Poisson summation formula

Let $f:ℝ\to ℝ$ be an integrable function and let

$$\widehat{f}(\xi )={\int }_{ℝ}{e}^{-2\pi i\xi x}f(x)𝑑x,\xi \in ℝ.$$

be its Fourier transform. The Poisson summation formula is the assertion that

$$\sum _{n\in \mathbb{Z}}f(n)=\sum _{n\in \mathbb{Z}}\widehat{f}(n).$$(1)

whenever $f$ is such that both of the above infinite sums areabsolutely convergent.

Equation (1) is useful because it establishes acorrespondence between Fourier series and Fourier integrals. To seethe connection, let

$$g(x)=\sum _{n\in \mathbb{Z}}f(x+n),x\in ℝ,$$

be the periodicfunction obtained by pseudo-averaging $f$ relative to $\mathbb{Z}$ actingas the discrete group of translations on $ℝ$. Since $f$ wasassumed to be integrable, $g$ is defined almost everywhere, and isintegrable over $[0,1]$ with

$${\parallel g\parallel }_{L^1[0,1]}\le {\parallel f\parallel }_{L^1(ℝ)}.$$

Since $f$ is integrable, we may interchange integration and summationto obtain

$$\widehat{f}(k)=\sum _{n\in \mathbb{Z}}{\int }_0^{1}f(x+n)e^{-2\pi ikx}𝑑x={\int }_0^1{e}^{-2\pi ikx}g(x)𝑑x$$

for every $k\in \mathbb{Z}$. In other words, the restriction of the Fouriertransform of $f$ to the integers gives the Fourier coefficients of theaveraged, periodic function $g$. Since we have assumed that the$\widehat{f}(k)$ form an absolutely convergent series, we have that

$$g(x)=\sum _{k\in \mathbb{Z}}\widehat{f}(k)e^{2\pi ikx}$$

in the sense ofuniform convergence. Evaluating the above equation at $x=0$, weobtain the Poisson summation formula (1).