $\sqrt{a}=0.2$. Many physical samples matter, but the dual zero sample already givesFor $a=25$, use the physical side instead:These are bounds on the full positive tails, obtained below, rather than estimates based only on the first omitted term. Ordinary floating-point arithmetic rounds these sums to $5$ and $1$; the exact real numbers are slightly larger. A Gaussian sum truncation bound lets us choose the cheaper side and certify its truncation error before evaluating it.

The identity between Jacobi theta functions has the oscillator-product form ${\prod }_{r\in \mathbb{N}-1/2}(1+q^r{)}^8-{\prod }_{r\in \mathbb{N}-1/2}(1-q^r{)}^8=16q^{1/2}{\prod }_{n\ge 1}(1+q^n{)}^8$ for $|q|<1$. In a critical RNS string, dividing this identity by the eight-boson oscillator product proves equality of GSO projection-retained Neveu–Schwarz sector and Ramond sector state multiplicities at every mass level.

Take the Jacobi theta functions at zero, with $q=e^{\pi i\tau }$:Define the modular lambda function byThis is an explicit holomorphic expression on the complex upper half-plane. The product formsshow that all three are nonzero there. The theta identity ${\theta }_3^4={\theta }_2^4+{\theta }_4^4$ gives $1-\lambda ={\theta }_4^4/{\theta }_3^4$, so $\lambda$ omits $0,1$ and infinity on the half-plane.

The theta-constant inversion and translation laws, obtained from Poisson summation and changing indices in the defining sums, giveThese formulas imply invariance under the projective groupgenerated by $\tau \mapsto \tau +2$ and $\tau \mapsto \tau /(2\tau +1)$. This is the effective level-two principal congruence subgroup.

Its action is properly discontinuous because it is a discrete subgroup of $PS{L}_2(\mathbb{R})$. It is also free. An elliptic integral matrix has trace $0$ or $\pm 1$. The odd traces are excluded because $a+d$ is even. Trace zero would give $d=-a$ and $bc=-a^2-1\equiv 2(\mathrm{mod}4)$, impossible when $b,c$ are even. Thus the only elements acting trivially are the already quotiented central matrices $\pm I$.

A fundamental domain is the ideal quadrilateralwith vertices $-1,0,1,\infty$. The vertical sides are paired by translation by two, and the semicircular sides by $\tau \mapsto \tau /(2\tau +1)$. Reduction using these pairings gives the domain; equivalently it is a union of six translates of the standard modular fundamental domain, consistent with the six elements of $PS{L}_2({\mathbb{F}}_2)$. On adjoining the cusp points, there are three vertex classes, two paired edges and one face. The resulting orientable compact surface has Euler characteristic $3-2+1=2$, hence is a sphere. This is the compactification of the level-two modular curve.

All three cusp widths are two, so their local parameters are obtained from $q=e^{\pi i\tau }$ after a modular change of coordinate. The defining products giveat infinity: there is a simple zero. At the cusp zero, inversion gives $1-\lambda =16q_0+O(q_0^2)$, so the value is $1$. At the cusp one, combine inversion and translation; the value is infinity and the pole is simple. The three cusp classes are precisely the parity classes of primitive rational pairs $(p,q)$: odd-even, even-odd and odd-odd. A Bézout identity, with parity adjusted by adding multiples of $(p,q)$, places any such cusp in the orbit of $\infty$, $0$ or $1$, respectively.

The descended meromorphic function on the compactified quotient thus has exactly one simple pole. Its degree is one, since the degree equals its total pole multiplicity; it is therefore a biholomorphic coordinate on that sphere. Removing the three cusps identifies the quotient with $\mathbb{C}\setminus \{0,1\}$. Consequentlyis a covering map with deck group $\overline{\Gamma (2)}$. The source half-plane is simply connected, so it is the required universal covering map. This proves the universal covering by the modular lambda function, including the absence of interior branching. Three arbitrary distinct punctures are handled by a target Möbius transformation.

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 Fourier transform convention $\hat{h}(\xi )={\int }_{\mathbb{R}}h(x)e^{-2\pi ix\xi }dx$. For a Schwartz function $h$, the Poisson summation formula isTo prove it, periodize $h$ as $P(x)={\sum }_{n\in \mathbb{Z}}h(x+n)$. Rapid decay of all derivatives makes this a smooth periodic function, with uniform termwise differentiability. Its $m$th Fourier coefficient iswhere absolute convergence justifies splitting the real-line integral into unit intervals. Integration by parts gives rapid decay of the coefficients of $P$, so its Fourier series converges absolutely and uniformly to $P$. Evaluate it at zero to obtain the formula.

For $\mathrm{Im}\tau >0$ and $z\in \mathbb{C}$, take the complex Gaussian function $h(x)=e^{\pi i\tau x^2+2\pi izx}$. It is a Schwartz function of the real variable $x$. Its Fourier transform isThe real Gaussian integral proves this first for $\tau =iy$, $y>0$, and real $z$. Normal convergence of the defining integrals makes both sides entire in $z$ and holomorphic in $\tau \in \mathbb{H}$, so the identity theorem extends it to the stated domain. Choose the square-root branch holomorphic for $-i\tau$ in the right half-plane and positive when $\tau =iy$.

Applying Poisson summation gives the transformation of the Jacobi theta function:The series on the right is ${\vartheta }_{00}(z/\tau ,-1/\tau )$. Rearranging therefore givesThe branch choice is part of the formula; at $\tau =i$ its square-root factor equals one.

Use the Fourier transform convention $\hat{f}(u)={\int }_{\mathbb{R}}f(x)e^{-2\pi iux}dx$. A version of the Poisson summation formula sufficient here isBoth sums converge absolutely. To prove it, form the periodization of a Schwartz function $P(x)={\sum }_{n\in \mathbb{Z}}f(x+n)$. The rapid decay of $f$ and all its derivatives makes this series and its differentiated series uniformly convergent on $[0,1]$, so $P$ is smooth and one-periodic. Its Fourier coefficients areRepeated integration by parts makes these coefficients rapidly decreasing. Thus ${\sum }_m\hat{f}(m)e^{2\pi imx}$ converges absolutely and uniformly and has the same coefficients as $P$. By uniqueness of Fourier series it equals $P$. Evaluation at zero proves the formula.

For $u>0$, the Gaussian $f(x)=e^{-\pi u{x}^2}$ has transform $u^{-1/2}{e}^{-\pi m^2/u}$. This follows from the Gaussian integral at frequency zero and integration by parts, which gives the differential equation ${\hat{f}}^{\prime }(v)=-(2\pi v/u)\hat{f}(v)$. Hence the Jacobi theta functionsatisfies $\Theta (u)=u^{-1/2}\Theta (1/u)$.

Use the normalization of the completed Riemann zeta functionInitially, for $\Re s>1$, the Gamma function integral and absolute convergence give its Mellin transform representationIndeed, the two equal terms for $n$ and $-n$ cancel the factor $1/2$, and the $n$th integral is ${\pi }^{-s/2}\Gamma (s/2)n^{-s}$.

Split at $u=1$ and use the theta transformation in the integral over $(0,1)$. Its elementary part isIn the remaining part substitute $v=1/u$. DefineThe pole-subtracted theta integral for the completed zeta function is thereforeSince $\Theta (u)-1=O(e^{-\pi u})$ for $u\ge 1$, both integrals and every derivative in $s$ converge uniformly on compact subsets of $\mathbb{C}$. Thus $J$ is an entire function. The boxed identity supplies the meromorphic continuation of $\Xi$ to the whole plane, with no other singularities. Its poles really are simple, because the residues of the displayed rational terms are nonzero:The identity also gives $\Xi (s)=\Xi (1-s)$, the completed functional equation of the Riemann zeta function.

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.

The Schwartz space $\mathcal{S}(\mathbb{R})$ consists of smooth functions for which every Schwartz seminorm ${\sup }_x|x^a{f}^{(b)}(x)|$ is finite, for nonnegative integers $a,b$. Differentiation under the integral sign and integration by parts giveAll differentiated integrands are integrable, and their boundary terms vanish because of rapid decay. The last transform is bounded by its integrand's $L^1$ norm. The same argument applies at every derivative order, so $\hat{f}\in \mathcal{S}(\mathbb{R})$.

For $f\in \mathcal{S}(\mathbb{R})$, the Poisson summation formula isTo prove it, form the periodization of a Schwartz function $F(x)={\sum }_{n}f(x+n)$. Rapid decay gives uniform convergence of this series and of every differentiated series on $[0,1]$, so $F$ is a smooth periodic function. Its Fourier coefficients areThe series ${\sum }_k\hat{f}(k)e(kx)$ converges absolutely and uniformly. Its Fourier coefficients equal those of $F$. The uniqueness theorem for Fourier series says that two integrable functions with the same coefficients agree almost everywhere; as both functions here are continuous, they agree everywhere. Evaluating at zero proves Poisson summation.

The Gaussian Fourier transform of $e^{-\pi t{x}^2}$, $t>0$, is $t^{-1/2}{e}^{-\pi {\xi }^2/t}$. Apply Poisson summation to obtain the transformation of the Jacobi theta functionFor $\Re s>1$, termwise integration in the Mellin transform gives the Mellin representation of the completed Riemann zeta functionSplit at one and substitute $t=1/u$ in the lower integral. The theta transformation givesThe integral is an entire function of $s$, since $\vartheta (t)-1$ decays exponentially. This pole-subtracted theta integral for the completed zeta function supplies a meromorphic continuation with simple poles at zero and one. Both the rational term and the integral are unchanged by $s\mapsto 1-s$, proving the functional equation of the Riemann zeta functionEquivalently the Riemann xi function $\xi (s)=\frac{1}{2}s(s-1)\Lambda (s)$ is entire and satisfies $\xi (s)=\xi (1-s)$. The reflection and duplication identities for the Gamma function also give $\zeta (s)=2^s{\pi }^{s-1}\sin (\pi s/2)\Gamma (1-s)\zeta (1-s)$ by meromorphic continuation.

An even Dirichlet character satisfies $\chi (-1)=1$, while an odd Dirichlet character satisfies $\chi (-1)=-1$. Write $a=0$ or $1$ for its character parity and, for $x>0$, define the Dirichlet character theta functionFor a primitive Dirichlet character whose conductor of a Dirichlet character is $q>1$, the term at zero is zero. Put $\tau (\chi )={\sum }_{r\mathrm{mod}q}\chi (r)e(r/q)$, using the positive exponential, and ${\epsilon }_{\chi }=\tau (\chi )/(i^a\sqrt{q})$. The primitive Gauss sum of a Dirichlet character has magnitude $\sqrt{q}$, so $|{\epsilon }_{\chi }|=1$. The theta transformation isThus the powers are $x^{-1/2}$ in the even case and $x^{-3/2}$ in the odd case; the odd root number contains $1/i$. The conjugate character is necessary for a nonreal character. These formulas also follow by applying Poisson summation to the Gaussian function on each residue class, and to its derivative for odd parity. For the primitive principal Dirichlet character whose conductor of a Dirichlet character is one, use the ordinary Jacobi theta function with constant term one; its transformation has root number one.

Use the Fourier transform normalization $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e^{-2\pi ix\xi }dx$. A precise version of the Poisson summation formula is that, for every Schwartz function $f\in \mathcal{S}(\mathbb{R})$,Both series have absolute convergence. Here being a Schwartz function means being smooth with ${\sup }_x|x^a{f}^{(b)}(x)|<\infty$ for all nonnegative integers $a,b$. This hypothesis makes the sums and the termwise operations below legitimate; the formula is not being asserted for arbitrary integrable functions.

Form the periodization of a Schwartz function $P_f(x)={\sum }_{n}f(x+n)$. Every differentiated series has uniform convergence on $[0,1]$, by rapid decay, so $P_f$ is a smooth periodic function of period one. Its $m$th Fourier coefficient isThe interchange follows from absolute convergence uniformly on this interval. Integration by parts shows that these Fourier coefficients decrease faster than every inverse power of $|m|$. Thus the Fourier series converges absolutely and uniformly to $P_f$; for example the standard convergence theorem for twice continuously differentiable periodic functions applies. Evaluating at zero proves the Poisson summation formula.

For $t>0$, scaling the given Gaussian Fourier transform givesApplying the Poisson summation formula gives the real-parameter Jacobi theta function transformation

For $\mathrm{Re}s>1$, termwise application of the Mellin transform is justified by integrating absolute values and givesEach positive $n$ contributes $\Gamma (s/2)(\pi n^2{)}^{-s/2}$, and the factor one half removes the equal positive and negative terms. This is the Mellin representation of the completed Riemann zeta function.

Split the integral at one. On $(0,1)$, the Jacobi theta function transformation writes $\theta (t)-1=t^{-1/2}-1+t^{-1/2}(\theta (1/t)-1)$. Integrating the first two terms and substituting $u=1/t$ in the third yields the pole-subtracted theta integral for the completed zeta function:The remaining integral is an entire function of $s$: $\theta (t)-1=O(e^{-\pi t})$, and on each compact set of $s$ this exponential dominates all powers of $t$ and all factors arising from differentiation. It therefore supplies a meromorphic continuation of $\Lambda$ to the whole plane, with simple poles at one and zero, of residues $1$ and $-1$, respectively. Its expression is unchanged by $s\mapsto 1-s$.

The reciprocal Gamma function is entire, with simple zeros at its nonpositive integer arguments. Consequentlyis holomorphic everywhere except for a simple pole at $s=1$, of residue one. The apparent pole at $s=0$ cancels; in fact $\Gamma (s/2{)}^{-1}\sim s/2$ gives $\zeta (0)=-1/2$. This proves the analytic continuation of the Riemann zeta function. The completed functional equation of the Riemann zeta function isThese are equalities of meromorphic functions; at apparent singularities they are interpreted by continuation. Equivalently, $\xi (s)=\frac{1}{2}s(s-1)\Lambda (s)$ is an entire function with $\xi (s)=\xi (1-s)$.

The Poisson summation formula for the Gaussian gives the transformations of the Jacobi theta functionwhere the square root is the holomorphic branch on $\mathfrak{h}$. Raising to the $k$th power gives invariance under $T^2$. If $k=8m$, thenso ${\theta }^k(-1/\tau )={\tau }^{k/2}{\theta }^k(\tau )$. Since $T^2$ and $S$ generate the theta group, this is the weight-$k/2$ transformation law on $\Gamma$.

The defining series converges locally uniformly, so ${\theta }^k$ is holomorphic on $\mathfrak{h}$, and its expansion at infinity has no negative powers of $e^{\pi i\tau }$. Applying Poisson summation to the shifted Gaussian $e^{-\pi t(n+1/2{)}^2}$ gives the corresponding nonnegative-power expansion at the remaining cusp. Hence ${\theta }^k$ is holomorphic at every cusp and is a modular form of weight $k/2$ and level $\Gamma$.

The specialization of the Jacobi triple product to the Jacobi theta function isSince $\tau \in \mathfrak{h}$, one has $|q|<1$, so every displayed factor is nonzero. MoreoverThe standard convergence criterion for infinite products therefore shows that the product converges to a nonzero value. This proves the nonvanishing of the Jacobi theta function throughout $\mathfrak{h}$.

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 $\zeta (s)$.

This convention for the Jacobi theta function is the usual theta constant evaluated at twice the argument. The Poisson summation formula gives $\theta (-1/(4\tau ))=\sqrt{-2i\tau }\theta (\tau )$, with the square root holomorphic on the half-plane and positive on the imaginary axis after evaluating its positive real argument.