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

A three-dimensional Bravais lattice is the set of all integer combinations of three linearly independent primitive vectors. Its reciprocal lattice isand the first Brillouin zone is the Wigner-Seitz cell of the origin in reciprocal space.

For the stated FCC lattice, the primitive-cell volume isThe reciprocal-basis formula givesA reciprocal primitive cell, and hence the Brillouin zone, has volume

In the first Born approximation, with momentum transfer $q=k-k^{\prime }$, the scattering amplitude isFor the spherical shell, align the polar axis with $q$ and use its radial Dirac delta function:Therefore

Translation of each atomic potential multiplies its transform by a phase, so the crystal lattice structure factor givesFor an infinite lattice, the lattice Poisson summation formula turns the sum intoThus the amplitude vanishes away from reciprocal-lattice momentum transfers.

In elastic scattering, $|k|=|k^{\prime }|$ and henceThe two shortest relevant reciprocal-vector lengths of this FCC lattice areBoth are kinematically available when $k>2\pi /a$. Assigning their scattering angles ${\theta }_2$ and ${\theta }_1$, respectively, gives

The functional equation isEquivalently,

For a proof, letThe Poisson summation formula applied to a Gaussian function gives the theta transformationThe standard Gamma function integral and termwise integration initially give, for $\Re s>1$,Split the integral at one, substitute $u\mapsto 1/u$ in the lower half, and use the theta transformation. The result isThe integral is an entire function of $s$ because $\Theta (u)-1$ decays exponentially. The right side is visibly invariant under $s\mapsto 1-s$, proving both the analytic continuation and the functional equation of the Riemann zeta function. This is the Mellin representation of the completed Riemann zeta function.

Choose a real lift $\Delta \theta ={\theta }_f-{\theta }_i$. Classical paths fall into winding number sectors $j\in \mathbb{Z}$ and areEach sector has the same fluctuation determinant, so the image-sum form of the propagator isSet $a=\hslash T/(2m{R}^2)$. Applying the Poisson summation formula to the Gaussian $e^{i{x}^2/(4a)}$ giveswhich is exactly the spectral propagator found in part i. Thus the angular-momentum sum is dual to a sum over homotopy classes of classical paths.

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 $\Gamma$.

For $F\in L^1(\mathbb{R})$ the periodization is absolutely convergent almost everywhere in a period by integrating the sum of absolute values. Its Fourier series coefficient is $c_n=\tilde{F}(n)/(2\pi )$, where $\tilde{F}(\omega )={\int }_{\mathbb{R}}F(x)e^{-i\omega x}dx$. Absolute integrability permits integrating term by term and changing variables across the disjoint period intervals. This coefficient identity does not by itself assert pointwise convergence of the ordinary Fourier series. Even uniform convergence of the original periodization cannot repair arbitrary point values: $F=1_{\{0\}}$ has zero transform and a nonzero periodization at multiples of $2\pi$. Additional regularity, such as a continuous piecewise continuously differentiable periodization, validates the pointwise Poisson summation formula in this setting.

// Target: fluid-mechanics.bigb

For a Schwartz function on the real numbers, its periodization $P_f$ is a smooth periodic function of period of a function one. Rapid decay gives uniform convergence on $[0,1]$ of the defining series and each differentiated series. Its $k$th Fourier coefficient isThe change of variable $u=x+n$ uses $e^{2\pi ikn}=1$. Rapidly decreasing Fourier coefficients give the absolutely and uniformly convergent Fourier series $P_f(x)={\sum }_k\hat{f}(k)e^{2\pi ikx}$. Evaluation at $x=0$ proves the Poisson summation formula.^[ref]

The same identity explains aliasing in the Nyquist–Shannon sampling theorem: applying the Poisson summation formula to the Schwartz function $x\mapsto f(x)e^{-2\pi i\xi x}$ givesThe sampled spectrum is periodic in frequency, and shifted copies of the continuous Fourier transform add together. When those copies overlap, the samples cannot separate their contributions. The Nyquist–Shannon sampling theorem adds a bandlimited function hypothesis to make reconstruction possible.

The Poisson summation formula equates sample sums of a function and its Fourier transform. Exchanging broad and narrow sums can save work. Use the conventionFor every Schwartz function $f\in \mathcal{S}(\mathbb{R})$,Both series have absolute convergence. A sufficient hypothesis is that $f$ is smooth and ${\sup }_x(1+|x|{)}^M|f^{(r)}(x)|<\infty$ for all nonnegative integers $M,r$; its Fourier transform is then also a Schwartz function.^[ref] This assumption controls both decay and regularity; mere membership of an L1 space does not control values at individual sampling points, as periodization of an integrable function illustrates.

Here $\xi$ counts cycles per unit length. In the angular convention used elsewhere in Fourier analysis, $\tilde{f}(\omega )=\int f(x)e^{-i\omega x}dx$, so $\hat{f}(\xi )=\tilde{f}(2\pi \xi )$ and the right-hand side becomes ${\sum }_k\tilde{f}(2\pi k)$. Keeping the Fourier transform convention explicit prevents stray $2\pi$ factors. Changing the sample spacing leads to the Poisson summation formula for a Euclidean lattice.

Let $\Lambda =B{\mathbb{Z}}^d$ be a full-rank Euclidean lattice, with covolume $m(\Lambda )=|\det B|$ and dual lattice ${\Lambda }^{\vee }=B^{-\mathsf{T}}{\mathbb{Z}}^d$. For a Schwartz function $f\in \mathcal{S}({\mathbb{R}}^d)$ use the Fourier transform conventionThen the Poisson summation formula becomesBoth series have absolute convergence. To see the normalization, apply integer-lattice Poisson summation to $g(u)=f(Bu)$. The change of variables formula gives $\hat{g}(k)=|\det B{|}^{-1}\hat{f}(B^{-\mathsf{T}}k)$, yielding both the reciprocal covolume factor and the dual lattice argument. The factor is the physical lattice-point density. In particular, for spacing $h>0$,For the Gaussian function $f(x)=e^{-\pi a\Vert x{\Vert }^2}$ with $a>0$, the product of the one-dimensional Gaussian Fourier transforms givesHere $\Vert \cdot \Vert$ is the Euclidean norm. As a normalization check, take $\Lambda =\mathrm{diag}(2,3){\mathbb{Z}}^2$. Its covolume is $6$, and its dual lattice is $\mathrm{diag}(1/2,1/3){\mathbb{Z}}^2$, soThis also follows by multiplying two Gaussian Poisson summation identities with parameters $4a$ and $9a$. If one instead uses an angular Fourier transform kernel $e^{-i⟨x,\omega ⟩}$, the reciprocal frequencies are $2\pi {\Lambda }^{\vee }$, while the prefactor remains $m(\Lambda {)}^{-1}$.

The quantum rotor with angle period $2\pi$ has both the spectral kernel $(2\pi {)}^{-1}{\sum }_n{e}^{in\Delta -i\hslash n^{2}T/(2I)}$ and the displayed winding representation. The Poisson summation formula converts one into the other by a Gaussian integral. Each winding sector has a constant-velocity classical path and action $I(\Delta +2\pi m{)}^2/(2T)$. Quadratic fluctuations give the same prefactor in every sector, so this is exact. Real time is understood by $T\to T-i0$, which fixes the square-root branch and convergence prescription.

For a particle with mass $m$ on $(0,L)$, put $K_0(x;\beta )=\sqrt{m/(2\pi {\hslash }^2\beta )}{e}^{-m{x}^2/(2{\hslash }^2\beta )}$. The method of images gives $K_D(q_f,q_i;\beta )={\sum }_r[K_0(q_f-q_i+2rL;\beta )-K_0(q_f+q_i+2rL;\beta )]$. The diagonal integral isThe reflected intervals tile the real line, giving the subtraction $1/2$. The Poisson summation formula converts this to ${\sum }_{n\ge 1}{e}^{-\beta {\hslash }^2{\pi }^2{n}^2/(2m{L}^2)}$, proving equality between the image and energy eigenstate calculations.

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.

A Villain model replaces the cosine Boltzmann factor of the XY model by a periodic Gaussian weight. The Poisson summation formula makes its bond weight equal to ${\sum }_{p\in \mathbb{Z}}\exp [-\beta (\psi -2\pi p{)}^2/2]$. The periodic Gaussian retains integer winding defects and permits an explicit separation into spin waves and vortices in two dimensions. Its replacement of the cosine model is an approximation; subsequent transformations within the Villain model can be exact.