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The $2\pi$ periodization of a Schwartz function isFor every nonnegative integer $j$, the Schwartz seminorms give $|{\varphi }^{(j)}(x+2\pi k)|\le C_j(1+|k|{)}^{-2}$ uniformly for $x\in [0,2\pi ]$. The Weierstrass M-test gives absolute and uniform convergence there of the defining series and all its differentiated series. Termwise differentiation proves that $P\varphi$ is a smooth function, with $(P\varphi {)}^{(j)}=P({\varphi }^{(j)})$. Reindexing the absolutely convergent series proves $P\varphi (x+2\pi )=P\varphi (x)$, so it is a periodic function.

For an explicit smooth periodic partition of unity, choose the bump functionIt is a test function, positive on $[-\pi ,\pi ]$. The smooth function $S(x)=P\eta (x)$ is $2\pi$ periodic and strictly positive, since every $x$ has a translate in $[-\pi ,\pi ]$. SetDivision preserves smoothness because $S>0$, and the support remains contained in $[-2\pi ,2\pi ]$.

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 $e(t)=e^{2\pi it}$ 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.

For a full-rank Euclidean lattice $\Lambda \subset {\mathbb{R}}^n$, its dual lattice isIf $\Lambda =B{\mathbb{Z}}^n$, then ${\Lambda }^{\prime }=B^{-\mathsf{T}}{\mathbb{Z}}^n$, $m(\Lambda )=|\det B|$ and $m({\Lambda }^{\prime })=m(\Lambda {)}^{-1}$. The characters of a real torus identify the additive dual lattice with the multiplicative character group byThe map is well defined precisely because $u\in {\Lambda }^{\prime }$, and is injective. For surjectivity, a continuous group homomorphism from the compact torus into ${\mathbb{C}}^{\times }$ has compact image. Its modulus has logarithm a homomorphism into $(\mathbb{R},+)$ with compact image, hence is zero, so the image lies in the unit circle. Pull the character back to ${\mathbb{R}}^n$. Its continuous real lift under $t\mapsto e^{2\pi it}$, normalized to zero at the origin, is additive: its additive defect is an integer-valued continuous function and vanishes at the origin. A continuous additive function is $⟨u,x⟩$ for a unique $u$. Triviality on $\Lambda$ says $u\in {\Lambda }^{\prime }$, proving surjectivity.

Use the Fourier transform convention $\hat{f}(u)={\int }_{{\mathbb{R}}^n}f(x)e^{-2\pi i⟨u,x⟩}dx$, and take $f$ to be a Schwartz function. The periodization of a Schwartz function $P_f(x)={\sum }_{\lambda \in \Lambda }f(x+\lambda )$ is smooth and $\Lambda$-periodic. On a fundamental cell $F$, the coefficient of the torus character ${\chi }_u$ isThe equality follows by translating each cell and using $e^{2\pi i⟨u,\lambda ⟩}=1$. The rapidly convergent Fourier series can be evaluated at zero, yielding the Poisson summation formula for a Euclidean latticeThis argument keeps track of the covolume factor rather than tacitly assuming a unit-volume lattice.

For $\tau$ in the complex upper half-plane, let $g_{\tau }(x)=e^{\pi i\tau |x{|}^2}$. Scaling the self-dual real Gaussian function gives its Fourier transform at $\tau =it$, $t>0$, and holomorphic continuation in $\tau$ gives the complex Gaussian Fourier transformThe branch is $\exp [-(n/2)\mathrm{Log}(-i\tau )]$ with the logarithm on the right half-plane; it is positive for $\tau =it$. Both the integrals and the lattice sums are locally normally convergent on the complex upper half-plane. Applying the Poisson summation formula proves the lattice theta functional equationNo integrality or self-duality hypothesis on the lattice is needed.

Put $a=n/2$, $m=m(\Lambda )$ and ${\theta }_{\Lambda }(t)={\Theta }_{\Lambda }(it)$. The Epstein zeta function converges absolutely for $\mathrm{Re}s>a$, and its Mellin transform representation isAt infinity, ${\theta }_{\Lambda }(t)-1$ decays exponentially. Define the entire functionOn $(0,1)$, substitute ${\theta }_{\Lambda }(t)=m^{-1}{t}^{-a}{\theta }_{{\Lambda }^{\prime }}(1/t)$ and then $u=1/t$. Isolate the two elementary terms before integrating; this gives the pole-subtracted theta integral for an Epstein zeta functionThe formula initially holds for $\mathrm{Re}s>a$ and continues the completed Epstein zeta function meromorphically to all $\mathbb{C}$. Apply the same formula to the dual lattice at $a-s$, using $m({\Lambda }^{\prime })=m^{-1}$ and ${\Lambda }^{\prime \prime }=\Lambda$. The entire terms and the two rational terms match, provingFinally $E_{\Lambda }(s)={\pi }^s{\mathcal{E}}_{\Lambda }(s)/\Gamma (s)$ also continues meromorphically. Its only pole is a simple one at $s=n/2$, with residue ${\pi }^{n/2}/[m(\Lambda )\Gamma (n/2)]$; the pole of the completed function at zero is cancelled by $1/\Gamma (s)$, and $E_{\Lambda }(0)=-1$. The residues and this cancellation make explicit why subtracting the constant theta term was necessary.

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 $\xi (s)=\xi (1-s)$.